1
$\begingroup$

$\DeclareMathOperator{\im}{im}$The First Isomorphism Theorem for rings is usually stated as follows:

Let $f\colon R\to S$ be a surjective homomorphism of rings with kernel $K$. Then the quotient ring $R/K$ is isomorphic to $S$.

In some other textbook, the theorem would be stated slightly differently as:

Let $f\colon R\to S$ be a surjective homomorphism of rings with kernel $K$. Then the quotient ring $R/K$ is isomorphic to $\im(f)$ where $\im(f)$ denotes the direct image of $f$.

In the second form, if $R/K \cong\im(f)$ and $\im(f)$ is a subring of $S$, then how would I show that $R/K\cong S$. I can prove the theorem in both formulations, but from the second formulation of the theorem, I don't know how to how to show $R/K \cong S$ from $R/K \cong\im(f)$. I know that $f$ is a surjective homomorphic mapping from $R$ to $S$ and that if $R/K\cong\im(f)$, can Is that enough to show that $\im(f)=S.$

Thank you in advance.

$\endgroup$
2
  • 5
    $\begingroup$ The second version is true without the assumption on surjectivity. The equivalence just using the fact that if $f$ is surjective, then by definition $\mathrm{Im}(f) = S$. $\endgroup$
    – Mathmo123
    Apr 19 '21 at 10:08
  • $\begingroup$ @Mathmo123 because of $f$ is a surjective mapping and in the proof of the theorem, by constructing a function $\phi(r+ker f)=f(r)$, then $\phi$ is surjective from $R/I$ to $S$ because $f$ is surjective from $R$ to $S$. Would that be a correct reasoning. $\endgroup$
    – Seth Mai
    Apr 19 '21 at 10:12
2
$\begingroup$

If $f$ is surjective then $\operatorname{im}f=S$ by definition. Hence the second formulation implies the first one when we consider a surjective map. Conversly, any map $f\colon R\to S$ "can be made surjective" by restricting the codomain $S$ to the subring $\operatorname{im}f$ onto which $f$ is trivially surjective. In this case the first formulation implies the second one.

Your argument in the comments is correct. Alternatively, if we write $\pi\colon R\to R/I$ for the canonical projection (defined by $r\mapsto r+I$) we have that $\phi\circ\pi=f$. By elementary set theory the surjectivity of $f$ then implies the surjectivity of $\phi$ (this is a nice standard fact to keep in mind).

$\endgroup$
4
  • $\begingroup$ thank you for your detailed explanation. $\endgroup$
    – Seth Mai
    Apr 19 '21 at 17:58
  • $\begingroup$ @SethMai Glad to help! Is there anything I can add? otherwise make sure to accept the answer to close the thread :) $\endgroup$
    – mrtaurho
    Apr 19 '21 at 19:57
  • $\begingroup$ I just added the check marks. What is the differences between upvoting an answer vs adding a checkmark to the answer. I have not been paying much attention about these features. I usually just write a thank you comment. $\endgroup$
    – Seth Mai
    Apr 20 '21 at 7:20
  • $\begingroup$ @SethMai See here for a short explanation. $\endgroup$
    – mrtaurho
    Apr 20 '21 at 10:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.