\chapter{Foundations} \label{foundations} The present foundational chapter will introduce the notion of a ring and, next, that of a module over a ring. These notions will be the focus of the present book. Most of the chapter will be definitions. We begin with a few historical remarks. Fermat's last theorem states that the equation \[ \label{ft} x^n + y^n = z^n \] has no nontrivial solutions in the integers, for $n \ge 3$. We could try to prove this by factoring the expression on the left hand side. We can write \[ (x+y)(x+ \zeta y) (x+ \zeta^2y) \dots (x+ \zeta^{n-1}y) = z^n, \] where $\zeta$ is a primitive $n$th root of unity. Unfortunately, the factors lie in $\mathbb{Z}[\zeta]$, not the integers $\mathbb{Z}$. Though $\mathbb{Z}[\zeta]$ is still a \emph{ring} where we have notions of primes and factorization, just as in $\mathbb{Z}$, we will see that prime factorization is not always unique in $\mathbb{Z}[\zeta]$. (If it were always unique, then we could at least one important case of Fermat's last theorem rather easily; see the introductory chapter of \cite{Wa97} for an argument.) For instance, consider the ring $\mathbb{Z}[\sqrt{-5}]$ of complex numbers of the form $a + b\sqrt{-5}$, where $a, b \in \mathbb{Z}$. Then we have the two factorizations \[ 6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}). \] Both of these are factorizations of 6 into irreducible factors, but they are fundamentally different. In part, commutative algebra grew out of the need to understand this failure of unique factorization more generally. We shall have more to say on factorization in the future, but here we just focus on the formalism. The basic definition for studying this problem is that of a \emph{ring}, which we now introduce. \section{Commutative rings and their ideals} \subsection{Rings} We shall mostly just work with commutative rings in this book, and consequently will just say ``ring'' for one such. \begin{definition} A \textbf{commutative ring} is a set $R$ with an addition map $+ : R \times R \to R$ and a multiplication map $\times : R \times R \to R$ that satisfy the following conditions. \begin{enumerate} \item $R$ is a group under addition. \item The multiplication map is commutative and distributes over addition. This means that $x \times (y+z) = x \times y + x\times z$ and $x \times y = y \times x$. \item There is a \textbf{unit} (or \textbf{identity element}), denoted by $1$, such that $1 \times x = x$ for all $x \in R$. \end{enumerate} We shall typically write $xy$ for $x \times y$. Given a ring, a \textbf{subring} is a subset that contains the identity element and is closed under addition and multiplication. \end{definition} A \emph{noncommutative} (i.e. not necessarily commutative) ring is one satisfying the above conditions, except possibly for the commutativity requirement $xy = yx$. For instance, there is a noncommutative ring of $2$-by-$2$ matrices over $\mathbb{C}$. We shall not work too much with noncommutative rings in the sequel, though many of the basic results (e.g. on modules) do generalize. \begin{example} $\mathbb{Z}$ is the simplest example of a ring. \end{example} \begin{exercise}\label{polynomial} Let $R$ be a commutative ring. Show that the set of polynomials in one variable over $R$ is a commutative ring $R[x]$. Give a rigorous definition of this. \end{exercise} \begin{example} For any ring $R$, we can consider the polynomial ring $R[x_1, \ldots, x_n]$ which consists of the polynomials in $n$ variables with coefficients in $R$. This can be defined inductively as $(R[x_1, \dots, x_{n-1}])[x_n]$, where the procedure of adjoining a single variable comes from the previous \cref{polynomial}. \end{example} We shall see a more general form of this procedure in \cref{groupring}. \begin{exercise} If $R$ is a commutative ring, recall that an \textbf{invertible element} (or, somewhat confusingly, a \textbf{unit}) $u \in R$ is an element such that there exists $v \in R$ with $uv = 1$. Prove that $v$ is necessarily unique. \end{exercise} \begin{exercise} \label{ringoffns} Let $X$ be a set and $R$ a ring. The set $R^X$ of functions $f:X \to R$ is a ring. \end{exercise} \subsection{The category of rings} The class of rings forms a category. Its morphisms are called ring homomorphisms. \begin{definition} A \textbf{ring homomorphism} between two rings $R$ and $S$ as a map $f : R \to S$ that respects addition and multiplication. That is, \begin{enumerate} \item $f(1_R) = 1_S$, where $1_R$ and $1_S$ are the respective identity elements. \item $f(a + b) = f(a) + f(b)$ for $a, b \in R$. \item $f(ab) = f(a)f(b)$ for $a, b \in R$. \end{enumerate} There is thus a \emph{category} $\mathbf{Ring}$ whose objects are commutative rings and whose morphisms are ring-homomorphisms. \end{definition} The philosophy of Grothendieck, as expounded in his EGA \cite{EGA}, is that one should always do things in a relative context. This means that instead of working with objects, one should work with \emph{morphisms} of objects. Motivated by this, we introduce: \begin{definition} Given a ring $A$, an \textbf{$A$-algebra} is a ring $R$ together with a morphism of rings (a \textbf{structure morphism}) $A \to R$. There is a category of $A$-algebras, where a morphism between $A$-algebras is a ring-homomorphism that is required to commute with the structure morphisms. \end{definition} So if $R$ is an $A$-algebra, then $R$ is not only a ring, but there is a way to multiply elements of $R$ by elements of $A$ (namely, to multiply $a \in A$ with $r \in R$, take the image of $a $ in $R$, and multiply that by $r$). For instance, any ring is an algebra over any subring. We can think of an $A$-algebra as an arrow $A \to R$, and a morphism from $A \to R$ to $A \to S$ as a commutative diagram \[ \xymatrix{ R \ar[rr] & & S \\ & \ar[lu] A \ar[ru] }\] This is a special case of the \emph{undercategory} construction. If $B$ is an $A$-algebra and $C$ a $B$-algebra, then $C$ is an $A$-algebra in a natural way. Namely, by assumption we are given morphisms of rings $A \to B$ and $B \to C$, so composing them gives the structure morphism $A \to C$ of $C$ as an $A$-algebra. \begin{example} Every ring is a $\mathbb{Z}$-algebra in a natural and unique way. There is a unique map (of rings) $\mathbb{Z} \to R$ for any ring $R$ because a ring-homomorphism is required to preserve the identity. In fact, $\mathbb{Z}$ is the \emph{initial object} in the category of rings: this is a restatement of the preceding discussion. \end{example} \begin{example} If $R$ is a ring, the polynomial ring $R[x]$ is an $R$-algebra in a natural manner. Each element of $R$ is naturally viewed as a ``constant polynomial.'' \end{example} \begin{example} $\mathbb{C}$ is an $\mathbb{R}$-algebra. \end{example} Here is an example that generalizes the case of the polynomial ring. \begin{example} \label{groupring} If $R$ is a ring and $G$ a commutative monoid,\footnote{That is, there is a commutative multiplication on $G$ with an identity element, but not necessarily with inverses.} then the set $R[G]$ of formal finite sums $\sum r_i g_i$ with $r_i \in R, g_i \in G$ is a commutative ring, called the \textbf{moniod ring} or \textbf{group ring} when $G$ is a group. Alternatively, we can think of elements of $R[G]$ as infinite sums $\sum_{g \in G} r_g g$ with $R$-coefficients, such that almost all the $r_g$ are zero. We can define the multiplication law such that \[ \left(\sum r_g g\right)\left( \sum s_g g\right) = \sum_h \left( \sum_{g g' = h} r_g s_{g'} \right) h. \] This process is called \emph{convolution.