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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?

TR-8R
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  • 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

1 Answers1

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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.

Here are breadcrumbs to follow to get to this idea:

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.

Community
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rschwieb
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  • 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
  • I'm sorry, but it isn't to me :( – TR-8R Jan 17 '16 at 13:49
  • I'm pretty sure that this is because, as the timestamps on your comments show, you haven't paused to think about anything I have said or read the link at all. If you are not going to think longer than a minute about anything someone says to you, you are not going to solve many problems in life. Now, try again to connect the dots... At least try for 10 minutes this time.. – rschwieb Jan 17 '16 at 13:51
  • No, $\mathrm{End}_R(M)$ is the set of all $R$-homomorphisms $M\to M$ Where do you get that the elements are elements of $R$? – Thomas Andrews Jan 17 '16 at 13:53
  • @ThomasAndrews $End(M_S)$ is a subring of $End(M_\Bbb Z)$ for any $S$ module structure on $M$. In this case, the second endomorphism ring is labeled $R$. – rschwieb Jan 17 '16 at 13:57
  • So the elements of End$ (M_R)$ are just $aI_n$ for $a\in\mathbb{Z}$? – TR-8R Jan 17 '16 at 13:57
  • @TR-8R Yes, that it what I'm suggesting. – rschwieb Jan 17 '16 at 14:01
  • Oh, I definitely see it, but this answer is useless because it just jumps into assuming all that. @rschwieb – Thomas Andrews Jan 17 '16 at 14:01
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    @ThomasAndrews Don't say useless, because it did help me understand my problem better – TR-8R Jan 17 '16 at 14:02
  • @ThomasAndrews This is a work in progress meant to evolve with the poster's responses, it is not written in stone. – rschwieb Jan 17 '16 at 14:07
  • Though I would not mind if you or someone else would post a more complete answer where everything said here combined – TR-8R Jan 17 '16 at 14:13
  • @TR-8R yes, after getting you to a certain point in the comments, that was my intention, and is what I've been doing for several minutes ( on my mobile device) – rschwieb Jan 17 '16 at 14:16