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I have the following problem:

Is it possible to construct ring homomorphism from $M_2(\mathbb Z)\to \mathbb Z$, or in other words, a homomorphism from ring of all $2\times2$ matrices over the integers into integers?

I tried determinant, trace and mapping that maps matrix to it's element in the position (1,1) but non of that obviously works, which led me to believe there might not be such homomorphism.

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    $\begingroup$ Does it have to respect $1$? $\endgroup$ – Hoot Jun 21 '15 at 22:42
  • $\begingroup$ Seems to be special case of this MO question, but perhaps there is simpler argument. $\endgroup$ – Prism Jun 21 '15 at 23:19
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No, not at all. A homomorphism must take nilpotent elements to zero, since $\Bbb Z$ has no proper nilpotents. The matrix that’s all zero except for a $1$ in the upper right corner must thus be taken to $0$. Similarly for the matrix with $1$ in the lower left. But their sum squares to the identity matrix, so your homomorphism is zero. (There are much better abstract proofs.)

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The kernel of a ring homomorphism is an ideal and the ideals of $M_n(\mathbb{Z})$ are of the form $M_n(k\mathbb{Z})$, for $k\ge0$. Since $$ M_n(\mathbb{Z})/M_n(k\mathbb{Z})\cong M_n(\mathbb{Z}/k\mathbb{Z}) $$ we see that the image of a homomorphism is either a finite subring of $\mathbb{Z}$ (if $k>0$) or the homomorphism is injective (if $k=0$).

The second possibility is ruled out, because $M_n(\mathbb{Z})$ is not commutative, for $n>1$. The second possibility only gives the zero homomorphism (if you don't require the identity is mapped to the identity).


The characterization of the ideals in the full matrix ring $M_n(R)$ over the (commutative) ring $R$ as being of the form $M_n(I)$, where $I$ is an ideal of $R$, is well known.

Once we accept it, we can generalize the statement. If $\varphi\colon M_n(R)\to R$ is a ring homomorphism, then $\ker\varphi=M_n(I)$ for some ideal $I$ of $R$. It's easy to see that $M_n(R)/M_n(I)\cong M_n(R/I)$, so we have an injective homomorphism $$ \hat{\varphi}\colon M_n(R/I)\to R $$ If $R$ is commutative, this forces $n=1$ or $I=R$, because $M_n(R)$ is not commutative for $n>1$ unless $R$ is the zero ring.

If we consider $R$ not the zero ring and ring homomorphisms to carry the unity to the unity, we conclude that, for every $n>1$, there is no ring homomorphism $M_n(R)\to R$.

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    $\begingroup$ This was the “much better” method I had in mind in my answer. $\endgroup$ – Lubin Jun 21 '15 at 23:51
  • $\begingroup$ @Lubin I added the obvious generalization, that shows nilpotent elements are an ad hoc trick. A very nice one, though. ;-) $\endgroup$ – egreg Jun 22 '15 at 6:32
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Let $S$ be a commutative unital ring and $R:=\text{Mat}_{n\times n}(S)$ where $n\in\mathbb{Z}_{>1}$. Suppose that $\phi:R\to T$ is a (not necessarily unitary) $S$-algebra homomorphism from $R$ to an $S$-algebra $T$ without zero divisors. (In the given problem, $S:=\mathbb{Z}$, $n:=2$, and $T:=\mathbb{Z}$.)

The ring $R$ is generated by the matrices $E_{i,j}$ for $i,j\in\{1,2,\ldots,n\}=:[n]$, where $E_{i,j}$ is the matrix with $1$ at the $(i,j)$-entry and $0$ everywhere else for every $i,j\in[n]$. As noted by Lubin, $E_{i,j}$ must be mapped to $0$ when $i\neq j$, as the matrix is nilpotent (this is where the assumption that $T$ have no zero divisors is used).

