Question about ring homomorphism within a ring.

Question

I confronted with a statement:

Given a ring homomorphism $f:A\to B$, with commutative rings with identity $A,B$. If $A,B$ are both subrings of a bigger commutative ring with identity $R$, then $A$ must be a subring of $B$.

I have thought about quotient ring $B/I\to B$, where $I$ is an ideal of $B$. Then $B/I$ is not a subring of $B$. Can we embed them in a bigger ring? Hope someone could help. Thanks!

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Thanks to Gae. S., the conclusion above is not true. Now, what if the completion of $A$ and $B$ are equal, i.e. $\widehat{A}\cong \widehat{B}$. Does the conclusion holds?

Answer 1

$A$ needs not be a subring of $B$. Consider $R=\Bbb R[x]$, $A=\Bbb R[x^2]\subseteq \Bbb R[x]$, $B=\Bbb R[x^3]\subseteq \Bbb R[x]$ and $f(p(x))=p(x^{3/2})$ (with the convention $(x^2)^{1/2}=x$).

Also consider $R=\Bbb R[x,y]$, $B=\Bbb R[x]$, $A=\Bbb R[x^3,y^3]$ and $f(p(x,y))=p(x^{1/3},0)$ for all $p\in A$ to see that $f$ needs not be injective.

Answer 2

Generally, a ring homomorphsim $f:A\to B$ within a bigger ring cannot deduce that $A$ is a subring of $B$. But under noetherian hypothesis, if $A$ and $B$ have the same completion, i.e. $\hat{A}=\hat{B}$, we can conclude that $A$ is a subring of $B$.

Let's write down the exact sequence $0\to \ker(f)\to A \xrightarrow{f} B\to 0$. Since completion preserves exactness, we have $0\to \widehat{\ker(f)}\to \hat{A}\to \hat{B}\to 0$. But $\hat{A}=\hat{B}$, hence $\widehat{\ker(f)}=0$, i.e. $\ker(f)=0$. Thus, $f$ is injective. So $A$ can be seen as a subring of $B$.