Commutative Algebra/Basics on prime and maximal ideals and local rings

Prime ideals
Lemma 12.2:

Let $$R$$ be a ring and $$I \le R$$ an ideal. $$I$$ is prime if and only if $$R/I$$ is an integral domain.

Proof:

$$I$$ prime is equivalent to $$ab \in I \Rightarrow a \in I \vee b \in I$$. This is equivalent to
 * $$ab + I = 0 + I \Rightarrow a + I = 0 + I \vee b + I = 0 + I$$.

Proof:

Order all ideals of $$R$$ not intersecting $$S$$ by set inclusion, and let a chain
 * $$I_1 \subseteq I_2 \subseteq \cdots \subseteq I_k \subseteq \cdots$$

be given. The ideal
 * $$I := \bigcup_{k \in \mathbb N} I_k$$

(this is an ideal, since $$a, b \in I \Rightarrow a \in I_m, b \in I_n \Rightarrow a, b \in I_{\max\{m,n\}}$$, hence $$a + b \in I$$, $$ra \in I$$) is an upper bound of the chain, since $$I$$ cannot intersect $$S$$ for else one of the $$I_k$$ would intersect $$S$$. Since the given chain was arbitrary, Zorn's lemma implies the existence of a maximal ideal among all ideals not intersecting $$S$$. This ideal shall be called $$J$$; we prove that it is prime.

Let $$ab \in J$$, and assume for contradiction that $$a \notin J$$ and $$b \notin J$$. Then $$\langle a \rangle + J$$, $$\langle b \rangle + J$$ are strict superideals of $$J$$ and hence intersect $$S$$, that is,
 * $$s = x a + y j$$,
 * $$t = z b + w j'$$,

$$s, t \in S$$, $$x, y, z, w \in R$$, $$j, j' \in J$$. Then $$S \ni st = xzab + yzbj + xawj' + ywjj' \in J$$, contradiction.

Projection to the quotient ring
In this section, we want to fix a notiation. Let $$R$$ be a ring and $$I \le R$$ an ideal. Then we may form the quotient ring $$R/I$$ consisting of the elements of the form $$r + I$$, $$r \in R$$. Throughout the book, we shall use the following notation for the canonical projection $$r \mapsto r + I$$:

Maximal ideals
Lemma 12.6:

An ideal $$I \le R$$ is maximal iff $$R/I$$ is a field.

Proof:

A ring is a field if and only if its only proper ideal is the zero ideal. For, in a field, every nonzero ideal contains $$1$$, and if $$R$$ is not a field, it contains a non-unit $$a$$, and then $$\langle a \rangle$$ does not contain $$1$$.

By the correspondence given by the correspondence theorem, $$R/I$$ corresponds to $$R$$, the zero ideal of $$R/I$$ corresponds to $$I$$, and any ideal strictly in between corresponds to an ideal $$K \le R$$ such that $$I \subsetneq K \subsetneq R$$. Hence, $$R/I$$ is a field if and only if there are no proper ideals strictly containing $$I$$.

Lemma 12.7:

Any maximal ideal is prime.

Proof 1:

If $$R$$ is a ring, $$m \le R$$ maximal, then $$R/m$$ is a field. Hence $$R/m$$ is an integral domain, hence $$m$$ is prime.

Proof 2:

Let $$m \le R$$ be maximal. Let $$ab \in m$$. Assume $$a, b \notin m$$. Then $$1 = ra + sn = tb + uk$$ for suitable $$n, k \in m$$, $$r,s,t,u \in R$$. But then $$1 = 1^2 = (ra +sn)(tb + uk) = rtab + stbn + rauk + sunk \in m$$.

Proof:

We order the set of all ideals $$J$$ such that $$I \subseteq J$$ and $$J \neq R$$ by inclusion. Let
 * $$J_1 \subseteq J_2 \subseteq \cdots \subseteq J_k \subseteq \cdots$$

be a chain of those ideals. Then set
 * $$J := \bigcup_{k \in \mathbb N} J_k$$.

Clearly, all $$J_k$$ are contained within $$J$$. Since $$I \subseteq J_1$$, $$I \subseteq J$$. Further, assume $$1 \in J$$. Then $$1 \in J_m$$ for some $$m$$, contradiction. Hence, $$J$$ is a proper ideal such that $$I \subseteq J$$, and hence an upper bound for the given chain. Since the given chain was arbitrary, we may apply Zorn's lemma to obtain the existence of a maximal element with respect to inclusion. This ideal must then be maximal, for any proper superideal also contains $$I$$.

Proof: From the correspondence theorem.

Local rings
Proof:

1. $$\Rightarrow$$ 2.: Assume $$a$$ and $$b$$ are both non-units. Then $$\langle a \rangle$$ and $$\langle b \rangle$$ are proper ideals of $$R$$ and hence they are contained in some maximal ideal of $$R$$ by theorem 12.7. But there is only one maximal ideal $$m$$ of $$R$$, and hence $$a, b \in m$$, thus $$a + b \in m$$. Maximal ideals can not contain units.

2. $$\Rightarrow$$ 3.: The sum of two non-units is a non-unit, and if $$a$$ is a non-unit and $$r \in R$$, $$ra$$ is a non-unit (for if $$sra = 1$$, $$sr$$ is an inverse of $$a$$). Hence, all non-units form an ideal. Any proper ideal of $$R$$ contains only non-units, hence this ideal is maximal.

3. $$\Rightarrow$$ 4.: Assume the $$r_j$$ are all non-units. Since the non-units form an ideal, $$r_1 + \cdots + r_n$$ is contained in that ideal of non-units, contradiction.

4. $$\Rightarrow$$ 5.: Assume $$r$$, $$1 - r$$ are non-units. Then $$1 = r + (1 - r)$$ is a non-unit, contradiction.

5. $$\Rightarrow$$ 1.: Let $$m, n \le R$$ two distinct maximal ideals. Then $$m + n = R$$, hence $$1 = s + t$$, $$s \in m$$, $$t \in n$$, that is, $$s = 1 - t$$. $$t$$ is not a unit, so $$1 - t = s$$ is, contradiction.

Localisation at prime ideals
In chapter 9, we had seen how to localise a ring at a multiplicatively closed subset $$S$$. An important special case is $$S = R \setminus p$$, where $$p$$ is a prime ideal.

Lemma 12.12:

Let $$p \le R$$ be a prime ideal of a ring. Then $$S := R \setminus p \subseteq R$$ is multiplicatively closed.

Proof: Let $$a, b \notin p$$. Then $$ab$$ can't be in $$p$$, hence $$ab \in S$$.

Theorem 12.14:

Let $$R$$ be a ring, $$p \le R$$ be prime. $$R_p$$ is a local ring.

Proof:

Set $$S := R \setminus p$$, then $$R_p = S^{-1} R$$. Set
 * $$m := \{r/s | r \in p, s \in S\} \subseteq S^{-1} R$$.

All elements of $$m$$ are non-units, and all elements of $$R_p \setminus m$$ are of the form $$r'/t$$, $$r' \notin p$$, $$t \in S$$ and thus are units. Further, $$m$$ is an ideal since $$p$$ is and by definition of addition and multiplication in $$S^{-1}R$$ and since $$S$$ is multiplicatively closed. Hence $$R_p$$ is a local ring.

This finally explains why we speak of localisation.