A maximal ideal among those avoiding a multiplicative set is prime

1. Let $S$ be a multiplicatively closed subset of a ring $R$, and let $I$ be an ideal of $R$ which is maximal among ideals disjoint from $S$. Show that $I$ is prime.

2. If $R$ is an integral domain, explain briefly how one may construct a field $F$ together with a ring homomorphism $R\to F$.

3. Deduce that if $R$ is an arbitrary ring, $I$ an ideal of $R$, and $S$ a multiplicatively closed subset disjoint from $I$, then there exists a ring homomorphism $f\colon R\to F$, where $F$ is a field, such that $f(x)=0$ for all $x\in I$ and $f(y)\neq 0$ for all $y\in S$.

[You may assume that if $T$ is a multiplicatively closed subset of a ring, and $0\notin T$, then there exists an ideal which is maximal among ideals disjoint from $T$.]

Here is a question I need to answer. For the first part I can show if $x, y$ are not in $I$ then nor does $xy$. The second part is just the field of fractions.

For the third part, I think I need to find an ideal $J$ containing $I$ such $J$ is prime, so that $R/J$ is an integral domain, and use the second part. To find $J$ prime, I think as suggested by the hint, I should go for a maximal ideal disjoint from $S$ containing $I$, but how can I do that?

Solution 1:

Let us slow down and first do $(1)$. By a standard application of Zorn's Lemma one can show that there is an ideal $P \supset I$ maximal subject to the condition that $P \cap S = \emptyset$.

Now the following steps lead to a solution: You want to show that for all $f,g \in R$ such that $f \notin P$, $g \notin P$ then $fg \notin P$.

1) If $f,g \notin P$ what can you say about the ideals $P + (f)$ and $P + (g)$? (Look at the maximality condition on $P$)

2) Recall that $S$ is a multiplicative set.

Conclude your result from here (also called Krull's Lemma).

Supplementary problem: Using Krull's Lemma prove that the set of zero - divisors in a ring is a union of prime ideals.

Now we come to actually proving $(3)$. By Krull's Lemma you know that you can find a prime ideal $P$ containing $I$ maximal with respect to the property that $P \cap S = \emptyset$. Now consider the following diagram

$$R \stackrel{h}{\longrightarrow} R/P \stackrel{g}{\longrightarrow} \textrm{Frac}(R/P)$$

where $h$ is a surjective map from $R$ onto $R/ P$ and $g$ is the canonical morphism from the integral domain $R/P$ into its fraction field. The canonical morphism is now injective because $R/P$ is an integral domain.

You should now be able to complete your problem by asking:

What is the kernel of the map $g \circ h$?

Is $I$ contained in the kernel of the map $g \circ h$?

Can you complete your problem from here?