# [ac/1] Ideals, Modules, Tensor products

## Ideals

An ideal \(I\) of \(A\) is defined to be a additive subgroup of \(A\) that is closed under multiplication with elements of \(A\); that is, \(Ia \in I \, \forall a \in A\). The `quotient ring` \(A/I\) inherits a uniquely defined multiplication from \(A\), making it into a subring, the quotient ring or the `residue-class ring`. \(\varphi : A \rightarrow A/I\) is a `ring homomorphism`, and it maps every \(x \in A\) to the `coset` \(x + I\).

There is a one-to-one order-preserving correspondence between ideals \(J\) of \(A\) which contain \(I\), and ideals \(\bar{J}\) of \(A/I\), given by \(J = \varphi^{-1}(\bar{J})\). Notice that the kernel of the homomorphism \(f : A \rightarrow B\) is \(I\), image \(f(A)\) is a subring \(C \subseteq B\), and \(f\) induces the homomorphism \(A/I \cong C\).

A `zero-divisor` of an element \(x \in A\) is, if it exists, an element \(y \neq 0 \implies xy = 0\). A ring with no zero-divisors is called an `integral domain`. A `unit` of an element \(x \in A\) is an element \(y : xy = 1\); then \(y = x^{-1}\). The units in \(A\) form a multiplicative abelian group, and a ring in which every element has an inverse is called a `field`; obviously every field is an integral domain. The multiples \(Ix, x \in A\) form the `principal ideal`, and is denoted by \((I)\).

\(P\) is a `prime ideal` of \(A\) if \(P \neq (1)\) and \(xy \in P \implies x \in P \vee y \in P\) for \(x, y \in A\). \(M\) is a `maximal ideal` of \(A\) if \(M \neq (1)\), and there exists no ideal \(I\) strictly between \(M\) and \(A\), that is no \(I : M \subset I \subset A\). In \(A[x]\), prime ideals consist of \((0)\), maximals, and irreducibles.

Three elementary results (proved in [ra/hw](/ra/hw)):

- \(M\) is maximal \(\iff A/M\) is a field.
- \(P\) is prime \(\iff A/P\) is an integral domain.
- Every maximal ideal is a prime ideal.

If \(f : A \rightarrow B\) is a ring homomorphism, and \(Q\) is a prime ideal of \(B\), then \(f^{-1}(Q)\) is a prime ideal of \(A\), for \(A/f^{-1}(Q)\) is isomorphic to a subring of \(B/Q\), and hence has no zero-divisors. But if \(N\) is a maximal ideal in \(B\), \(A/f^{-1}(N)\) need not be a maximal ideal in \(A\); all we can say is that it is prime.

Every ring \(A \neq 0\) has a maximal ideal, and this can be proved using `Zorn's lemma`. If \(A \neq (1)\) is an ideal of \(A\), then there is some maximal ideal containing \(A\). It follows that every non-unit of \(A\) is contained in some maximal ideal. A ring \(A\) with exactly one maximal ideal \(M\) is called a `local ring`, and the field \(A/M\) is called the `residue field`.

Let \(A\) be a ring, then:

- Let \(M\) be an ideal such that every \(x \in A - M\) is a unit in \(A\). Then, \(A\) is a local ring, and \(M\) is a maximal ideal.
- Let \(M\) be a maximal ideal such that every \(1 + M \in A\) is a unit in \(A\). Then, \(A\) is a local ring.

To prove (i), notice that every ideal \(\neq (1)\) consists of non-units and is hence contained within \(M\).

To prove (ii), let \(x \in A - M\). Since \(M\) is maximal, the ideal generated by \(x\) and \(M\) is \((1)\), hence there exists \(m \in M, y \in A\) such that \(xy + m = 1\) or \(xy = 1 - m\); this is contained in \(1 + M\) and is hence unit.

What follows is a bunch of examples:

- Let \(A = k[x_0, \ldots, x_n]\), the ring of polynomials. If \(f\) is an irreducible polynomial, then, by unique factorization, \((f)\) is a prime ideal. Similarly, if \(A\) is a ring of polynomials with constant term zero, then every ideal in \(A\) is a maximal ideal; it is the kernel of the homomorphism \(f : A \rightarrow k\), which maps every \(f \in A\) to \(f(0)\).
- Let \(A = \mathbb{Z}\) with ideals of the form \((m)\). Then \((m)\) is a prime ideal \(\iff\) \(m\) is prime number or \(0\). All prime ideals \((p)\) are maximal: \(A/(p)\) is a field.
- A `principal ideal domain` is an integral domain where all ideals are principal. In such a ring, every nonzero prime ideal is a maximal ideal.
- Principal ideal domain \(\subset\) unique factorization domain \(\subset\) commutative rings.

## Nilradicals

The set of all nilpotent elements \(\mathfrak{N}\) of \(A\) is an ideal, and \(A/\mathfrak{N}\) does not have any nilpotent elements. To prove this, let \(x, y \in \mathfrak{N}\) be nilpotent elements of order \(m, n\). By the binomial theorem, \((x + y)^{m + n - 1}\) has terms of the form \(x^r y^s\); but we cannot have both \(r < m\) and \(s < n\), so each of these terms vanish, and \((x + y)^{m + n - 1} = 0, x + y \in \mathfrak{N}\). Let \(\bar{x^n} \in A/\mathfrak{N}\) be a representation of \(x^n\). Then \((x^n)^p = 0\), and \(x \in \mathfrak{N} \implies \bar{x} = 0\). Ideal \(\mathfrak{N}\) is called the `nilradical` of \(A\).

