$\DeclareMathOperator{\im}{im}$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.

  • 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

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).

  • $\begingroup$ 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

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