# How do I show that $\text{End}_R\mathbb{Z}^n\cong\mathbb{Z}$ for $R=\text{Mat}_n(\mathbb{Z})$?

Let $R=\text{Mat}_n(\mathbb{Z})$ and $M=\mathbb{Z}^n$ the (left) $R$-module with action the matrix multiplication.

How do I prove that $\text{End}_RM\cong\mathbb{Z}$?

Should I find an explicit isomorphism?

• Sorry that I havent tried anything yet, but I dont see it at all :( – TR-8R Jan 17 '16 at 13:27
• Here is a hint: every linear transformation from $\Bbb{Z}^n$ to itself has the matrix representation (under standard basis, of course.) Which condition holds for such matrix representation? – Hanul Jeon Jan 17 '16 at 13:43
• @HanulJeon I am not sure... – TR-8R Jan 17 '16 at 13:46
• Which point is a point you have not understood? Let me know what it is. Possible statements you have not gotten are: First, every $R$-module endomorphism over $M$ is also a linear transformation over $M$ to itself. Second, every linear transformation from finite-dimensional free module to itself has corresponding matrix representation. If you get such steps then you are almost done; next steps are described in @rschwieb's answer. – Hanul Jeon Jan 17 '16 at 14:03

To rephrase your question, the elements of $End(M_R)$ are those elements of $R$ which commute with all other elements of $R$.

So, you are just looking for the center of a matrix ring.

For any $S$-module $M$, we have $S\subseteq End(M_\Bbb Z)$ in a natural way (multiplication by elements of $S$ make additive maps.)

For any ring $S$, $End(S^n_S)\cong Mat_n(S)$.

Finally, $End(M_S)$ is, by definition, the subring of $End(M_\Bbb Z)$ whose elements all commute with the elements of $S$.

By slotting these with the specific situation you were given, you arrive at my original suggestion.

• Why would that be? – TR-8R Jan 17 '16 at 13:38
• $End(M_R)$ is just the subring of $End(M_\Bbb Z)=R$ which centralizes $R$. That is what it means to be $R$ linear, in this case. – rschwieb Jan 17 '16 at 13:44
• And why do I have an isomorphism when i find the center? – TR-8R Jan 17 '16 at 13:45
• @TR-8R if you know what the center of $R$ looks like, then the isomorphism is obvious. – rschwieb Jan 17 '16 at 13:48
• @ThomasAndrews Don't say useless, because it did help me understand my problem better – TR-8R Jan 17 '16 at 14:02