The nilradical \(\mathfrak{N}\) of \(A\) is the intersection of all prime ideals of \(A\). To prove this, let \(\mathfrak{N}'\) be the intersection of prime ideals. If \(f \in A\) is nilpotent, and \(P\) is a prime ideal, \(f^n = 0 \in P\) for some \(n\), and hence \(f \in P\) (because \(P\) is prime) and \(f \in \mathfrak{N}'\). Conversely, suppose that \(f\) is not nilpotent. Let \(\Sigma\) be the set of ideals \(I\) such that \(n \gt 0 \implies f^n \notin I\). \(\Sigma\) is not zero, and by Zorn's lemma, there is some maximal element in \(\Sigma\). Let \(P\) be this maximal element: it remains to be shown that \(P\) is a prime ideal. Let \(x, y \notin P\), so \(P + (x), P + (y)\) contain \(P\), and therefore do not belong to \(\Sigma\). If: \[f^m = P + (x) \quad f^n = P + (y)\]

then, \(f^{m + n} \in P + (xy)\), hence the ideal \(P + (xy) \notin \Sigma\), and \(xy \notin P\). Hence, we have a prime ideal \(P\), and \(f \notin P, f \notin \mathfrak{N}'\).

The `Jacobson radical` \(\mathfrak{R}\) of \(A\) is defined to be the intersection of all maximal ideals of \(A\).

\(x \in \mathfrak{R} \iff 1 - xy\) is a unit in \(A\) for all \(y \in A\). To prove this, let is first assume that \(1 - xy\) is not a unit. But \(1 - xy\) belongs to some maximal ideal \(M\); and \(x \in \mathfrak{R} \subseteq M\), so \(xy \in M\), and hence \(1 \in M\), which is absurd. To prove the converse, let \(x \notin M\), for some maximal ideal \(M\). Then, \(M\) and \(x\) generate \((1)\), and \(u + xy \in M\), for some \(u \in M, y \in A\). Hence, \(1 - xy \in M\) is not unit.

## Operations on ideals

The sum of two ideals \(I, J\), denoted \(I + J\), is the smallest set, \(x + y, x \in I, y \in J\), containing both ideals. The product, denoted \(IJ\) is the ideal generated by \(xy\). In \(\mathbb{Z}\), let \(I = (m), J = (n)\); then the sum \(I + J\) is the l.c.m of \(m, n\), and \(IJ\) is the h.c.f of \(m, n\). In this case, \(IJ = I \cap J \iff m, n\) are coprime; more generally, \((I + J)(I \cap J) = IJ\).

In \(A = k[x_0, \ldots, x_n]\), we have the distributive law \(I (J + K) = IJ + IK\) as in the case of \(\mathbb{Z}\). However, unlike the \(\mathbb{Z}\) case,

\[I \cap (J + K) = I \cap J + I \cap K \quad \text{if } J \subseteq I \vee K \subseteq I\] \[I \cap J = IJ \quad \text{if } I + J = (1)\]

Ideals \(I, J\) are said to be `coprime` if \(I + J = (1)\).

Define the following homomorphism on direct products of ideals: \[\varphi : A \rightarrow \prod_q (A/I_q)\]

by the rule \(\varphi(x) = (x + I_1, \ldots, x + I_n)\). Then:

- If \(I_q, I_r\) are coprime whenever \(q \neq r\), then \(\prod_q I_q = \bigcap_q I_q\).
- \(\varphi\) is surjective \(\iff I_q, I_r\) are coprime whenever \(q \neq r\).
- \(\varphi\) is injective \(\iff \bigcap_q I_q = 0\).

In the ring \(A\):

- Let \(P_1, \ldots, P_n\) be prime ideals, and \(I\) any ideal such that \(I = \bigcup_q I_q\). Then, \(I \subseteq P_q\) for some \(q\).
- Let \(I_1, \ldots, I_n\) be ideals, and \(P\) a prime ideal containing \(\bigcap_q I_q\). Then, \(I_q \subseteq P\). In particular, if \(P = \bigcap_q I_q\), then \(P = I_q\) for some \(q\).

To prove (i), use induction of the form:

\[I \not\subseteq P_q (1 \leq q \leq n) \implies I \not\subseteq \bigcup_q P_q\]

Certainly, this is true for \(n = 1\). It holds for \(n \gt 1\), if it holds for \(n - 1\). Let \(x_i \in I\) such that \(x_i \notin P_j, i \neq j\). If \(x_i \notin P_i\), we are done. Otherwise, \(x_i \in P_i\) for all \(i\). Consider an element \(y \in I_q\) such that \(y \notin P_q\). Hence, we have \(I \not\subseteq P_q\).

To prove (ii), let \(I_q \not\subseteq P \, \forall q\), and \(\exists x_q : x_q \in I_q, x_i \notin P\). Then, \(x_q \subseteq \bigcap I_q\). But \(\prod_q x_q \notin P\), since \(P\) is prime. Hence, \(I_q \subseteq P\). Finally, if \(P = \bigcap_q I_q\), then \(P \subseteq I_q\), and hence \(P = I_q\), for some \(q\).

If \(I, J\) are ideals on ring \(A\), the `ideal quotient` is defined as: \[(I : J) = \{x \in A : xJ \subseteq I\}\]

which is an ideal in itself. In particular \((0 : J)\) is called the `annihilator` of \(J\), also written as \(Ann(J)\).

As an example, in \(\mathbb{Z}\), if \(I = (m), J = (n)\), then \((I : J) = (Q)\), where \(Q = m/(m, n)\), with \((m, n)\) denoting the h.c.f of \(m, n\).

