$\text{Mat}_n(R)$ is $K$-algebra $\Rightarrow$ $R$ is $K$-algebra

Question

Let $R$ be a ring and $K$ be a field.

Claim. The centre of $\text{Mat}_n(R)$ is $Z(\text{Mat}_n(R))=Z(R)\cdot I_n$ and if $\text{Mat}_n(R)$ is a $K$-algebra, so is $R$.

Remark. We have defined a $K$-algebra as a ring $S$ together with a ringhomomorphism $K\rightarrow Z(S)$.

I am done with the first part of the claim.

My question is the second part of the claim. I have a ringhomomorphism $K\rightarrow\text{Mat}_n(R)$ and I want to get a ringhomomorphism $K\rightarrow R$. That would show $\text{Mat}_n(R)$ is $K$-algebra $\Rightarrow$ $R$ is $K$-algebra.

Do you have any ideas?

Answer 1

By assumption, we have a morphism $$K\xrightarrow\phi Z(\text{Mat}_n(R))=Z(R)\cdot I_n$$ and we want a morphism $$K\to Z(R).$$ It suffices to find a morphism $$Z(R)\cdot I_n\xrightarrow\psi Z(R).$$ We can take the map that sends a matrix to the element in its left upper corner. (This is indeed a ring morphism, as is easily checked.)