On the structure of subrings of $R/I$

Question

I was thinking about the following problem: Let $I$ be an ideal of $R$ (commutative and contains $1$). Now consider a subring $S < R/I$. Can we say that the subring $S$ is of the form $S'/I$ where $S'$ is a subring of $R$? Can somebody add other conditions to make this true?

Intuitively I think that the answer to the first question is negative because it would make the theorem about the correspondence of the ideals of $R$ and $R/I$ redundant. So in order to find a counterexample we need to find $S$ which is not an ideal in $R/I$. But I cannot think of any right now. The second question seems more tricky but I don't know how to think about that.

The correspondence between ideals is not made redundant by a correspondence between subrings. The condition that $Rf^{-1}(S)\subseteq f^{-1}(S)$ is not automatically satisfied when one has $(R/I)S\subseteq S$, even though it is actually true. — Karl Kroningfeld, Dec 17 '13 at 01:31
@KarlKronenfeld, I understand. More work is needed to prove the correspondence of ideals. — user53970, Dec 17 '13 at 01:39

Karl Kroningfeld · Accepted Answer · 2013-12-17T02:00:17.100

2

Let's look at this a little more abstractly. Given a pair of monoids $M, N$ and a homomorphism $f:M\to N$, we note that for any submonoid $N'\subseteq N$, $M'=f^{-1}(N')$ is a submonoid of $M$. For, if $a,b\in M'$, then $f(ab)=f(a)f(b)\in N'$ implies $ab\in M'$, and $f(1_M)=1_N\in N'$ means $1_M\in M'$.

There are three things to observe. Firstly, a ring is a monoid with respect to each of its operations. Secondly, ring homomorphisms are monoid homomorphisms when considering either of the operations. Thirdly, a subset of a ring $R$ that is a submonoid of both $(R,+)$ and $(R,\cdot)$ is a subring of $R$ (and conversely).

Therefore, given a homomorphism $f:R\to S$ of rings and a subring $S'\subseteq S$, the set $R'=f^{-1}(S')$ is a submonoid of $(R,+)$ and a submonoid of $(R,\cdot)$. We conclude that $R'$ is a subring of $R$.

Of course, this isn't a complete answer to your question, since you still need to verify that $f(R')=S'$ when $f$ is surjective--it simply does not hold when $f$ is not surjective.

edited Dec 17 '13 at 02:00

answered Dec 17 '13 at 01:45

Karl Kroningfeld

5,204

In the second to last sentence do you mean to define $R'=f^{-1}(S')$? – user53970 Dec 17 '13 at 01:50
@user53970 Yes, thank you. – Karl Kroningfeld Dec 17 '13 at 01:50
That result was unexpected to me. I thought that it would not be true. Thanks for the clear answer. – user53970 Dec 17 '13 at 01:56
@user53970 You're welcome, I was hoping to clarify this with a tool you can use in other settings. Also, be sure to take into account my (new) final paragraph. It's the reason why there is a counterexample in your other question about subrings of localizations. – Karl Kroningfeld Dec 17 '13 at 02:00

Berci · Answer 2 · 2013-12-17T01:33:19.043

1

$R/I$ is the image of the canonical surjection $R\to R/I$, so basically your question translates as

Having a surjective ring homomorphism $f:A\to B$ and a subring $S<B$, can we always find an $S'< A$ such that $f(S')=S$?

$\,$

And the answer is yes: $S':=f^{-1}(S)$.

edited Dec 17 '13 at 01:33

answered Dec 17 '13 at 01:27

Berci

90,745

You are correct. But I wrote the problem in that way because I assume that the general case (ie. the case of a surjective ring homomorphism) is not true (it might be but I cannot show it). So I concentrated on that specific case to find additional conditions on $I$ or $R$ that might make the hypothesis true. – user53970 Dec 17 '13 at 01:44

On the structure of subrings of $R/I$

2 Answers2