What follows is a bunch of elementary results about ideal quotients:

- \((I : J)J \subseteq I\).
- \((I : (J : K)) = ((I : J) : K)\).
- \((\bigcap_q I_q : J) = \bigcap_q (I_q : J)\).
- \((I : \sum_q J_q) = \bigcap_q (I : J_q)\).

If \(I\) is any ideal of \(A\), then the `radical` of \(I\) is defined as: \[\sqrt{I} = \{x \in A \mid \exists n \gt 0 : x^n \in I\}\]

If \(\varphi : A \rightarrow A/I\) is a ring homomorphism, then \(\sqrt{I} = \varphi^{-1}(\mathfrak{N}_{A/I})\), hence the radical of an ideal is an ideal.

What follows is a set of results about radicals:

- \(I \subseteq \sqrt{I}\).
- \(\sqrt{\sqrt{I}} = \sqrt{I}\).
- \(\sqrt{IJ} = \sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}\).
- \(\sqrt{I} = (1) \iff I = (1)\).
- \(\sqrt{I + J} = \sqrt{(\sqrt{I} + \sqrt{J})}\).
- For prime ideal \(P\), \(\sqrt{P^n} = P \, \forall n\).

The radical of ideal \(I\) is the intersection of prime ideals containing \(I\). In general, we can define a radical over a set \(E \subseteq A\), but this is not an ideal in general, and \(\bigcap \sqrt{E_\alpha} = \sqrt{\bigcap E_\alpha}\).

Set of zerodivisors of \(A\) is equal to \(\sqrt{\bigcup_{x \neq 0} \text{Ann } x}\), and hence, \(\bigcup_{x \neq 0} \sqrt{\text{Ann } x}\).

Let \(I, J\) be ideals on ring \(A\), such that \(\sqrt{I}, \sqrt{J}\) are coprime. Then, \(I, J\) are coprime. This can be proved by observing \(\sqrt{I + J} = \sqrt{(\sqrt{I} + \sqrt{J})} = \sqrt{1}\).

## Extensions and contractions

Let \(f : A \rightarrow B\) be a ring homomorphism. Then for ideal \(I\), \(f(I)\) is not necessarily an ideal in \(B\). We define `extension` \(I^e\) to be the set \(Bf(I)\), generated by \(B\). Alternatively, if \(J\) is an ideal of \(B\), then \(f^{-1}(J)\) is necessarily an ideal in \(A\), and this is termed `contraction` \(J^c\). We can hence factorize \(f\) as follows: \[A \xrightarrow{p} f(A) \xrightarrow{q} B\]

Here, \(p\) is surjective and \(q\) is injective. For \(p\), the situation is very simple: there is a one-to-one correspondence between ideals of \(f(A)\) and ideals of \(A\) that contain \(\text{ker }(f)\), and prime ideals correspond to prime ideals. For \(q\), however, the situation is complicated.

As an example, consider \(f : \mathbb{Z} \rightarrow \mathbb{Z}[i]\). Primes in \(\mathbb{Z}\) may not stay prime in \(\mathbb{Z}[i]\). In fact, \(\mathbb{Z}[i]\) is a principal ideal domain, and the situation is as follows:

- \((2)^e ((1 + i)^2)\), the square of a prime ideal in \(\mathbb{Z}[i]\).
- If \(p \equiv 1 \text{ mod } 4\), then \((p)^e\) is the product of two distinct prime ideals. Example: \((5)^e = (2 + i)(2 - i)\).
- If \(p \equiv 3 \text{ mod } 4\), then \((p)^e\) is prime in \(\mathbb{Z}[i]\).

Let \(f : A \rightarrow B\), and \(I, J\) as before. Then:

- \(a \subseteq a^{ec}, b^{ce} \subseteq b\).
- \(J^c = J^{cec}, I^e = I^{ece}\).
- If \(C\) is the set of contracted ideals in \(A\), and \(E\) is the set of extended ideals in \(B\), then \(C = \{I \mid I^{ec} = I\}, E = \{J \mid J^{ce} = J\}\), and \(I \mapsto I^e\) is a bijective map that maps \(C\) onto \(E\) with inverse \(J \mapsto J^c\).

Let us prove (iii), since (i) and (ii) are trivial. If \(I \in C\), then \(I = J^c = J^{cec} = I^{ec}\). Conversely, if \(I = I^{ec}\), then \(I\) is the contraction of \(I^e\). A similar argument holds for \(E\).

## Modules

Modules let us do linear algebra on general rings, even non-commutative ones. For instances of non-commutative rings, there's no need to look further than matrices over \(\mathbb{R}\), the field of real numbers. Another instance of non-commutative algebra can be found in Quaternions, that have different properties when cycling clockwise and counter-clockwise. The ideal \(I\) and quotient ring \(A/I\) are both examples of modules.

A module \(M\) over ring \(A\) is formally defined as the multiplication map \(A \times M \rightarrow AM\), or \((f, m) \mapsto fm\) satisfying the following axioms:

- \((f + g) m = fm + gm\)
- \(f (m + n) = fm + fn\)
- \((fg) m = f (gm)\)
- \(1_A m = m\)

Theory of modules \(\leftrightarrow\) Theory of linear algebra over \(A\).

Examples:

- If \(A\) is a field in \(k\), then \(A\)-module \(= k\)-vector space.
- If \(A = \mathbb{Z}\), then \(\mathbb{Z}\)-module \(=\) abelian group.
- If \(A = k[x]\) where \(k\) is a field, then the \(A\)-module is a \(k\)-vector space with linear transformation.
- If \(A = k[G]\), where \(G\) is a group, then \(A\)-module \(= k\)-representation of \(G\).