} We can think of the multiplication law as extended the group multiplication law (because the product of the ring-elements corresponding to $g, g'$ is the ring element corresponding to $gg' \in G$). The case of $G = \mathbb{Z}_{\geq 0}$ is the polynomial ring. In some cases, we can extend this notion to formal infinite sums, as in the case of the formal power series ring; see \cref{powerseriesring} below. \end{example} \begin{exercise} \label{integersinitial} The ring $\mathbb{Z}$ is an \emph{initial object} in the category of rings. That is, for any ring $R$, there is a \emph{unique} morphism of rings $\mathbb{Z} \to R$. We discussed this briefly earlier; show more generally that $A$ is the initial object in the category of $A$-algebras for any ring $A$. \end{exercise} \begin{exercise} The ring where $0=1$ (the \textbf{zero ring}) is a \emph{final object} in the category of rings. That is, every ring admits a unique map to the zero ring. \end{exercise} \begin{exercise} \label{corepresentable} Let $\mathcal{C}$ be a category and $F: \mathcal{C} \to \mathbf{Sets}$ a covariant functor. Recall that $F$ is said to be \textbf{corepresentable} if $F$ is naturally isomorphic to $X \to \hom_{\mathcal{C}}(U, X)$ for some object $U \in \mathcal{C}$. For instance, the functor sending everything to a one-point set is corepresentable if and only if $\mathcal{C}$ admits an initial object. Prove that the functor $\mathbf{Rings} \to \mathbf{Sets}$ assigning to each ring its underlying set is representable. (Hint: use a suitable polynomial ring.) \end{exercise} The category of rings is both complete and cocomplete. To show this in full will take more work, but we can here describe what certain cases (including all limits) look like. As we saw in \cref{corepresentable}, the forgetful functor $\mathbf{Rings} \to \mathbf{Sets}$ is corepresentable. Thus, if we want to look for limits in the category of rings, here is the approach we should follow: we should take the limit first of the underlying sets, and then place a ring structure on it in some natural way. \begin{example}[Products] The \textbf{product} of two rings $R_1, R_2$ is the set-theoretic product $R_1 \times R_2$ with the multiplication law $(r_1, r_2)(s_1, s_2) = (r_1 s_1, r_2 s_2)$. It is easy to see that this is a product in the category of rings. More generally, we can easily define the product of any collection of rings. \end{example} To describe the coproduct is more difficult: this will be given by the \emph{tensor product} to be developed in the sequel. \begin{example}[Equalizers] Let $f, g: R \rightrightarrows S$ be two ring-homomorphisms. Then we can construct the \textbf{equalizer} of $f,g$ as the subring of $R$ consisting of elements $x \in R$ such that $f(x) = g(x)$. This is clearly a subring, and one sees quickly that it is the equalizer in the category of rings. \end{example} As a result, we find: \begin{proposition} $\mathbf{Rings}$ is complete. \end{proposition} As we said, we will not yet show that $\mathbf{Rings}$ is cocomplete. But we can describe filtered colimits. In fact, filtered colimits will be constructed just as in the set-theoretic fashion. That is, the forgetful functor $\mathbf{Rings} \to \mathbf{Sets}$ commutes with \emph{filtered} colimits (though not with general colimits). \begin{example}[Filtered colimits] Let $I$ be a filtering category, $F: I \to \mathbf{Rings}$ a functor. We can construct $\varinjlim_I F$ as follows. An object is an element $(x,i)$ for $i \in I$ and $x \in F(i)$, modulo equivalence; we say that $(x, i)$ and $(y, j)$ are equivalent if there is a $k \in I$ with maps $i \to k, j \to k$ sending $x,y$ to the same thing in the ring $F(k)$. To multiply $(x, i)$ and $(y,j)$, we find some $k \in I$ receiving maps from $i, j$, and replace $x,y$ with elements of $F(k)$. Then we multiply those two in $F(k)$. One easily sees that this is a well-defined multiplication law that induces a ring structure, and that what we have described is in fact the filtered colimit. \end{example} \subsection{Ideals} An \emph{ideal} in a ring is analogous to a normal subgroup of a group. As we shall see, one may quotient by ideals just as one quotients by normal subgroups. The idea is that one wishes to have a suitable \emph{equivalence relation} on a ring $R$ such that the relevant maps (addition and multiplication) factor through this equivalence relation. It is easy to check that any such relation arises via an ideal. \begin{definition} Let $R$ be a ring. An \textbf{ideal} in $R$ is a subset $I \subset R$ that satisfies the following. \begin{enumerate} \item $0 \in I$. \item If $x, y \in I$, then $x + y \in I$. \item If $x \in I$ and $y \in R$, then $xy \in I$. \end{enumerate} \end{definition} There is a simple way of obtaining ideals, which we now describe. Given elements $x_1, \ldots, x_n \in R$, we denote by $(x_1, \ldots, x_n) \subset R$ the subset of linear combinations $\sum r_i x_i$, where $r_i \in R$. This is clearly an ideal, and in fact the smallest one containing all $x_i$. It is called the ideal \textbf{generated} by $x_1, \ldots, x_n$. A \textbf{principal ideal} $(x)$ is one generated by a single $x \in R$. \begin{example} Ideals generalize the notion of divisibility. Note that in $\mathbb{Z}$, the set of elements divisible by $n \in \mathbb{Z}$ forms the ideal $I = n\mathbb{Z} = (n)$. We shall see that every ideal in $\mathbb{Z}$ is of this form: $\mathbb{Z}$ is a \emph{principal ideal domain.} \end{example} Indeed, one can think of an ideal as axiomatizing the notions that ``divisibility'' ought to satisfy. Clearly, if two elements are divisible by something, then their sum and product should also be divisible by it. More generally, if an element is divisible by something, then the product of that element with anything else should also be divisible. In general, we will extend (in the chapter on Dedekind domains) much of the ordinary arithmetic with $\mathbb{Z}$ to arithmetic with \emph{ideals} (e.g. unique factorization). \begin{example} We saw in \cref{ringoffns} that if $X$ is a set and $R$ a ring, then the set $R^X$ of functions $X \to R$ is naturally a ring. If $Y \subset X$ is a subset, then the subset of functions vanishing on $Y$ is an ideal. \end{example} \begin{exercise} Show that the ideal $(2, 1 + \sqrt{-5}) \subset \mathbb{Z}[\sqrt{-5}]$ is not principal. \end{exercise} \subsection{Operations on ideals} There are a number of simple operations that one may do with ideals, which we now describe. \begin{definition} The sum $I + J$ of two ideals $I, J \subset R$ is defined as the set of sums \[ \left\{ x + y : x \in I, y \in J \right\}. \] \end{definition} \begin{definition} The product $IJ$ of two ideals $I, J \subset R$ is defined as the smallest ideal containing the products $xy$ for all $x \in I, y \in J$. This is just the set \[ \left\{ \sum x_i y_i : x_i \in I, y_i \in J \right\}. \] \end{definition} We leave the basic verification of properties as an exercise: \begin{exercise} Given ideals $I, J \subset R$, verify the following. \begin{enumerate} \item $I + J$ is the smallest ideal containing $I$ and $J$. \item $IJ$ is contained in $I$ and $J$. \item $I \cap J$ is an ideal. \end{enumerate} \end{exercise} \begin{example} In $\mathbb{Z}$, we have the following for any $m, n$. \begin{enumerate} \item $(m) + (n) = (\gcd\{ m, n \})$, \item $(m)(n) = (mn)$, \item $(m) \cap (n) = (\mathrm{lcm}\{ m, n \})$. \end{enumerate} \end{example} \begin{proposition} For ideals $I, J, K \subset R$, we have the following. \begin{enumerate} \item Distributivity: $I(J + K) = IJ + IK$. \item $I \cap (J + K) = I \cap J + I \cap K$ if $I \supset J$ or $I \supset K$. \item If $I + J = R$, $I \cap J = IJ$. \end{enumerate} \begin{proof} 1 and 2 are clear. For 3, note that $(I + J)(I \cap J) = I(I \cap J) + J(I \cap J) \subset IJ$. Since $IJ \subset I \cap J$, the result follows. \end{proof} \end{proposition} \begin{exercise} There is a \emph{contravariant} functor $\mathbf{Rings} \to \mathbf{Sets}$ that sends each ring to its set of ideals. Given a map $f: R \to S$ and an ideal $I \subset S$, we define an ideal $f^{-1}(I) \subset R$; this defines the functoriality. This functor is not representable, as it does not send the initial object in $\mathbf{Rings} $ to the one-element set. We will