Commutative Algebra/Fractions, annihilator

Fractions within rings
The following two lemmata ensure that everything is correctly defined.

Lemma 9.3:

$$\sim_S$$ is an equivalence relation.

Proof:

For reflexivity and symmetry, nothing interesting happens. For transitivity, there is a little twist. Assume
 * $$r/s \sim_S u/t$$ and $$u/t \sim_S v/w$$.

Then there are $$i, j \in S$$ such that
 * $$i(rt - su) = 0$$ and $$j(uw - tv) = 0$$.

But in this case, we have
 * $$ijt(rw - vs) = ij(rwt - vst) = ij(rwt - suw + suw - vst) = 0$$;

note $$ijt \in S$$ because $$S$$ is multiplicatively closed.

Lemma 9.4:

The addition and multiplication given above turn $$S^{-1}R$$ into a ring.

Proof:

We only prove well-definedness; the other rules follow from the definition and direct computation.

Let thus $$r/s \sim_S u/t$$ and $$a/p \sim_S b/q$$.

Thus, we have $$i(rt - su) = 0$$ and $$j(aq - bp) = 0$$ for suitable $$i, j \in S$$.

We want
 * $$(rp + as) \big/ sp = (uq + bt) \big/ tq$$

and
 * $$ra \big/ sp = ub \big/ tq$$.

These translate to
 * $$x((rp+as)tq - (uq + bt)sp) = 0$$

and
 * $$y(ratq - ubsp) = 0$$

for suitable $$x,y \in S$$. We get the desired result by picking $$x = y = ij$$ and observing
 * $$ij(ratq - ubsp) = ij(ratq - sauq + sauq - ubsp) = 0$$

and
 * $$ij((rp+as)tq - (uq + bt)sp) = ij(rptq + astq - uqsp - btsp) = 0$$.

Note that we were heavily using commutativity here.

We will see further properties like 4. when we go to modules, but we can't phrase it in full generality because in modules, we may not have a product of two module elements.

Proof:

1.:

If $$s \in S$$, then the rules for multiplication for $$S^{-1}R$$ indicate that $$1/s$$ is an inverse for $$\pi_S(s) = s/1$$.

2.:

Assume $$r/1 = 0 = 0/1$$. Then there exists $$s \in S$$ such that $$s(r - 0) = sr = 0$$.

3.:

Let $$r/s$$ be an arbitrary element of $$S^{-1}R$$. Then $$r/s = r/1 \cdot 1/s = \pi_S(r) \pi_S(s)^{-1}$$.

4.


 * $$\begin{align}

r/s \in S^{-1}(I \cdot J) & \Leftrightarrow r/s = ij/t, i \in I, j \in J, t \in S \\ & \Leftrightarrow r/s = (i/t)(j/1), i \in I, j \in J, t \in S \\ & \Leftrightarrow r/s \in S^{-1}I \cdot S^{-1}J \end{align}$$

5.

Let $$I \cap S \neq \emptyset$$, that is, $$s \in S \cap I$$. Then $$\pi_S(s) \in \pi_S(I)$$, where $$\pi_S(s)$$ is a unit in $$S^{-1}R$$. Further, $$\pi_S(I)$$ is an ideal within $$S^{-1}R$$ since $$\pi_S$$ is a morphism. Thus, $$\pi_S(I) = S^{-1}R$$.

Proof:

We first prove uniqueness. Assume there exists another such morphism $$g'$$. Then we would have
 * $$g'(r/s) = g'(r/1) g'(s/1)^{-1} = f(r) f^{-1} = g(r/1) g(s/1)^{-1}$$.

Then we prove existence; we claim that
 * $$g(r/s) := f(r) (f(s))^{-1}$$

defines the desired morphism.

First, we show well-definedness.

Firstly, $$f(s)^{-1}$$ exists for $$s \in S$$.

Secondly, let $$r/s \sim_S u/t$$, that is, $$i(rt - su) = 0$$. Then
 * $$\begin{align}

g(r/s) & = g(itr/its) \\ & = f(itr) (f(its))^{-1} \\ & = f(isu) (f(its))^{-1} \\ & = g(isu/its) = g(u/t). \end{align}$$

The multiplicativity of this morphism is visually obvious (use that $$f \circ \pi_S$$ is a morphism and commutativity); additivity is proven as follows:


 * $$\begin{align}

g(r/s + u/t) & = g\left( (rt + su) \big/ st \right) \\ & = f(rt + su) (f(st))^{-1} \\ & = f(rt) (f(st))^{-1} + f(su) (f(st))^{-1} \\ & = g(r/s) + g(u/t). \end{align}$$

It is obvious that the unit is mapped to the unit.

Theorem 9.7:

Category theory context

Fractions within modules
Note that applying this construction to a ring $$R$$ that is canonically an $$R$$-module over itself, we obtain nothing else but $$S^{-1}R$$ canonically seen as an $$S^{-1}R$$-module over itself, since multiplication and addition coincide. Thus, we have a generalisation here!

That everything is well-defined is seen exactly as in the last section; the proofs carry over verbatim.

Proof:

1.