Let \(M, N\) be \(A\)-modules, and \(f : M \rightarrow N\) be an \(A\)-module homomorphism, or \(A\)-linear. Then, \(f(x + y) = f(x) + f(y)\), and \(f(ax) = a . f(x)\), for all \(a \in A; x, y \in M\). If \(A\) is a field, then an \(A\)-module homomorphism is the same thing as linear transformation over vector spaces.

The composition of any two \(A\)-module homomorphisms is again an \(A\)-module homomorphism. The set of all \(A\)-module homomorphisms from \(M\) to \(N\) can be turned into an \(A\)-module with the axioms: \[(f + g)(x) = f(x) + g(x) \quad (af)(x) = a . f(x)\]

This \(A\)-module is denoted by \(\text{Hom}(M, N)\). Homomorphisms \(u : M' \rightarrow M\) and \(v : N \rightarrow N''\) induce the homomorphisms:

\[\bar{u} : \text{Hom}(M, N) \rightarrow \text{Hom}(M', N) \\ \bar{v} : \text{Hom}(M, N) \rightarrow \text{Hom}(M, N'')\]

where: \[\bar{u}(f) = u \circ f \quad \bar{v}(f) = f \circ v\]

For any module \(M\), there is the natural isomorphism \(\text{Hom}(A, M) \cong M\): any \(A\)-module homomorphism \(f : A \rightarrow M\) is determined uniquely determined by \(f(1)\), which can be any element of \(M\).

## Submodules

Submodule \(M'\) of \(M\) is a subgroup of \(M\) closed under multiplication by elements of \(A\). The abelian group \(M/M'\) inherits the \(A\)-module structure from \(M\), given by \(a(x + M') = ax + M'\). The \(A\)-module \(M/M'\) is termed `quotient` of \(M\) by \(M'\). There is a natural homomorphism between \(M\) and \(M/M'\), and this is an \(A\)-module homomorphism. There is a one-to-one order-preserving bijection from submodules of \(M\) containing \(M'\) and submodules of \(M''\), as in the case of ideals.

If \(f : M \rightarrow N\) is an \(A\)-module homomorphism, then its kernel and image are also submodules defined by:

\[\text{Ker}(f) = \{x \in M \mid f(x) = 0\}\] \[Im(f) = f(M)\]

The cokernel of \(f\) is a quotient module of \(N\), defined by: \[\text{Coker}(f) = N/Im(f)\]

If \(M' \subseteq \text{Ker}(f)\), then \(f\) gives rise to the homomorphism \(\bar{f} : M/M' \rightarrow N\), where \(\text{Ker}(\bar{f}) = \text{Ker}(f)/M'\). Taking \(M' = \text{Ker}(f)\), we then have the following isomorphism of \(A\)-modules: \[M/Ker(f) \cong Im(f)\]

## Operations on submodules

Let \(M\) be a module, and let \(M_i\) be the family of submodules of \(M\). Then, the `direct sum` is defined as \(\sum M_i\), the smallest submodule of \(M\) containing all \(M_i\). \(\bigcap M_i\) is again a submodule.

What follows is a result on modules:

- If \(L \supseteq M \supseteq N\) are \(A\)-modules, then \((L/N)/(M/N) \cong (L/M)\).
- For submodules \(M_1, M_2\), \((M_1 + M_2)/M_1 = M_2/(M_1 \cap M_2)\).

To prove (i), define homomorphism \(\theta : L/N \rightarrow L/M\) by \(\theta(x + N) = x + M\). Then, \(\theta\) is a well-defined \(A\)-module homomorphism with kernel \(M/N\).

To prove (ii), consider the composition \(M_2 \rightarrow M_1 + M_2 \rightarrow (M_1 + M_2)/M_1\). It is surjective with kernel \(M_1 \cap M_2\).

We cannot define a product of submodules. However, we can define the product \(IM\), where \(I\) is an ideal. If \(N, P\) are two \(A\)-modules, we can define \(N : P\) as the set of all \(a \in A\) such that \(aP \subseteq N\); it is an ideal. In particular \((0 : M)\) is termed `annihilator`, and written as \(\text{Ann}(M)\). An \(A\)-module is `faithful` if \(\text{Ann}(M) = 0\); if \(\text{Ann}(M) = I\), then the \(A\)-module is faithful as an \((A/I)\)-module.

Two elementary results on annihilators:

- \(\text{Ann}(N + P) = \text{Ann}(N) \cap \text{Ann}(P)\).
- \((N : P) = \text{Ann}(N + P)/N\).

If \(M = \sum_i Ax_i\), then the \(x_i\) are termed the set of `generators` of the module \(M\). This means that every element of \(M\) can be expressed as finite linear combinations (not necessarily unique) of \(x_i\)s with coefficients in \(A\). \(M\) is said to be `finitely generated` if this is the case.

## Direct sum and product

If \(M, N\) are modules, then their `direct sum` \(M \oplus N\) is defined as the pairs \((x, y)\) for \(x \in M, y \in N\). Sum and scalar multiplication are defined in the usual way: \[(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2)\] \[a (x, y) = (ax, ay)\]

More generally, we can define the direct sum as \(\bigoplus_{q \in Q} M_q\) to have elements in the family \((x_q)_{q \in Q}\). For a finite index set \(Q\), direct sum and `direct product`, \(\prod_{q \in Q} M_q\), are the same thing, but not otherwise.