later use a \emph{subfunctor} of this functor, the $\spec$ construction, when we replace ideals with ``prime'' ideals. \end{exercise} \subsection{Quotient rings} We next describe a procedure for producing new rings from old ones. If $R$ is a ring and $I \subset R$ an ideal, then the quotient group $R/I$ is a ring in its own right. If $a+I, b+I$ are two cosets, then the multiplication is $(a+I)(b+I) = ab + I$. It is easy to check that this does not depend on the coset representatives $a,b$. In other words, as mentioned earlier, the arithmetic operations on $R$ \emph{factor} through the equivalence relation defined by $I$. As one easily checks, this becomes to a multiplication \[ R/I \times R/I \to R/I \] which is commutative and associative, and whose identity element is $1+I$. In particular, $R/I$ is a ring, under multiplication $(a+I)(b+I) = ab+I$. \begin{definition} $R/I$ is called the \textbf{quotient ring} by the ideal $I$. \end{definition} The process is analogous to quotienting a group by a normal subgroup: again, the point is that the equivalence relation induced on the algebraic structure---either the group or the ring---by the subgroup (or ideal)---is compatible with the algebraic structure, which thus descends to the quotient. The reduction map $\phi \colon R \to R/I$ is a ring-homomorphism with a \emph{universal property}. Namely, for any ring $B$, there is a map \[ \hom(R/I, B) \to \hom(R, B) \] on the hom-sets by composing with the ring-homomorphism $\phi$; this map is injective and the image consists of all homomorphisms $R \to B$ which vanish on $I$. Stated alternatively, to map out of $R/I$ (into some ring $B$) is the same thing as mapping out of $R$ while killing the ideal $I \subset R$. This is best thought out for oneself, but here is the detailed justification. The reason is that any map $R/I \to B$ pulls back to a map $R \to R/I \to B$ which annihilates $I$ since $R \to R/I$ annihilates $I$. Conversely, if we have a map \[ f: R \to B \] killing $I$, then we can define $R/I \to B$ by sending $a+I$ to $f(a)$; this is uniquely defined since $f$ annihilates $I$. \begin{exercise} If $R$ is a commutative ring, an element $e \in R$ is said to be \textbf{idempotent} if $e^2 = e$. Define a covariant functor $\mathbf{Rings} \to \mathbf{Sets}$ sending a ring to its idempotents. Prove that it is corepresentable. (Answer: the corepresenting object is $\mathbb{Z}[X]/(X - X^2)$.) \end{exercise} \begin{exercise} Show that the functor assigning to each ring the set of elements annihilated by 2 is corepresentable. \end{exercise} \begin{exercise} If $I \subset J \subset R$, then $J/I$ is an ideal of $R/I$, and there is a canonical isomorphism \[ (R/I)/(J/I) \simeq R/J. \] \end{exercise} \subsection{Zerodivisors} Let $R$ be a commutative ring. \begin{definition} If $r \in R$, then $r$ is called a \textbf{zerodivisor} if there is $s \in R, s \neq 0$ with $sr = 0$. Otherwise $r$ is called a \textbf{nonzerodivisor.} \end{definition} As an example, we prove a basic result on the zerodivisors in a polynomial ring. \begin{proposition} Let $A=R[x]$. Let $f=a_nx^n+\cdots +a_0\in A$. If there is a non-zero polynomial $g\in A$ such that $fg=0$, then there exists $r\in R\smallsetminus\{0\}$ such that $f\cdot r=0$. \end{proposition} So all the coefficients are zerodivisors. \begin{proof} Choose $g$ to be of minimal degree, with leading coefficient $bx^d$. We may assume that $d>0$. Then $f\cdot b\neq 0$, lest we contradict minimality of $g$. We must have $a_i g\neq 0$ for some $i$. To see this, assume that $a_i\cdot g=0$, then $a_ib=0$ for all $i$ and then $fb=0$. Now pick $j$ to be the largest integer such that $a_jg\neq 0$. Then $0=fg=(a_0 + a_1x + \cdots a_jx^j)g$, and looking at the leading coefficient, we get $a_jb=0$. So $\deg (a_jg)}[ld] \ar[d] \\ K \ar[r] & M \ar[r]^f & N }\] if and only if the composite $T \to N$ is zero. In particular, if we think of the hom-sets as abelian groups (i.e. $\mathbb{Z}$-modules) \[ \hom_R( T,K) = \ker\left( \hom_R(T, M) \to \hom_R(T, N) \right). \] \end{exercise} In other words, one may think of the kernel as follows. If $X \stackrel{f}{\to} Y$ is a morphism, then the kernel $\ker(f)$ is the equalizer of $f$ and the zero morphism $X \stackrel{0}{\to} Y$. \begin{exercise} What is the universal property of the cokernel? \end{exercise} \begin{exercise} \label{moduleunderlyingsetrepresentable} On the category of modules, the functor assigning to each module $M$ its underlying set is corepresentable (cf. \rref{corepresentable}). What is the corepresenting object? \end{exercise} We shall now introduce the notions of \emph{direct sum} and \emph{direct product}. Let $I$ be a set, and suppose that for each $i \in I$, we are given an $R$-module $M_i$. \begin{definition} The \textbf{direct product} $\prod M_i$ is set-theoretically the cartesian product. It is given the structure of an $R$-module by addition and multiplication pointwise on each factor. \end{definition} \begin{definition} The \textbf{direct sum} $\bigoplus_I M_i$ is the set of elements in the direct product such that all but finitely many entries are zero. The direct sum is a submodule of the direct product. \end{definition} \begin{example} \label{productcoproduct} The direct product is a product in the category of modules, and the direct sum is a coproduct. This is easy to verify: given maps $f_i: M \to M_i$, then we get get a unique map $f: M \to \prod M_i$ by taking the product in the category of sets. The case of a coproduct is dual: given maps $g_i: M_i \to N$, then we get a map $\bigoplus M_i \to N$ by taking the \emph{sum} $g$ of the $g_i$: on a family $(m_i) \in \bigoplus M_i$, we take $g(m_i) = \sum_I g_i(m_i)$; this is well-defined as almost all the $m_i$ are zero. \end{example} \cref{productcoproduct} shows that the category of modules over a fixed commutative ring has products and coproducts. In fact, the category of modules is both complete and cocomplete (see \cref{completecat} for the definition). To see this, it suffices to show that (by \cref{coprodcoequalsufficeforcocomplete} and its dual) that this category admits equalizers and coequalizers. The equalizer of two maps \[ M \stackrel{f,g}{\rightrightarrows} N \] is easily checked to be the submodule of $M$ consisting of $m \in M$ such that $f(m) = g(m)$, or, in other words, the kernel of $f-g$. The coequalizer of these two maps is the quotient module of $N$ by the submodule $\left\{f(m) - g(m), m \in M\right\}$, or, in other words, the cokernel of $f-g$. Thus: \begin{proposition} If $R$ is a ring, the category of $R$-modules is complete and cocomplete. \end{proposition} \begin{example} Note that limits in the category of $R$-modules are calculated in the same way as they are for sets, but colimits are not. That is, the functor from $R$-modules to $\mathbf{Sets}$, the forgetful functor, preserves limits but not colimits. Indeed, we will see that the forgetful functor is a right adjoint (\cref{freeadj}), which implies it preserves limits (by \cref{adjlimits}). \end{example} \subsection{Exactness} Finally, we introduce the notion of \emph{exactness}. \begin{definition} \label{exactness} Let $f: M \to N$ be a morphism of $R$-modules. Suppose $g: N \to P$ is another morphism of $R$-modules. The pair of maps is a \textbf{complex} if $g \circ f = 0: M \to N \to P$. This is equivalent to the condition that $\im(f) \subset \ker(g)$. This complex is \textbf{exact} (or exact at $N$) if $\im(f) = \ker(g)$. In other words, anything that is killed when mapped to $P$ actually comes from something in $M$. \end{definition} We shall often write pairs of maps as sequences \[ A \stackrel{f}{\to} B \stackrel{g}{\to} C \] and say that the sequence is exact if the pair of maps is, as in \rref{exactness}. A longer (possibly infinite) sequence of modules \[ A_0 \to A_1 \to A_2 \to \dots \] will be called a \textbf{complex} if each set of three consecutive terms is a complex, and \textbf{exact} if it is exact at each step. \begin{example} The sequence $0 \to A \stackrel{f}{\to} B$ is exact if and only if the map $f$ is injective. Similarly, $A \stackrel{f}{\to} B \to 0$ is exact if and only if $f$ is surjective. Thus, $0 \to A \stackrel{f}{\to} B \to 0$ is exact if and only if $f$ is an isomorphism. \end{example} One typically sees this definition applied to sequences of the form \[ 0 \to M'\stackrel{f}{ \to} M \stackrel{g}{\to} M'' \to 0, \] which, if exact, is called a \textbf{short exact sequence}. Exactness here means that $f$ is injective, $g$ is surjective, and $f$ maps onto the kernel of $g$. So $M''$ can be thought of as the quotient $M/M'$. \begin{example} Conversely, if $M$ is a module and $M' \subset M$ a submodule, then there is a short exact sequence \[ 0 \to M' \to M \to M/M' \to 0. \] So every short exact sequence is of this form. \end{example} Suppose $F$ is a functor from the category of $R$-modules to the category of $S$-modules, where $R, S$ are rings. Then: \begin{definition} \begin{enumerate} \item $F$ is called \textbf{additive} if $F$ preserves direct sums. \item $F$ is called \textbf{exact} if $F$ is additive and preserves exact sequences. \item $F$ is called \textbf{left exact} if $F$ is additive and preserves exact sequences of the form $0 \to M' \to M \to M''$. Equivalently, $F$ preserves kernels. \item $F$ is \textbf{right exact} if $F$ is additive and $F$ preserves exact sequences of the form $M' \to M \to M'' \to 0$, i.e. $F$ preserves cokernels. \end{enumerate} \end{definition} The reader should note that much of homological algebra can be developed using the more general setting of an \emph{abelian category,} which axiomatizes much of the standard properties of the category of modules over a ring. Such a generalization turns out to be necessary when many natural categories, such as the category of chain complexes or the category of sheaves on a topological space, are not naturally categories of modules. We do not go into this here, cf. \cite{Ma98}. A functor $F$ is exact if and only if it is both left and right exact. This actually requires proof, though it is not hard. Namely, right-exactness implies that $F$ preserves cokernels. Left-exactness implies that $F$ preserves kernels. $F$ thus preserves images, as the image of a morphism is the kernel of its cokernel. So if \[ A \to B \to C \] is a short exact sequence, then the kernel of the second map is equal to the image of the first; we have just seen that this is preserved under $F$. From this, one can check that left-exactness is equivalent to requiring that $F$ preserve finite limits (as an additive functor, $F$ automatically preserves products, and we have just seen that $F$ is left-exact iff it preserves kernels). Similarly, right-exactness is equivalent to requiring that $F$ preserve finite colimits. So, in \emph{any} category with finite limits and colimits, we can talk about right or left exactness of a functor, but the notion is used most often for categories with an additive structure (e.g. categories of modules over a ring). \begin{exercise} Suppose whenever $0 \to A' \to A \to A'' \to 0$ is short exact, then $FA' \to FA \to FA'' \to 0$ is exact. Prove that $F$ is right-exact. So we get a slightly weaker criterion for right-exactness. Do the same for left-exact functors. \end{exercise} \subsection{Split exact sequences} Let $f: A \to B$ be a map of sets which is injective. Then there is a map $g: A \to B$ such that the composite $g \circ f: A \stackrel{f}{\to} B \stackrel{g}{\to} A$ is the identity. Namely, we define $g$ to be the inverse of $f$ on $f(A)$ and arbitrarily on $B-f(A)$. Conversely, if $f: A \to B$ admits an element $g: B \to A$ such that $g \circ f = 1_A$, then $f$ is injective. This is easy to see, as any $a \in A$ can be ``recovered'' from $f(a)$ (by applying $g$). In general, however, this observation does not generalize to arbitrary categories. \begin{definition} Let $\mathcal{C}$ be a category. A morphism $A \stackrel{f}{\to} B$ is called a \textbf{split injection} if there is $g: B \to A$ with $g \circ f = 1_A$. \end{definition} \begin{exercise}[General nonsense] Suppose $f: A \to B$ is a split injection. Show that $f$ is a categorical monomorphism. (Idea: the map $\hom(C,A) \to \hom(C,B)$ becomes a split injection of sets thanks to $g$.) \end{exercise} \add{what is a categorical monomorphism? Maybe omit the exercise} In the category of sets, we have seen above that \emph{any} monomorphism is a split injection. This is not true in other categories, in general. \begin{exercise} Consider the morphism $\mathbb{Z} \to \mathbb{Z}$ given by multiplication by 2. Show that this is not a split injection: no left inverse $g$ can exist. \end{exercise} We are most interested in the case of modules over a ring. \begin{proposition} A morphism $f: A \to B$ in the category of $R$-modules is a split injection if and only if: \begin{enumerate} \item $f$ is injective. \item $f(A)$ is a direct summand in $B$. \end{enumerate} \end{proposition} The second condition means that there is a submodule $B' \subset B$ such that $B = B' \oplus f(A)$ (internal direct sum). In other words, $B = B' + f(A)$ and $B' \cap f(A) = \left\{0\right\}$. \begin{proof} Suppose the two conditions hold, and we have a module $B'$ which is a complement to $f(A)$. Then we define a left inverse \[ B \stackrel{g}{\to} A \] by letting $g|_{f(A)} = f^{-1}$ (note that $f$ becomes an \emph{isomorphism} $A \to f(A)$) and $g|_{B'}=0$. It is easy to see that this is indeed a left inverse, though in general not a right inverse, as $g$ is likely to be non-injective. Conversely, suppose $f: A \to B$ admits a left inverse $g: B \to A$. The usual argument (as for sets) shows that $f$ is injective. The essentially new observation is that $f(A) $ is a direct summand in $B$. To define the complement, we take $\ker(g) \subset B$. It is easy to see (as $g \circ f = 1_A$) that $\ker(g) \cap f(A) = \left\{0\right\}$. Moreover, $\ker(g) +f(A)$ fills $B$: given $b \in B$, it is easy to check that \[ b - f(g(b)) \in \ker(g). \] Thus we find that the two conditions are satisfied. \end{proof} \add{further explanation, exactness of filtered colimits} \subsection{The five lemma} The five lemma will be a useful tool for us in proving that maps are isomorphisms. Often this argument is used in inductive proofs. Namely, we will see that often ``long exact sequences'' (extending infinitely in one or both directions) arise from short exact sequences in a natural way. In such events, the five lemma will allow us to prove that certain morphisms are isomorphisms by induction on the dimension. \begin{theorem} Suppose given a commutative diagram \[ \xymatrix{ A \ar[d] \ar[r] & B \ar[d] \ar[r] & C \ar[d] \ar[r] & D \ar[d] \ar[r] & E \ar[d] \\ A' \ar[r] & B' \ar[r] & C' \ar[r] & D' \ar[r] & E' }\] such that the rows are exact and the four vertical maps $A \to A', B \to B', D \to D', E \to E'$ are isomorphisms. Then $C \to C'$ is an isomorphism. \end{theorem} This is the type of proof that goes by the name of ``diagram-chasing,'' and is best thought out visually for oneself, even though we give a complete proof. \begin{proof} We have the diagram \[ \xymatrix{ A \ar[r]^k \ar[d]^\a & B \ar[r]^l \ar[d]^\b & C \ar[r]^m \ar[d]^g & D \ar[r]^n \ar[d]^\d & E \ar[d]^\e \\ F \ar[r]_p & G \ar[r]_q & H \ar[r]_r & I \ar[r]_s & J } \] where the rows are exact at $B, C, D, G, H, I$ and the squares commute. In addition, suppose that $\a, \b, \d, \e$ are isomorphisms. We will show that $\g$ is an isomorphism. \emph{We show that $\g$ is surjective:} Suppose that $h \in H$. Since $\d$ is surjective, there exists an element $d \in D$ such that $r(h) = \d(d) \in I$. By the commutativity of the rightmost square, $s(r(h)) = \e(n(d))$. The exactness at $I$ means that $\im r = \ker s$, so hence $\e(n(d)) = s(r(h)) = 0$. Because $\e$ is injective, $n(d) = 0$. Then $d \in \ker(n) = \im(m)$ by exactness at $D$. Therefore, there is some $c \in C$ such that $m(c) = d$. Now, $\d(m(c)) = \d(d) = r(h)$ and by the commutativity of squares, $\d(m(c)) = r(\g(c))$, so therefore $r(\g(c)) = r(h)$. Since $r$ is a homomorphism, $r(\g(c) - h) = 0$. Hence $\g(c) - h \in \ker r = \im q$ by exactness at $H$. Therefore, there exists $g \in G$ such that $q(g) = \g(c) - h$. $\b$ is surjective, so there is some $b \in B$ such that $\b(b) = g$ and hence $q(\b(b)) = \g(c) - h$. By the commutativity of squares, $q(\b(b)) = \g(l(b)) = \g(c) - h$. Hence $h = \g(c) - \g(l(b)) = \g(c - l(b))$, and therefore $\g$ is surjective. So far, we've used that $\b$ and $\g$ are surjective, $\e$ is injective, and exactness at $D$, $H$, $I$. \emph{We show that $\g$ is injective:} Suppose that $c \in C$ and $\g(c) = 0$. Then $r(\g(c)) = 0$, and by the commutativity of squares, $\d(m(c)) = 0$. Since $\d$ is injective, $m(c) = 0$, so $c \in \ker m = \im l$ by exactness at $C$. Therefore, there is $b \in B$ such that $l(b) = c$. Then $\g(l(b)) = \g(c) = 0$, and by the commutativity of squares, $q(\b(b)) = 0$. Therefore, $\b(b) \in \ker q$, and by exactness at $G$, $\b(b) \in \ker q = \im p$. There is now $f \in F$ such that $p(f) = \b(b)$. Since $\a$ is surjective, this means that there is $a \in A$ such that $f = \a(a)$, so then $\b(b) = p(\a(a))$. By commutativity of squares, $\b(b) = p(\a(a)) = \b(k(a))$, and hence $\b(k(a) - b) = 0$. Since $\b$ is injective, we have $k(a) -b = 0$, so $k(a) = b$. Hence $b \in \im k = \ker l$ by commutativity of squares, so $l(b) = 0$. However, we defined $b$ to satisfy $l(b) = c$, so therefore $c = 0$ and hence $\g$ is injective. Here, we used that $\a$ is surjective, $\b, \d$ are injective, and exactness at $B, C, G$. Putting the two statements together, we see that $\g$ is both surjective and injective, so $\g$ is an isomorphism. We only used that $\b, \d$ are isomorphisms and that $\a$ is surjective, $\e$ is injective, so we can slightly weaken the hypotheses; injectivity of $\a$ and surjectivity of $\e$ were unnecessary. \end{proof} \section{Ideals} The notion of an \emph{ideal} has already been defined. Now we will introduce additional terminology related to the theory of ideals. \subsection{Prime and maximal ideals} Recall that the notion of an ideal generalizes that of divisibility. In elementary number theory, though, one finds that questions of divisibility basically reduce to questions about primes. The notion of a ``prime ideal'' is intended to generalize the familiar idea of a prime number. \begin{definition} An ideal $I \subset R$ is said to be \textbf{prime} if \begin{enumerate}[\textbf{P} 1] \item $1 \notin I$ (by convention, 1 is not a prime number) \item If $xy \in I$, either $x \in I$ or $y \in I$. \end{enumerate} \end{definition} \begin{example} \label{integerprimes} If $R = \mathbb{Z}$ and $p \in R$, then $(p) \subset \mathbb{Z}$ is a prime ideal iff $p$ or $-p$ is a prime number in $\mathbb{N}$ or if $p$ is zero. \end{example} If $R$ is any commutative ring, there are two obvious ideals. These obvious ones are the zero ideal $(0)$ consisting only of the zero element, and the unit element $(1)$ consisting of all of $R$. \begin{definition} \label{maximalideal} An ideal $I \subset R$ is called \textbf{maximal}\footnote{Maximal with respect to not being the unit ideal.} if \begin{enumerate}[\textbf{M} 1] \item $1 \notin I$ \item Any larger ideal contains $1$ (i.e., is all of $R$). \end{enumerate} \end{definition} So a maximal ideal is a maximal element in the partially ordered set of proper ideals (an ideal is \textbf{proper} if it does not contain 1). \begin{exercise} Find the maximal ideals in $\mathbb{C}[t]$. \end{exercise} \begin{proposition} A maximal ideal is prime. \end{proposition} \begin{proof} First, a maximal ideal does not contain 1. Let $I \subset R$ be a maximal ideal. We need to show that if $xy \in I$, then one of $x,y \in I$. If $x \notin I$, then $(I,x) = I + (x)$ (the ideal generated by $I$ and $x$) strictly contains $I$, so by maximality contains $1$. In particular, $1 \in I+(x)$, so we can write \[ 1 = a + xb \] where $a \in I, b \in R$. Multiply both sides by $y$: \[ y = ay + bxy. \] Both terms on the right here are in $I$ ($a \in I$ and $xy \in I$), so we find that $y \in I$. \end{proof} Given a ring $R$, what can we say about the collection of ideals in $R$? There are two obvious ideals in $R$, namely $(0)$ and $ (1)$. These are the same if and only if $0=1$, i.e. $R$ is the zero ring. So for any nonzero commutative ring, we have at least two distinct ideals. Next, we show that maximal ideals always \emph{do} exist, except in the case of the zero ring. \begin{proposition} \label{anycontainedinmaximal} Let $R$ be a commutative ring. Let $I \subset R$ be a proper ideal. Then $I$ is contained in a maximal ideal. \end{proposition} \begin{proof} This requires the axiom of choice in the form of Zorn's lemma. Let $P$ be the collection of all ideals $J \subset R$ such that $I \subset J$ and $J \neq R$. Then $P$ is a poset with respect to inclusion. $P$ is nonempty because it contains $I$. Note that given a (nonempty) linearly ordered collection of ideals $J_{\alpha} \in P$, the union $\bigcup J_{\alpha} \subset R$ is an ideal: this is easily seen in view of the linear ordering (if $x,y \in \bigcup J_{\alpha}$, then both $x,y$ belong to some $J_{\gamma}$, so $x+y \in J_{\gamma}$; multiplicative closure is even easier). The union is not all of $R$ because it does not contain $1$. This implies that $P$ has a maximal element by Zorn's lemma. This maximal element may be called $\mathfrak{M}$; it's a proper element containing $I$. I claim that $\mathfrak{M}$ is a maximal ideal, because if it were contained in a larger ideal, that would be in $P$ (which cannot happen by maximality) unless it were all of $R$. \end{proof} \begin{corollary} Let $R $ be a nonzero commutative ring. Then $R$ has a maximal ideal. \end{corollary} \begin{proof} Apply the lemma to the zero ideal. \end{proof} \begin{corollary} Let $R$ be a nonzero commutative ring. Then $x \in R$ is invertible if and only if it belongs to no maximal ideal $\mathfrak{m} \subset R$. \end{corollary} \begin{proof} Indeed, $x$ is invertible if and only if $(x) = 1$. That is, if and only if $(x)$ is not a proper ideal; now \rref{anycontainedinmaximal} finishes the argument. \end{proof} \subsection{Fields and integral domains} Recall: \begin{definition} A commutative ring $R$ is called a \textbf{field} if $1 \neq 0$ and for every $x \in R - \left\{0\right\}$ there exists an \textbf{inverse} $x^{-1} \in R$ such that $xx^{-1} = 1$. \end{definition} This condition has an obvious interpretation in terms of ideals. \begin{proposition} A commutative ring with $1 \neq 0$ is a field iff it has only the two ideals $(1), (0)$. \end{proposition} Alternatively, a ring is a field if and only if $(0)$ is a maximal ideal. \begin{proof} Assume $R$ is a field. Suppose $I \subset R$. If $I \neq (0)$, then there is a nonzero $x \in I$. Then there is an inverse $x^{-1}$. We have $x^{-1} x =1 \in I$, so $I = (1)$. In a field, there is thus no room for ideals other than $(0)$ and $(1)$. To prove the converse, assume every ideal of $R$ is $(0)$ or $(1)$. Then for each $x \in R$, $(x) = (0)$ or $(1)$. If $x \neq 0$, the first cannot happen, so that means that the ideal generated by $x$ is the unit ideal. So $1$ is a multiple of $x$, implying that $x$ has a multiplicative inverse. \end{proof} So fields also have an uninteresting ideal structure. \begin{corollary} \label{maximalfield} If $R$ is a ring and $I \subset R$ is an ideal, then $I$ is maximal if and only if $R/I$ is a field. \end{corollary} \begin{proof} The basic point here is that there is a bijection between the ideals of $R/I$ and ideals of $R$ containing $I$. Denote by $\phi: R \to R/I$ the reduction map. There is a construction mapping ideals of $R/I$ to ideals of $R$. This sends an ideal in $R/I$ to its inverse image. This is easily seen to map to ideals of $R$ containing $I$. The map from ideals of $R/I$ to ideals of $R$ containing $I$ is a bijection, as one checks easily. It follows that $R/I$ is a field precisely if $R/I$ has precisely two ideals, i.e. precisely if there are precisely two ideals in $R$ containing $I$. These ideals must be $(1)$ and $I$, so this holds if and only if $I$ is maximal. \end{proof} There is a similar characterization of prime ideals. \begin{definition} A commutative ring $R$ is an \textbf{integral domain} if for all $ x,y \in R$, $x \neq 0 $ and $y \neq 0$ imply $xy \neq 0$. \end{definition} \begin{proposition} \label{primeifdomain} An ideal $I \subset R$ is prime iff $R/I$ is a domain. \end{proposition} \begin{exercise} Prove \rref{primeifdomain}. \end{exercise} Any field is an integral domain. This is because in a field, nonzero elements are invertible, and the product of two invertible elements is invertible. This statement translates in ring theory to the statement that a maximal ideal is prime. Finally, we include an example that describes what \emph{some} of the prime ideals in a polynomial ring look like. \begin{example} Let $R$ be a ring and $P$ a prime ideal. We claim that $PR[x] \subset R[x]$ is a prime ideal. Consider the map $\tilde{\phi}:R[x]\rightarrow(R/P)[x]$ with $\tilde{\phi}(a_0+\cdots+a_nx^n)=(a_0+P)+\cdots+(a_n+P)x^n$. This is clearly a homomorphism because $\phi:R\rightarrow R/P$ is, and its kernel consists of those polynomials $a_0+\cdots+a_nx^n$ with $a_0,\ldots,a_n\in P$, which is precisely $P[x]$. Thus $R[x]/P[x]\simeq (R/P)[x]$, which is an integral domain because $R/P$ is an integral domain. Thus $P[x]$ is a prime ideal. However, if $P$ is a maximal ideal, then $P[x]$ is never a maximal ideal because the ideal $P[x]+(x)$ (the polynomials with constant term in $P$) always strictly contains $P[x]$ (because if $x\in P[x]$ then $1\in P$, which is impossible). Note that $P[x]+(x)$ is the kernel of the composition of $\tilde{\phi}$ with evaluation at 0, i.e $(\text{ev}_0\circ\tilde{\phi}):R[x]\rightarrow R/P$, and this map is a surjection and $R/P$ is a field, so that $P[x]+(x)$ is the maximal ideal in $R[x]$ containing $P[x]$. \end{example} \begin{exercise} \label{quotfld1} Let $R$ be a domain. Consider the set of formal quotients $a/b, a, b \in R$ with $b \neq 0$. Define addition and multiplication using usual rules. Show that the resulting object $K(R)$ is a ring, and in fact a \emph{field}. The natural map $R \to K(R)$, $r \to r/1$, has a universal property. If $R \hookrightarrow L$ is an injection of $R$ into a field $L$, then there is a unique morphism $K(R) \to L$ of fields extending $R \to L$. This construction will be generalized when we consider \emph{localization.} This construction is called the \textbf{quotient field.} Note that a non-injective map $R\to L$ will \emph{not} factor through the quotient field! \end{exercise} \begin{exercise}\label{Jacobson} Let $R$ be a commutative ring. Then the \textbf{Jacobson radical} of $R$ is the intersection $\bigcap \mathfrak{m}$ of all maximal ideals $\mathfrak{m} \subset R$. Prove that an element $x$ is in the Jacobson radical if and only if $1 - yx$ is invertible for all $y \in R$. \end{exercise} \subsection{Prime avoidance} The following fact will come in handy occasionally. We will, for instance, use it much later to show that an ideal consisting of zerodivisors on a module $M$ is contained in associated prime. \begin{theorem}[Prime avoidance] \label{primeavoidance} Let $I_1,\dots, I_n \subset R$ be ideals. Let $A\subset R$ be a subset which is closed under addition and multiplication. Assume that at least $n-2$ of the ideals are prime. If $A\subset I_1\cup \cdots \cup I_n$, then $A\subset I_j$ for some $j$. \end{theorem} The result is frequently used in the following specific case: if an ideal $I$ is contained in a finite union $\bigcup \mathfrak{p}_i$ of primes, then $I \subset \mathfrak{p}_i$ for some $i$. \begin{proof} Induct on $n$. If $n=1$, the result is trivial. The case $n=2$ is an easy argument: if $a_1\in A\smallsetminus I_1$ and $a_2\in A\smallsetminus I_2$, then $a_1+a_2\in A\smallsetminus (I_1\cup I_2)$. Now assume $n\ge 3$. We may assume that for each $j$, $A\not\subset I_1\cup \cdots \cup \hat I_j\cup \cdots I_n$.\footnote{The hat means omit $I_j$.} Fix an element $a_j\in A\smallsetminus (I_1\cup \cdots \cup \hat I_j\cup \cdots I_n)$. Then this $a_j$ must be contained in $I_j$ since $A\subset \bigcup I_j$. Since $n\ge 3$, one of the $I_j$ must be prime. We may assume that $I_1$ is prime. Define $x=a_1+a_2a_3\cdots a_n$, which is an element of $A$. Let's show that $x$ avoids \emph{all} of the $I_j$. If $x\in I_1$, then $a_2a_3\cdots a_n\in I_1$, which contradicts the fact that $a_i\not\in I_j$ for $i\neq j$ and that $I_1$ is prime. If $x\in I_j$ for $j\ge 2$. Then $a_1\in I_j$, which contradicts $a_i\not\in I_j$ for $i\neq j$. \end{proof} \subsection{The Chinese remainder theorem} Let $m,n$ be relatively prime integers. Suppose $a, b \in \mathbb{Z}$; then one can show that the two congruences $x \equiv a \mod m$ and $x \equiv b \mod n$ can be solved simultaneously in $x \in \mathbb{Z}$. The solution is unique, moreover, modulo $mn$. The Chinese remainder theorem generalizes this fact: \begin{theorem}[Chinese remainder theorem] Let $I_1, \dots I_n$ be ideals in a ring $R$ which satisfy $I_i + I_j = R$ for $i \neq j$. Then we have $I_1 \cap \dots \cap I_n = I_1 \dots I_n$ and the morphism of rings \[ R \to \bigoplus R/I_i \] is an epimorphism with kernel $I_1 \cap \dots \cap I_n$. \end{theorem} \begin{proof} First, note that for any two ideals $I_1$ and $I_2$, we have $I_1I_2\subset I_1\cap I_2$ and $(I_1+I_2)(I_1\cap I_2)\subset I_1I_2$ (because any element of $I_1+I_2$ multiplied by any element of $I_1\cap I_2$ will clearly be a sum of products of elements from both $I_1$ and $I_2$). Thus, if $I_1$ and $I_2$ are coprime, i.e. $I_1+I_2=(1)=R$, then $(1)(I_1\cap I_2)=(I_1\cap I_2)\subset I_1I_2\subset I_1\cap I_2$, so that $I_1\cap I_2=I_1I_2$. This establishes the result for $n=2$. If the ideals $I_1,\ldots,I_n$ are pairwise coprime and the result holds for $n-1$, then $$\bigcap_{i=1}^{n-1} I_i=\prod_{i=1}^{n-1}I_i.$$ Because $I_n+I_i=(1)$ for each $1\leq i\leq n-1$, there must be $x_i\in I_n$ and $y_i\in I_i$ such that $x_i+y_i=1$. Thus, $z_n=\prod_{i=1}^{n-1}y_i=\prod_{i=1}^{n-1}(1-x_i)\in \prod_{i=1}^{n-1} I_i$, and clearly $z_n+I_n=1+I_n$ since each $x_i\in I_n$. Thus $I_n+\prod_{i=1}^{n-1}I_i=I_n+\bigcap_{i=1}^{n-1}I_i=(1)$, and we can now apply the $n=2$ case to conclude that $\bigcap_{i=1}^n I_i=\prod_{i=1}^n I_i$. Note that for any $i$, we can construct a $z_i$ with $z_i\in I_j$ for $j\neq i$ and $z_i+I_i=1+I_i$ via the same procedure. Define $\phi:R\rightarrow\bigoplus R/I_i$ by $\phi(a)=(a+I_1,\ldots,a+I_n)$. The kernel of $\phi$ is $\bigcap_{i=1}^n I_i$, because $a+I_i=0+I_i$ iff $a\in I_i$, so that $\phi(a)=(0+I_1,\ldots,0+I_n)$ iff $a\in I_i$ for all $i$, that is, $a\in\bigcap_{i=1}^n I_i$. Combined with our previous result, the kernel of $\phi$ is $\prod_{i=1}^n I_i$. Finally, recall that we constructed $z_i\in R$ such that $z_i+I_i=1+I_i$, and $z+I_j=0+I_j$ for all $j\neq i$, so that $\phi(z_i)=(0+I_1,\ldots,1+I_{i},\ldots,0+I_n)$. Thus, $\phi(a_1z_1+\cdots+a_nz_n)=(a_1+I_1,\ldots,a_n+I_n)$ for all $a_i\in R$, so that $\phi$ is onto. By the first isomorphism theorem, we have that $R/I_1\cdots I_n\simeq \bigoplus_{i=1}^nR/I_i$. \\ \end{proof} \section{Some special classes of domains} \subsection{Principal ideal domains} \begin{definition} A ring $R$ is a \textbf{principal ideal domain} or \textbf{PID} if $R \neq 0$, $R$ is not a field, $R$ is a domain, and every ideal of $R$ is principal. \end{definition} These have the next simplest theory of ideals. Each ideal is very simple---it's principal---though