# 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?

• Wait, please slow down. You just asked a question that turned out to be trivially false, and some people there (including me!) have been putting some effort into trying to figure out what you meant to ask. Could you please address this before moving on to a new question? May 20, 2012 at 0:12
• In fact, the first question you ask is answered in a link I gave in a comment to your last question. This link was provided based on educated guesswork by @Benjamin Lim and myself on what you might have meant. So again, please slow down. Also, you have reproduced verbatim what looks to be a portion of a problem set. Is this homework of some sort? What is the source you are quoting from? May 20, 2012 at 0:16
• @PeteL.Clark This is from a set of practice questions given to us for exam preparation. May 20, 2012 at 0:18
• @Zhang: thanks for the clarification. I don't really feel comfortable helping you solve practice problems for an exam. I don't find it ethically problematic per se, but I strongly recommend that you discuss them with your instructor and/or your classmates first. If you cannot do the practice problems, that's useful information for the instructor to have. Besides, getting other people to solve your practice problems may not be such good practice. May 20, 2012 at 0:23
• @BenjaminLim Yes. If $x, y$ are outside $I$ then by maximality, $(x, I), (y, I)$ both intersect $S$. So for some $r, s\in R$, $i, j\in I$, $xr+i, ys+j\in S$. So $(xr+i)( ys+j)\in S$, and on expanding one sees $(xy, I)$ intersects $S$, so $xy$ is outside $I$. Hence $I$ is prime. May 20, 2012 at 0:24

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.

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?

• Thank you. But I have not learnt any set theory yet, is there a way to tackle the question without Zorn's lemma? May 20, 2012 at 0:34
• @Zhang No I don't believe so. By the way you CANNOT do commutative algebra without knowing Zorn's Lemma.
– user38268
May 20, 2012 at 0:36
• @Zhang Can you complete your problem based on what I've put out above?
– user38268
May 20, 2012 at 0:41