Suppose ring \(A\) is a direct product \(\prod_q A_q\), then the set of elements of \(A\) of the form \((0, \ldots a_q, \ldots 0)\) where \(a_q \in A\), form an ideal \(I_q\). This is not a subring in general, because it does not contain identity. By considering ring \(A\) as a module, we have its decomposition into the direct sum of ideals. Conversely, given a ideal decomposition of a ring: \[A = I_1 \oplus \ldots \oplus I_n\]

we have: \[A \cong \prod_q A/J_q\]

where \(J_q = \bigoplus_{q \neq r} I_r\). Each \(I_q\) is a ring isomorphic to \(A/J_q\).

## Finitely generated modules

A `free` \(A\)-module, \(A^{(f)}\), is one which is isomorphic to an \(A\)-module of the form \(\bigoplus_q M_q\), where each \(M_q \cong A\), as an \(A\)-module. A finitely generated \(A\)-module is therefore isomorphic \(A \oplus \ldots \oplus A\), with \(n\) summands, written as \(A^n\).

\(M\) is a finitely generated \(A\) module \(\iff M\) is isomorphic to the quotient of \(A^n\) for some \(n \gt 0\). To prove this, let \(x_1, \ldots, x_n\) generate \(M\), and \(\varphi : A^n \rightarrow M\) be defined by \(\varphi(x_1, \ldots, x_n) = a_1 x_1 + \ldots + a_n x_n\). Then, \(\varphi\) is an \(A\)-module homomorphism onto \(M\) and \(M \cong A^n/\text{Ker}(\varphi)\). Conversely, if \(e_i = (0, \ldots, 1, \ldots, 0)\), with \(1\) being in the \(i^{\text{th}}\) place, then \(e_i\)s generate \(A^n\), and hence \(\varphi(e_i)\) generate \(M\).

Let \(M\) be a finitely generated \(A\)-module, \(I\) be an ideal of \(A\), and \(\varphi\) be an \(A\)-module endomorphism such that \(\varphi(M) \subseteq IM\). Then, \[\varphi^n + a_1 \varphi^{n - 1} + \ldots + a_n = 0\]

where \(a_q \in I\). To prove this, let \(x_1, \ldots, x_n\) be the generators of \(M\). Then, each \(\varphi(x_q) \in IM\), so that \(\varphi(x_q) = \sum_{r = 1}^n a_{qr} x_r\). That is: \[\sum_{r = 1}^n (\delta_{qr} \varphi - a_{qr}) x_r = 0\]

where \(\delta_{qr}\) is the Kronecker delta. Multiplying on the left by the adjoint of the matrix \((\delta_{qr} \varphi - a_{qr})\), we see that \(\text{det}(\delta_{qr} \varphi - a_{qr})\) annihilates each \(x_i\); expanding out this determinant gives us the required result.

Nakayama's lemma, corollary: Let \(M\) be a finitely generated \(A\)-module, and \(I\) be an ideal in \(A\). If \(M = IM\), then \(\exists x \in A \text{ such that } x \equiv 1 \text{ mod } I\) and \(xM = 0\). This can be proved by taking \(\varphi =\) identity and \(x = 1 + a_1 + \ldots + a_n\) in the previous result.

`Nakayama's lemma`: Let \(M\) be a finitely generated \(A\)-module, and \(I\) be an ideal contained in the Jacobson radical \(\mathfrak{R}\) of \(A\). Then \(M = IM \implies M = 0\). To prove this, notice that we have \(xM = 0\) for some \(x \equiv 1 \text{ mod } \mathfrak{R}\), by the corollary. But \(x\) is a unit in \(A\), and hence \(M = x^{-1}xM = 0\).

Let \(M\) be a finitely generated \(A\)-module, \(N\) a submodule of \(M\), and \(I\) an ideal \(I \subseteq \mathfrak{R}\). Then, \(M = IM + N \implies M = N\). This can be proved by observing that \(I(M + N) = (IM + N)/N\).

Let \(A\) be a local ring, \(M\) its maximal ideal, \(k = A/M\) its residue field, and \(N\) a finitely generated module. Then, \(N/MN\) is annihilated by \(M\), hence is an \(A/M\)-module i.e. a \(k\)-vector space that is finite-dimensional. Let \(x_i\)s be the elements of \(N\) whose image is in \(N/MN\). Then, \(x_i\)s generate \(N\). To prove this, let \(S\) be the submodule of \(N\) generated by \(x_i\), so it remains to be shown that \(S = N\). The composite \(S \rightarrow N \rightarrow N/MN\) maps \(S\) onto \(N/MN\), and hence \(S + MN = 1\), and \(N = S\).

## Homomorphism theorems of modules

Proofs for these correspond exactly to those in vector spaces.

- \(\text{ker } \varphi \subset M\), and \(\text{im } \varphi \subset N\) are submodules.
- Let \(N \subset M\) be a submodule; then \(\ni\) a surjective quotient homomorphism \(\varphi : M \rightarrow M/N\), ker \(\varphi = N\). Elements of \(M/N\) can be constructed either as equivalence classes \(m \in M \bmod N\), or as cosets \(m + N\) in \(M\).
- \(M/\text{ker } \varphi \cong \text{im } \varphi\).

## Exact sequences

If \(L\), \(M\), \(N\) are \(A\)-Modules, and \(\alpha\), \(\beta\) are homomorphisms, then:

\[0 \rightarrow L \xrightarrow{\alpha} M \xrightarrow{\beta} N \rightarrow 0\]

is `exact` if \(\alpha\) is injective, \(\beta\) is surjective, and \(\beta\) induces the isomorphism \(\text{Coker}(f) = M/f(L)\) onto \(N\).

Alternatively, the condition for a `short exact sequence` can be written as: \(M \cong L \oplus N\), \(L \subset M\) and \(N = M/L\), such that \(\alpha\) maps \(m \mapsto (m, 0)\) and \(\beta\) maps \((m, n) \mapsto m\).

These two results are easily proved:

- The sequence \[L \xrightarrow{\alpha} M \xrightarrow{\beta} N \rightarrow 0\] is exact \(\iff\) for any \(S\), the sequence \[0 \rightarrow \text{Hom}(N, S) \xrightarrow{\gamma} \text{Hom}(M, S) \xrightarrow{\delta} \text{Hom}(L, S)\] is exact.
- The sequence \[0 \rightarrow L \xrightarrow{\alpha} M \xrightarrow{\beta} N\] is exact \(\iff\) for any \(S\), the sequence \[0 \rightarrow \text{Hom}(S, L) \xrightarrow{\gamma} \text{Hom}(S, M) \xrightarrow{\delta} \text{Hom}(S, N)\] is exact.

Define the following commutative diagram of \(A\)-modules, and \(A\)-module homomorphisms:

\[ \begin{xy} \xymatrix{ 0\ar[r] & M'\ar[r]^u\ar[d]^{f'} & M\ar[r]^v\ar[d]^f & M''\ar[r]\ar[d]^{f''} & 0 \\ 0\ar[r] & N'\ar[r]_{u'} & N\ar[r]_{v'} & N''\ar[r] & 0 } \end{xy} \]

Then, there exists an exact sequence:

\[0 \rightarrow \text{Ker}(f') \xrightarrow{\bar{u}} \text{Ker}(f) \xrightarrow{\bar{v}} \text{Ker}(f'') \rightarrow \\ \quad\quad\quad \text{Coker}(f') \xrightarrow{\bar{u}'} \text{Coker}(f) \xrightarrow{\bar{v}'} \text{Coker}(f'') \rightarrow 0\]

where \(\bar{u}, \bar{v}\) are restrictions of \(u, v\), and \(\bar{u}', \bar{v}'\) are induced by \(u', v'\).

The boundary homomorphism \(d\) is defined as follows: if \(x'' = \text{Ker}(f'')\), then \(x'' = v(x)\) for some \(x \in M\), and \(v'(f(x)) = f'(v(x))\), hence \(f(x) \in \text{Ker}(v') = \text{Im}(u')\) so that \(f(x) = u'(y')\) for some \(y' in N'\). Then \(d(x'')\) is defined to be the image of \(y'\) in \(\text{Coker}(f')\).

Let \(C\) be a class of \(A\)-modules, and \(\lambda\) a function with values in abelian group \(G\). \(\lambda\) is `additive` if, for each s.e.s with terms in \(C\), we have \(\lambda(L) - \lambda(M) + \lambda(N) = 0\). As an example, let \(A\) be a vector space \(k\), and \(C\) be the class of all vector spaces \(V\). Then \(V \mapsto \text{dim}(V)\) is additive.

Let \(0 \rightarrow M_1 \rightarrow \ldots \rightarrow M_n\) be an exact sequence with modules and kernels of all homomorphisms in class \(C\). Then, for any additive function \(\lambda\), we have: \[\sum_i (-1)^i \lambda(M_i) = 0\]

This is proved by splitting up the long exact sequence into short exact sequences.

The Koszul complex of pair \((x, y)\) is an example of an s.e.s:

\[0 \rightarrow A \xrightarrow{(-y, x)} A^2 \xrightarrow{\begin{pmatrix} x \\ y \end{pmatrix}} I \rightarrow 0\]

## Tensor product

Let \(M, N, P\) be three \(A\)-modules. The mapping \(f : M \times N \rightarrow P\) is said to be \(A\)-bilinear if, for each \(x \in M, y \in N\), the mapping \(y \rightarrow (x, y)\) of \(N\) onto \(P\), and the mapping \(x \rightarrow (x, y)\) of \(M\) onto \(P\), are both \(A\)-linear.

We construct an \(A\)-module \(T\), called the `tensor product`, if the \(A\)-bilinear mapping \(M \times N \rightarrow P\) is in one-to-one correspondence with \(A\)-linear mappings \(T \rightarrow P\), for all \(A\)-modules \(P\). More precisely, there exists a pair \((T, g)\) consisting of \(A\)-module \(T\) and \(A\)-bilinear mapping \(g : M \times N \rightarrow T\), such that \(f = f' \circ g\), where \(f' : T \rightarrow P\) is a unique \(A\)-linear mapping. In other words, every mapping \(M \times N\) factors uniquely through \(T\).

Note that tensor product notation \(x \otimes y\) is ambiguous; it may be zero in \(M' \times N'\), submodules of \(M, N\), while being zero in modules \(M, N\). For example, consider \(M = \mathbb{Z}, N = \mathbb{Z}/2\mathbb{Z}, M' =\) submodule \(2 \mathbb{Z}\) of \(\mathbb{Z}\) while \(N' = N\). Then, \(2 \otimes x = 1 \otimes 2x = 1 \otimes 0 = 0\), but is nonzero as an element of \(M' \otimes N'\).

Let \(x_i \in M, y_i \in N\), and \(\sum x_i \otimes y_i = 0\) in \(M \times N\). Then, there exist finitely generated submodules \(M_0, N_0\) of \(M, N\) such that \(\sum x_i \otimes y_i = 0\) in \(M_0 \otimes N_0\). To prove this, consider \(\sum (x_i, y_i) = D\); then \(\sum (x_i, y_i)\) is a finite sum of generators of \(D\). Let \(M_0\) be the submodule of \(M\) generated by the \(x_i\) and all the elements of \(M\) that occur as the first coordinates in these generators of \(D\). Define \(N_0\) similarly. Then, \(\sum x_i \otimes y_i = 0\) in \(M_0 \otimes N_0\), as required.