there might be a lot of ideals. \begin{example} $\mathbb{Z}$ is a PID. The only nontrivial fact to check here is that: \begin{proposition} Any nonzero ideal $I \subset \mathbb{Z}$ is principal. \end{proposition} \begin{proof} If $I = (0)$, then this is obvious. Else there is $n \in I - \left\{0\right\}$; we can assume $n>0$. Choose $n \in I$ as small as possible and positive. Then I claim that the ideal $I$ is generated by $(n)$. Indeed, we have $(n) \subset I$ obviously. If $m \in I$ is another integer, then divide $m$ by $n$, to find $m = nb + r$ for $r \in [0, n)$. We find that $r \in I$ and $0 \leq r < n$, so $r=0$, and $m$ is divisible by $n$. And $I \subset (n)$. So $I = (n)$. \end{proof} \end{example} A module $M$ is said to be \emph{finitely generated} if there exist elements $x_1, \dots, x_n \in M$ such that any element of $M$ is a linear combination (with coefficients in $R$) of the $x_i$. (We shall define this more formally below.) One reason that PIDs are so convenient is: \begin{theorem}[Structure theorem] \label{structurePID} If $M$ is a finitely generated module over a principal ideal domain $R$, then $M$ is isomorphic to a direct sum \[ M \simeq \bigoplus_{i=1}^n R/a_i, \] for various $a_i \in R$ (possibly zero). \end{theorem} \add{at some point, the proof should be added. This is important!} \subsection{Unique factorization domains} The integers $\mathbb{Z}$ are especially nice because of the fundamental theorem of arithmetic, which states that every integer has a unique factorization into primes. This is not true for every integral domain. \begin{definition} An element of a domain $R$ is \textbf{irreducible} if it cannot be written as the product of two non-unit elements of $R$. \end{definition} \begin{example} Consider the integral domain $\mathbb{Z}[\sqrt{-5}]$. We saw earlier that \[ 6 = 2 \cdot 3 = (1 + \sqrt{-5})(1 - \sqrt{-5}), \] which means that $6$ was written as the product of two non-unit elements in different ways. $\mathbb{Z}[\sqrt{-5}]$ does not have unique factorization. \end{example} \begin{definition} \label{earlyUFD} A domain $R$ is a \textbf{unique factorization domain} or \textbf{UFD} if every non-unit $x \in R$ satisfies \begin{enumerate} \item $x$ can be written as a product $x = p_1 p_2 \cdots p_n$ of irreducible elements $p_i \in R$ \item if $x = q_1 q_2 \cdots q_m$ where $q_i \in R$ are irreducible then the $p_i$ and $q_i$ are the same up to order and multiplication by units. \end{enumerate} \end{definition} \begin{example} $\mathbb{Z}$ is a UFD, while $\mathbb{Z}[\sqrt{-5}]$ is not. In fact, many of our favorite domains have unique factorization. We will prove that all PIDs are UFDs. In particular, in \rref{gaussianintegersareprincipal} and \rref{polyringisprincipal}, we saw that $\mathbb{Z}[i]$ and $F[t]$ are PIDs, so they also have unique factorization. \end{example} \begin{theorem} \label{PIDUFD} Every principal ideal domain is a unique factorization domain. \end{theorem} \begin{proof} Suppose that $R$ is a principal ideal domain and $x$ is an element of $R$. We first demonstrate that $x$ can be factored into irreducibles. If $x$ is a unit or an irreducible, then we are done. Therefore, we can assume that $x$ is reducible, which means that $x = x_1 x_2$ for non-units $x_1, x_2 \in R$. If there are irreducible, then we are again done, so we assume that they are reducible and repeat this process. We need to show that this process terminates. Suppose that this process continued infinitely. Then we have an infinite ascending chain of ideals, where all of the inclusions are proper: $(x) \subset (x_1) \subset (x_{11}) \subset \cdots \subset R$. We will show that this is impossible because any infinite ascending chain of ideals $I_1 \subset I_2 \subset \cdots \subset R$ of a principal ideal domain eventually becomes stationary, i.e. for some $n$, $I_k = I_n$ for $k \geq n$. Indeed, let $I = \bigcup_{i=1}^\infty I_i$. This is an ideal, so it is principally generated as $I = (a)$ for some $a$. Since $a \in I$, we must have $a \in I_N$ for some $N$, which means that the chain stabilizes after $I_N$. It remains to prove that this factorization of $x$ is unique. We induct on the number of irreducible factors $n$ of $x$. If $n = 0$, then $x$ is a unit, which has unique factorization up to units. Now, suppose that $x = p_1 \cdots p_n = q_1 \cdots q_m$ for some $m \ge n$. Since $p_1$ divides $x$, it must divide the product $q_1 \cdots q_m$ and by irreducibility, one of the factors $q_i$. Reorder the $q_i$ so that $p_1$ divides $q_1$. However, $q_1$ is irreducible, so this means that $p_1$ and $q_1$ are the same up to multiplication by a unit $u$. Canceling $p_1$ from each of the two factorizations, we see that $p_2 \cdots p_n = u q_2 \cdots q_m = q_2' \cdots q_m$. By induction, this shows that the factorization of $x$ is unique up to order and multiplication by units. \end{proof} \subsection{Euclidean domains} A euclidean domain is a special type of principal ideal domain. In practice, it will often happen that one has an explicit proof that a given domain is euclidean, while it might not be so trivial to prove that it is a UFD without the general implication below. \begin{definition} An integral domain $R$ is a \textbf{euclidean domain} if there is a function $|\cdot |:R\to \mathbb \mathbb{Z}_{\geq 0}$ (called the norm) such that the following hold. \begin{enumerate} \item $|a|=0$ iff $a=0$. \item For any nonzero $a,b\in R$ there exist $q,r\in R$ such that $b=aq+r$ and $|r|<|a|$. \end{enumerate} In other words, the norm is compatible with division with remainder. \end{definition} \begin{theorem}\label{EDPID} A euclidean domain is a principal ideal domain. \end{theorem} \begin{proof} Let $R$ be an euclidean domain, $I\subset R$ and ideal, and $b$ be the nonzero element of smallest norm in $I$. Suppose $ a\in I$. Then we can write $ a = qb + r$ with $ 0\leq r < |b|$, but since $ b$ has minimal nonzero absolute value, $ r = 0$ and $ b|a$. Thus $ I=(b)$ is principal. \end{proof} As we will see, this implies that any euclidean domain admits \emph{unique factorization.} \begin{proposition} \label{polyringED} Let $F$ be a field. Then the polynomial ring $F[t]$ is a euclidean domain. In particular, it is a PID. \end{proposition} \begin{proof} We define \add{} \end{proof} \begin{exercise} \label{gaussianintegersareprincipal} Prove that $\mathbb{Z}[i]$ is principal. (Define the norm as $N(a+ib) = a^2 + b^2$.) \end{exercise} \begin{exercise} \label{polyringisprincipal} Prove that the polynomial ring $F[t]$ for $F$ a field is principal. \end{exercise} It is \emph{not} true that a PID is necessarily euclidean. Nevertheless, it was shown in \cite{Gre97} that the converse is ``almost'' true. Namely, \cite{Gre97} defines the notion of an \textbf{almost euclidean domain.