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Suppose $f:R \rightarrow R$ is an homomorphism between rings. Consider the function $\hat f:\mathcal{M}_2 (R) \rightarrow \mathcal{M}_2 (R)$ such that $\hat{f}(a_{ij})=(f(a_{ij}))$. Show $\hat{f}$ is an homomorphism. If $f$ is an epimorphism, does $\hat{f}$ have to be an epimorphism?

$\hat{f}$ is an homomorphism: We have to verify two things:

  1. $\hat{f}(a_{ij} + b_{ij})=\hat{f}(a_{ij}) + \hat{f}(b_{ij})$:

$\hat{f}(a_{ij} + b_{ij})= \begin{bmatrix}f(a_{11}+b_{11}) & f(a_{12}+b_{12})\\f(a_{21}+b_{21}) & f(a_{22}+b_{22})\end{bmatrix}=\begin{bmatrix}f(a_{11})+f(b_{11}) & f(a_{12})+f(b_{12})\\f(a_{21})+f(b_{21}) & f(a_{22})+f(b_{22})\end{bmatrix}=\hat{f}(a_{ij}) + \hat{f}(b_{ij})$

  1. $\hat{f}(a_{ij} b_{ij})=\hat{f}(a_{ij}) \hat{f}(b_{ij})$:

$\hat{f}(a_{ij} b_{ij})=\begin{bmatrix}f(a_{11}b_{11}+a_{12}b_{21}) & f(a_{11}b_{12}+a_{12}b_{22})\\f(a_{21}b_{11}+a_{22}b_{21}) & f(a_{21}b_{12}+a_{22}b_{22})\end{bmatrix}=\begin{bmatrix}f(a_{11})f(b_{11})+f(a_{12})f(b_{21}) & f(a_{11})f(b_{12})+f(a_{12})f(b_{22})\\f(a_{21})f(b_{11})+f(a_{22})f(b_{21}) & f(a_{21})f(b_{12})+f(a_{22})f(b_{22})\end{bmatrix}=\hat{f}(a_{ij}) \hat{f}(b_{ij})$

$\hat{f}$ is an epimorphism: We have already proved it is homomorphism, so we want to show $\hat{f}$ is surjective. This is, every element of $\mathcal{M}_2 (R)$ is the image of an element of $\mathcal{M}_2 (R)$ under $\hat{f}$.

Let $ \begin{bmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{bmatrix} \in \mathcal{M}_2 (R)$. We have that $b_{ij}\in R \ \forall \ i,j \in{ \{1,2\}}$ and because $f$ is an epimorphism there exist $a_{ij}$ such that $f(a_{ij})=b_{ij} \ \forall \ i,j \in{ \{1,2\}}$. So then $ \begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix} \in \mathcal{M}_2 (R)$ verifies that $\hat{f}\bigg( \begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}\bigg) = \begin{bmatrix}b_{11} & b_{12}\\b_{21} & b_{22}\end{bmatrix}$ and $\hat{f}$ is an epimorphism.

Is my reasoning correct?

Yagger
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    I’m sorry, but why does it follow that it’s monic? – Randall Aug 23 '18 at 01:29
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    "Epimorphism" and "surjective" are not equivalent in the category of rings, so your reasoning is not correct. Indeed, for any commutative ring $A$ and any multiplicative set $S \subset A$, the canonical localization morphism $A \to S^{-1}A$ is an epimorphism by the universal property of localization, for example. – Alex Wertheim Aug 23 '18 at 01:39
  • @Randall I meant that I have already proved it is monomorphism (or at least I think I have) but I'm not sure about the epimorphism part. Anyway I've edited the post to include my attempt in showing it is monomorphism as well. – Yagger Aug 23 '18 at 02:00
  • That’s not what mono means. – Randall Aug 23 '18 at 02:03
  • @Randall wow I did a huge mistake. I meant homomorphism. Thanks for pointing it out. – Yagger Aug 23 '18 at 02:04
  • This appears right if your book defines epimorphism to be surjective homomorphism. But then @AlexWertheim ‘s comment still stands. – Randall Aug 23 '18 at 02:07
  • Another way to prove that $\hat f$ is a homomorphism: note that $M_2(R) \cong R \otimes_{\Bbb Z} M_2(\Bbb Z)$. We can then see that $f \otimes \operatorname{id}$ is a homomorphism of $\Bbb Z$-algebras. – Ben Grossmann Aug 23 '18 at 04:47

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