Let \(M, N, P\) be \(A\)-modules. Then, there exist unique isomorphisms:

- \(M \otimes N \rightarrow N \otimes M\).
- \(M \otimes (N \otimes P) \rightarrow (M \otimes N) \otimes P \rightarrow M \otimes N \otimes P\).
- \((M \oplus N) \otimes P \rightarrow (M \otimes P) \oplus (N \otimes P)\).
- \(A \otimes M \rightarrow M\).

Let \(A, B\) be a rings, \(M\) an \(A\)-module, \(P\) a \(B\)-module, and \(N\) an \((A, B)\)-module (i.e. it is simultaneously an \(A\)-module and \(B\)-module, and the two structures are compatible in that \((ax)b = a(xb)\) for all \(a \in A, b \in B, x \in N\)). Then \(M \otimes_A P\) is naturally a \(B\)-module, \(N \otimes_B P\) is an \(A\)-module, and we have: \[(M \otimes_A N) \otimes_B P \cong M \otimes_A (N \otimes_B P)\]

Let \(f : M \rightarrow M', g : N \rightarrow N'\) be two \(A\)-module homomorphisms. Define \(h : M \times N \rightarrow M' \otimes N'\) by \(h(x, y) = f(x) \otimes g(y)\). Then, \(h\) is bilinear and induces the homomorphism: \[f \otimes g : M \times N \rightarrow M' \otimes N'\]

such that: \[(f \otimes g)(x \otimes y) = f(x) \otimes g(y) \quad x \in M, y \in N\]

Let \(f' : M' \rightarrow M'', g' : N' \rightarrow N''\) be two \(A\)-module homomorphisms. Then, \((f' \circ f) \otimes (g' \circ g)\) and \((f' \otimes g') \circ (f \otimes g)\) agree on all elements of the form \(x \otimes y\) in \(M \otimes N\). Since the elements generate \(M \otimes N\), it follows that: \[(f' \circ f) \otimes (g' \circ g) = (f' \otimes g') \circ (f \otimes g)\]

## Restriction and extension of scalars

Let \(f : A \rightarrow B\) be a homomorphism of rings, and let \(N\) be a \(B\)-module. Then \(N\) has an \(A\)-module structure defined as follows: if \(a \in A, x \in N\), \(ax\) is defined to be \(f(a)x\). This \(A\)-module is derived from \(N\) by `restriction of scalars`. In particular, \(f\) defines an \(A\)-module structure on \(B\).

Suppose \(M\) is a finitely generated \(B\)-module, and \(B\) is a finitely generated \(A\)-module, then \(M\) is finitely generated as an \(A\)-module. To prove this, let \(y_1, \ldots, y_n\) generate \(N\) over \(B\), \(x_1, \ldots, x_n\) generate \(B\) as an \(A\)-module. Then the \(mn\) products \(x_i y_j\) generate \(M\) over \(A\).

Let \(N\) be an \(A\)-module. Since an \(B\) can be regarded as an \(A\)-module, we can form the \(A\)-module \(N_B = B \otimes_A N\). In fact, \(M_B\) carries a \(B\)-module structure since \(b(b' \otimes x) = bb' \otimes x\), for all \(b, b' \in B, x \in M\). \(N_B\) is said to be obtained from \(N\) by `extension of scalars`.

If \(M\) is a finitely generated \(A\)-module, \(M_B\) is finitely generated as a \(B\)-module. To prove this, observe that, if \(x_1, \ldots, x_n\) generate \(M\) over \(A\), then \(1 \otimes x_i\) generates \(M_B\) over \(B\).

## Exactness of the tensor product

Let \(f : M \otimes N \rightarrow P\) be an \(A\)-bilinear mapping. Then, the mapping \(N \rightarrow P\) given by \(y \mapsto (x, y)\) is \(A\)-linear, and hence \(f\) gives rise to the mapping \(M \rightarrow \text{Hom}(N, P)\), which is also \(A\)-linear, because \(f\) is linear in the variable \(x\). Conversely, the \(A\)-homomorphism \(\varphi : M \rightarrow \text{Hom}_A(N, P)\) is a bilinear map given by \((x, y) \rightarrow \varphi(x) \varphi(y)\). Hence the set \(S\) of bilinear mappings \(M \times N \rightarrow P\) is in natural one-to-one correspondence with \(\text{Hom}(M, \text{Hom}(N, P))\). On the other hand, \(S\) is in one-to-one correspondence with \(\text{Hom}(M \otimes N, P)\), by definition of tensor product. Hence: \[\text{Hom}(M \otimes N, P) \cong \text{Hom}(M, \text{Hom}(N, P))\]

Consider the following exact sequence of \(A\)-module homomorphisms: \[M' \xrightarrow{f} M \xrightarrow{g} M'' \rightarrow 0\]

Let \(N\) be any \(A\)-module. Then, the following sequence is also exact:

\[M' \otimes N \xrightarrow{f \otimes 1} M \otimes N \xrightarrow{g \otimes 1} M'' \otimes N \rightarrow 0\]

To prove this, let \(E\) denote the first sequence, \(E \otimes N\) the second sequence, and \(P\) any \(A\)-module. Since the first sequence is exact, \(\text{Hom}(E, \text{Hom}(N, P))\) is exact \(\implies \text{Hom}(E \otimes N, P)\) is exact \(\implies E \otimes N\) is exact.