} A domain $R$ is almost euclidean if there is a function $d: R \to \mathbb{Z}_{\geq 0}$ such that \begin{enumerate} \item $d(a) = 0$ iff $a = 0$. \item $d(ab) \geq d(a)$ if $b \neq 0$. \item If $a,b \in R - \left\{0\right\}$, then either $b \mid a$ or there is $r \in (a,b)$ with $d(r)}[d] & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K \ar[r] & R^m \ar[r] & M \ar[r] & 0 } \] The dotted arrow $F \to R^m$ exists as $F$ is projective. There is induced a map $F' \to K$. We get a commutative and exact diagram \[ \xymatrix{ & F' \ar[r]\ar[d]^f & F \ar[r] \ar[d]^g & M \ar[r] \ar[d] & 0 \\ 0 \ar[r] & K \ar[r] & R^m \ar[r] & M \ar[r] & 0 }, \] to which we can apply the snake lemma. There is an exact sequence \[ 0 \to \coker(f) \to \coker(g) \to 0, \] which gives an isomorphism $\coker(f) \simeq \coker(g)$. However, $\coker(g)$ is finitely generated, as a quotient of $R^m$. Thus $\coker(f)$ is too. Since we have an exact sequence \[ 0 \to \im(f) \to K \to \coker(f) \to 0, \] and $\im(f)$ is finitely generated (as the image of a finitely generated object, $F'$), we find by \rref{exact-fingen} that $\coker(f)$ is finitely generated. \end{proof} \begin{proposition} \label{exact-finpres} Given an exact sequence \[ 0 \to M' \to M \to M'' \to 0, \] if $M', M''$ are finitely presented, so is $M$. \end{proposition} In general, it is not true that if $M$ is finitely presented, then $M'$ and $M''$ are. For instance, it is possible that a submodule of the free, finitely generated module $R$ (i.e. an ideal), might fail to be finitely generated. We shall see in \rref{noetherian} that this does not happen over a \emph{noetherian} ring. \begin{proof} Indeed, suppose we have exact sequences \[ F_1' \to F_0' \to M' \to 0 \] and \[ F_1'' \to F_0'' \to M'' \to 0 \] where the $F$'s are finitely generated and free. We need to get a similar sequence for $M$. Let us stack these into a diagram \[ \xymatrix{ & F_1' \ar[d] & & F_1'' \ar[d] \\ & F_0' \ar[d] & & F_0'' \ar[d] \\ 0 \ar[r] & M' \ar[r] & M \ar[r] & M'' \ar[r] & 0 }\] However, now, using general facts about projective modules (\rref{}), we can splice these presentations into a resolution \[ F_1' \oplus F_1'' \to F_0' \oplus F_0'' \to M \to 0, \] which proves the assertion. \end{proof} \begin{corollary} The (finite) direct sum of finitely presented modules is finitely presented. \end{corollary} \begin{proof} Immediate from \rref{exact-finpres} \end{proof} \subsection{Modules of finite length} A much stronger condition on modules that of finite generation is that of \emph{finite length}. Here, basically any operation one does will eventually terminate. Let $R$ be a commutative ring, $M$ an $R$-module. \begin{definition} $M$ is \textbf{simple} if $M \neq 0$ and $M$ has no nontrivial submodules. \end{definition} \begin{exercise} A torsion-free abelian group is never a simple $\mathbb{Z}$-module. \end{exercise} \begin{proposition} $M$ is simple if and only if it is isomorphic to $R/\mathfrak{m}$ for $\mathfrak{m} \subset R$ a maximal ideal. \end{proposition} \begin{proof} Let $M$ be simple. Then $M$ must contain a cyclic submodule $Rx$ generated by some $x \in M - \left\{0\right\}$. So it must contain a submodule isomorphic to $R/I$ for some ideal $I$, and simplicity implies that $M \simeq R/I$ for some $I$. If $I$ is not maximal, say properly contained in $J$, then we will get a nontrivial submodule $J/I$ of $R/I \simeq M$. Conversely, it is easy to see that $R/\mathfrak{m}$ is simple for $\mathfrak{m}$ maximal. \end{proof} \begin{exercise}[Schur's lemma] Let $f: M \to N$ be a module-homomorphism, where $M, N$ are both simple. Then either $f =0$ or $f$ is an isomorphism. \end{exercise} \begin{definition} $M$ is of \textbf{finite length} if there is a finite filtration $0 \subset M^0 \subset \dots \subset M^n = M$ where each $M^i/M^{i-1}$ is simple. \end{definition} \begin{exercise} Modules of finite length are closed under extensions (that is, if $0 \to M' \to M \to M'' \to 0$ is an exact sequence, then if $M', M''$ are of finite length, so is $M$). \end{exercise} In the next result (which will not be used in this chapter), we shall use the notions of a \emph{noetherian} and an \emph{artinian} module. These notions will be developed at length in \cref{chnoetherian}, and we refer the reader there for more explanation. A module is \emph{noetherian} if every ascending chain $M_1 \subset M_2 \subset \dots$ of submodules stabilizes, and it is \emph{artinian} if every descending chain stabilizes. \begin{proposition} $M$ is finite length iff $M$ is both noetherian and artinian. \end{proposition} \begin{proof} Any simple module is obviously both noetherian and artinian: there are two submodules. So if $M$ is finite length, then the finite filtration with simple quotients implies that $M$ is noetherian and artinian, since these two properties are stable under extensions (\rref{exactnoetherian} and \rref{exactartinian} of \rref{noetherian}). Suppose $M \neq 0$ is noetherian and artinian. Let $M_1 \subset M$ be a minimal nonzero submodule, which exists as $M$ is artinian. This is necessarily simple. Then we have a filtration \[ 0 = M_0 \subset M_1. \] If $M_1 = M$, then the filtration goes up to $M$, and we have that $M$ is of finite length. If not, find a submodule $M_2$ that contains $M_1$ and is minimal among submodules containing $M_1$; then the quotient $M_2/M_1$ is simple. We have the filtration \[ 0 = M_0 \subset M_1 \subset M_2, \] which we can keep continuing until at some point we reach $M$. Note that since $M$ is noetherian, we cannot continue this strictly ascending chain forever. \end{proof} \begin{exercise} In particular, any submodule or quotient module of a finite length module is of finite length. Note that the analog is not true for finitely generated modules unless the ring in question is noetherian. \end{exercise} Our next goal is to show that the length of a filtration of a module with simple quotients is well-defined. For this, we need: \begin{lemma} \label{simplefiltrationint} Let $0 = M_0 \subset M_1 \subset \dots \subset M_n = M$ be a filtration of $M$ with simple quotients. Let $N \subset M$. Then the filtration $0 = M_0 \cap N \subset M_1 \cap N \subset \dots \subset N$ has simple or zero quotients. \end{lemma} \begin{proof} Indeed, for each $i$, $(N \cap M_i)/(N \cap M_{i-1})$ is a submodule of $M_i / M_{i-1}$, so is either zero or simple. \end{proof} \begin{theorem}[Jordan-H\"older]\label{lengthexists} Let $M$ be a module of finite length. In this case, any two filtrations on $M$ with simple quotients have the same length. \end{theorem} \begin{definition} This number is called the \textbf{length} of $M$ and is denoted $\ell(M)$. \end{definition} \begin{proof}[Proof of \rref{lengthexists}] Let us introduce a temporary definition: $l(M)$ is the length of the \emph{minimal} filtration on $M$. We will show that any filtration of $M$ (with simple quotients) is of length $l(M)$. This is the proposition in another form. The proof of this claim is by induction on $l(M)$. Suppose we have a filtration \[ 0 = M_0 \subset M_1 \subset \dots \subset M_n = M \] with simple quotients. We would like to show that $n = l(M)$. By definition of $l(M)$, there is another filtration \[ 0 = N_0 \subset \dots \subset N_{l(M)} = M. \] If $l(M) = 0,1$, then $M$ is zero or simple, which will necessarily imply that $n=0,1$ respectively. So we can assume $l(M) \geq 2$. We can also assume that the result is known for strictly smaller submodules of $M$. There are two cases: \begin{enumerate} \item $M_{n-1} = N_{l(M) -1 } $. Then $M_{n-1} = N_{l(M)-1}$ has $l$ at most $l(M)-1$. Thus by the inductive hypothesis any two filtrations on $M_{n-1}$ have the same length, so $n-1 = l(M) -1$, implying what we want. \item We have $M_{n-1} \cap N_{l(M) - 1} \subsetneq M_{n-1}, N_{l(M)-1}$. Call this intersection $K$. Now we have two filtrations of these modules $M_{n-1}, N_{l(M)-1}$ whose quotients are simple. We can replace them such that the next term before them is $K$. To do this, consider the filtrations \[ 0 = M_0 \cap K \subset M_1 \subset K \subset \dots M_{n-1} \cap K = K \subset M_{n-1} \] and \[ 0 = N_0 \cap K \subset M_1 \subset K \subset \dots N_{l(M)-1} \cap K = K \subset N_{l(M)-1} . \] These filtrations have simple or zero quotients by \rref{simplefiltrationint}, and since $ M_{n-1}/K = M_{n-1}/M_{n-1} \cap N_{l(M)-1} = M/M_{n-1}$ is simple, and similarly for $N_{l(M)-1}/K$. We can throw out redundancies to eliminate the zero terms. So we get two new filtrations of $M_{n-1}$ and $N_{l(M)-1}$ whose second-to-last term is $K$. By the inductive hypothesis any two filtrations on either of these proper submodules $M_{n-1}, N_{l(M)-1} $ have the same length. Thus the lengths of the two new filtrations are $n-1$ and $l(M)-1$, respectively. So we find that $n-1 = l(K) +1$ and $l(M)-1 = l(K)+1$ by the inductive hypothesis. This implies what we want. \end{enumerate} \end{proof} \begin{exercise} Prove that the successive quotients $M_i/M_{i-1}$ are also determined (up to permutation). \end{exercise}