Let $u_i$ be the image of $E_{i,i}$ under $\phi$ for $i\in [n]$. Then, for a matrix $A=\sum\limits_{i,j\in[n]}\,a_{i,j}E_{i,j} \in R$, where $a_{i,j}\in S$ for all $i,j\in[n]$, we get $$\phi(A)=\sum_{i=1}^n\,a_{i,i}u_i\,.$$ As $\phi$ is multiplicative, we must have $$0=0\cdot \phi(A)=\phi(E_{i,j})\cdot \phi(A)=\phi\left(E_{i,j}\cdot A\right)=a_{j,i}u_i$$ whenever $i\neq j$. As $a_{i,j}$ for $i,j\in [n]$ are arbitrary, $u_i=0$ for all $i\in[n]$.

Hence, the zero map is the only possible ring homomorphism from $R$ to $T$. If you require the homomorphism to be unitary (i.e., the multiplicative identity of $R$ must be sent to $1\in T$), then there are no such homomorphisms.

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    $\begingroup$ I voted to undelete this to preserve the nice answers - including yours. It needs a couple more votes. $\endgroup$ – Bill Dubuque Jan 6 at 17:33
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I show here that there is no non-trivial ring homomorphism $\phi:M_2(\mathbb{Z})\to \mathbb{Z}$. To start, let us denote

\begin{align*} A=\begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}& &B=\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\\ C=\begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}& &D=\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix} \end{align*}

Assuming that there is a non-trivial ring homomorphism $\phi:M_2(\mathbb{Z})\to\mathbb{Z}$, we would have that $\phi(I)=1$. Notice that we have the following sums and products:

\begin{align*} \begin{bmatrix}1 & 0\\ 0 & 0\end{bmatrix}\cdot \begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}=\begin{bmatrix}0 & 0\\ 0 & 0\end{bmatrix} &\Rightarrow A\cdot D=0. &\text{(i)}\\ \begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}\cdot \begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix} &\Rightarrow B\cdot C=A.&\text{(ii)}\\ \begin{bmatrix}0 & 0 \\ 1 & 0\end{bmatrix}\cdot \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix} &\Rightarrow C\cdot B=D.&\text{(iii)}\\ \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}+\begin{bmatrix}0 & 0\\ 0 & 1\end{bmatrix}=\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix} &\Rightarrow A+D=I.&\text{(iv)} \end{align*}

Hence, from (i) we have that $\phi(A)\cdot \phi(D)=\phi(A\cdot D)=\phi(0)=0$, so one of the values $\phi(A),\phi(D)$ must be equal to $0$.

From (iv), we have that $\phi(A)+\phi(D)=\phi(A+D)=\phi(I)=1$, and we conclude that while of the values $\phi(A),\phi(D)$ is equal to 0, and the other one is equal to 1.

Without loss of generality let us assume that $\phi(A)=0$ and $\phi(D)=1$. Then, from (ii) we have $0=\phi(A)=\phi(B)\cdot \phi(C)$, and we have that one of the values $\phi(B),\phi(C)$ is equal to zero.

This leads to $1=\phi(D)=\phi(C)\cdot \phi(B)=0$ using (iii), a contradiction.

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    $\begingroup$ If a ring is not an integral domain, then $\phi(AB)=0$ does not necessarily imply that $\phi(A)=0$ or $\phi(B)=0$. $\endgroup$ – sequence Apr 12 '17 at 17:17
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    $\begingroup$ @sequence since $\phi$ is a ring homomorphism, $\phi(AB)=\phi(A)\phi(B)$. Also notice that $\phi(A),\phi(B)\in \mathbb{Z}$, and the latter actually is an integral domain. $\endgroup$ – Darío G Apr 12 '17 at 18:18
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Let $i$ be the image of the identity matrix. Then $i^2=i$ and so $i=0$ or $i=1$.

If $i=0$, then the map is the zero homomorphism because $A=AI$.

If $i=1$, let $J=\pmatrix{0&-1\\1&0}$. Then $J^2=-I$ translates to $j^2=-1$, which cannot happen in $\mathbb Z$.

Therefore, the only ring homomorphism $M_2(\mathbb Z)\to \mathbb Z$ is the zero map. If you require that a ring homomorphism must preserve the multiplicative identity, then there is none.

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