Let \(T(M) = M \otimes N, U(P) = \text{Hom}(N, P)\). Then, \(\text{Hom}(T(M), P) = \text{Hom}(M, U(P))\) for all \(M, P\). \(T\) is termed `left adjoint` of \(U\), and \(U\) is termed `right adjoint` of \(T\). From [category theory](/ct/3#adjunction), any functor which is left adjoint is `right exact`, and right adjoint is `left exact`.

If tensoring with \(N\) translates all exact sequences into exact sequences, then \(N\) is said to be a `flat` \(A\)-module. This is not true in general, and can be illustrated with an example: consider the exact sequence \(0 \rightarrow \mathbb{Z} \xrightarrow{f} \mathbb{Z}\), where \(f(x) = 2x\). Tensoring with \(N = \mathbb{Z}/2\mathbb{Z}\) yields \(0 \rightarrow \mathbb{Z} \otimes N \xrightarrow{f \otimes 1} \mathbb{Z} \otimes N\), which is not exact because, for any \(x \otimes y \in \mathbb{Z}\): \[(f \otimes 1)(x \otimes y) = 2x \otimes y = x \otimes 2y = 0\]

So \(f \otimes 1\) is the zero mapping, while \(\mathbb{Z} \otimes N\) is nonzero.

The following statements about an \(A\)-module \(N\) are equivalent:

- \(N\) is flat.
- The exact sequence \(0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0\), when tensored with \(N\), is exact.
- If \(f : M' \rightarrow M\) is injective, so is \(f \otimes 1 : M' \otimes N \rightarrow M \otimes N\).
- If \(f : M' \rightarrow M\) is injective, and \(M', M\) are finitely generated, then \(f \otimes 1\) is injective.

If \(f : A \rightarrow B\) is a ring homomorphism, and \(M\) is a flat \(A\)-module, then \(M_B = B \otimes_A M\) is a flat \(B\)-module.

## Algebras

Let \(f : A \rightarrow B\) be a ring homomorphism, \(a \in A, b \in B\). Define the operation \(ab = f(a)b\). This definition makes the ring \(B\) into an \(A\)-module. Thus, \(B\) has \(A\)-module structure as well as ring structure. Ring \(B\) equipped with \(f : A \rightarrow B\) is said to be an `\(A\)-algebra`.

What follows are two examples:

- If \(A = K\), a vector space, then \(f\) is injective, and therefore \(K\) can be identified with its image in \(B\). Thus a \(K\)-algebra is simply a ring containing \(K\) as a subring.
- Let \(A\) be any ring. Since \(A\) has identity, there is a unique homomorphism \(\mathbb{Z} \rightarrow A\), given by \(n \mapsto n . 1\). Thus every ring is automatically a \(\mathbb{Z}\)-algebra.

Let \(f : A \rightarrow B\) and \(g : A \rightarrow C\) be two ring homomorphisms. Then, \(h : B \rightarrow C\) is an \(A\)-algebra homomorphism, which is both a ring homomorphism and an \(A\)-module homomorphism. We also have \(f \circ h = g\).

A ring homomorphism \(f : A \rightarrow B\) is of `finite`, and \(B\) is a finite \(A\)-algebra if \(B\) is finitely generated as an \(A\)-module. \(f\) is of `finite type`, and \(B\) is a `finitely generated \(A\)-algebra`, if there is a finite set of elements \(x_1, \ldots, x_n\) in \(B\) such that every element of \(B\) can be written as a polynomial in \(x_1, \ldots, x_n\) with coefficients in \(f(A)\); equivalently, there is an \(A\)-algebra homomorphism from the polynomial ring \(A[t_1, \ldots, t_n]\) onto \(B\).

Ring \(A\) is said to be finitely generated, if it is finitely generated as a \(\mathbb{Z}\)-algebra. This means that there is a finite set \(x_1, \ldots, x_n\) of elements in \(A\), such that every element in \(A\) can be written as a polynomial in \(x_i\)s with rational coefficients.

## Tensor product of algebras

Let \(B, C\) be two \(A\)-algebras, and \(f : A \rightarrow B, g : A \rightarrow C\) be two homomorphisms. Since \(B, C\) are two \(A\)-modules, we may form the tensor product \(D = B \otimes_A C\), which is an \(A\)-module. Now, consider \(B \times C \times B \times C \rightarrow D\) defined by: \[(b, c, b', c') \rightarrow bb' \otimes cc'\]

This is \(A\)-linear, and therefore induces the \(A\)-homomorphism: \[B \otimes C \otimes B \otimes C \rightarrow D\]

And hence by \(A\)-module homomorphism, \[D \otimes D \rightarrow D\]

This corresponds to the bilinear mapping: \[\mu : D \times D \rightarrow D\]

which is such that: \[\mu(b \otimes c, b' \otimes c') = bb' \otimes cc'\]

We have therefore defined multiplication on the tensor product \(D = B \otimes_A C\); for elements of the form \(b \otimes c\), this is given by: \[(b \otimes c) (b' \otimes c') \rightarrow bb' \otimes cc'\]

Furthermore \(D\) is an \(A\)-algebra, and the mapping \(a \mapsto f(a) \otimes g(a)\) is a ring homomorphism \(A \rightarrow D\). We hence end up with the following commutative diagram:

\[ \begin{xy} \xymatrix{ & B\ar[dr]^u & \\ A\ar[ur]^f\ar[dr]_g & & D \\ & C\ar[ur]_v & } \end{xy} \]