 * $$\begin{align}

m/s \in S^{-1}(N+K) & \Leftrightarrow m/s = (n+k)/t, t \in S, n \in N, k \in K \\ & \Leftrightarrow m/s = n/t + k/t, t \in S, n \in N, k \in K \\ & \Leftrightarrow m/s \in S^{-1}N + S^{-1}K; \end{align}$$ note that to get from the third row back to the second, we used that submodules are closed under multiplication by an element of $$R$$ to equalize denominators and thus get a suitable $$t \in S$$ ($$S$$ is closed under multiplication).

2.


 * $$\begin{align}

m/s \in S^{-1}(N \cap K) & \Leftrightarrow m/s = l/t, t \in S, l \in N \cap K \\ & \Leftrightarrow m/s = n/u = k/v, u, v \in S, n \in N, k \in K \\ & \Leftrightarrow m/s \in S^{-1}N \cap S^{-1}K; \end{align}$$ to get from the second to the first row, we note $$n/u = k/v \Leftrightarrow w(vn-uk) = 0$$ for a suitable $$w \in S$$, and in particular for example
 * $$m/s = wvn/wvu$$,

where $$wvn = wuk \in N \cap K$$.

3.

We set
 * $$\varphi: S^{-1}(M/N) \to (S^{-1}M)/(S^{-1}N), \varphi((m + N)/s) := m/s + S^{-1}N$$

and prove that this is an isomorphism.

First we prove well-definedness. Indeed, if $$m + N = m' + N$$, then $$m - m' \in N$$, hence $$(m - m')/s \in S^{-1}N$$ and thus $$m/s + S^{-1}N = m'/s + S^{-1}N$$.

Then we prove surjectivity. Let $$m/s + S^{-1}N$$ be given. Then obviously $$(m+N)/s$$ is mapped to that element.

Then we prove injectivity. Assume $$m/s \in S^{-1}N$$. Then $$m/s = n/t$$, where $$n \in N$$ and $$t \in S$$, that is $$u(tm - sn) = 0$$ for a suitable $$u \in S$$. Then $$utm \in N$$ and therefore $$(m + N)/s = (utm + N)/uts = 0$$.

Proof:

Exercises

 * Exercise 9.2.1: Let $$M, N$$ be $$R$$-modules and $$I \le R$$ an ideal. Prove that $$IM := \{i m | i \in I, m \in M\}$$ is a submodule of $$M$$ and that $$(M \otimes_R N) / I(M \otimes_R N) \cong (M/I) \otimes_{R/I} (N/I)$$ (this exercise serves the purpose of practising the proof technique employed for theorem 9.11).

The annihilator, faithfulness
Proof:

Let $$a, b \in \text{Ann}_R(S)$$ and $$r \in R$$. Then for all $$s \in S$$, $$(ra - b) s = r(as) - bs = 0$$. Hence the theorem by lemma 5.3.

Proof: Let $$s \in R$$ such that $$\forall r \in R : rs = 0$$. Then in particular $$s1 = 0$$.

Proof:

From the definition it is clear that $$\text{Ann}_R N \subseteq \text{Ann}_R S$$, since annihilating all elements of $$N$$ is a stronger condition than only those of $$S$$.

Let now $$t \in \text{Ann}_R S$$ and $$x_1 s_1 + \cdots x_n s_n \in N$$, where $$x_j \in R$$ and $$s_j \in S$$. Then $$t(x_1 s_1 + \cdots x_n s_n) = x_1 t s_1 + \cdots x_n t s_n = 0 + 0 + \cdots + 0 = 0$$.

Local properties
Definition 9.17:

Let $$M$$ be an $$R$$-module (where $$R$$ is a ring) and let $$p \le R$$ be a prime ideal. Then the localisation of $$M$$ with respect to $$p$$, denoted by
 * $$M_p$$,

is defined to be $$S^{-1}M$$ with $$S := R \setminus p$$; note that $$S$$ is multiplicatively closed because $$p$$ is a prime ideal.

Theorem 9.19:

Being equal to zero is a local-global property.

Proof:

We check the equivalence of 1. - 4. from definition 9.12. Clearly, 4. $$\Rightarrow$$ 1. suffices.

Assume that $$N$$ is a nonzero module, that is, we have $$n \in N$$ such that $$n \neq 0$$. By theorem 9.11, $$\operatorname{Ann}_R(n) := \operatorname{Ann}_R(\{n\})$$ is an ideal of $$R$$. Therefore, it is contained within some maximal ideal of $$R$$, call $$m$$ (unfortunately, we have to refer to a later chapter, since we wanted to separate treatments of different algebraic objects. The required theorem is theorem 12.2). Then for $$s \in R \setminus m$$ we have $$sn \neq 0$$ and therefore $$n/1 \neq 0/1$$ in $$M_m$$.

The following theorems do not really describe local-global properties, but are certainly similar and perhaps related to those.

Theorem 9.20:

If $$f: M \to N$$ is a morphism, then the following are equivalent:
 * $$f: M \to N$$ surjective.
 * 1) $$f_S: S^{-1}M \to S^{-1}N$$ surjective for all $$S \subseteq R$$ multiplicatively closed.
 * 2) $$f_p: M_p \to N_p$$ surjective for all $$p \le R$$ prime.
 * 3) $$f_m: M_m \to N_m$$ surjective for all $$m \le R$$ maximal.

Proof: