Prove that the image of an ideal under a ring homomorphism $\mathbb Z[x]/I\to M_2(\mathbb Z)$ is not an ideal

Question

Let $R=\mathbb Z[x]/I$ where $I=(x^2-5x-2)$ and $S=M_2(\mathbb Z)$. Let $B$ be the matrix with rows $(1,2),\ (3,4)$. The kernel of the ring homomorphism $\mathbb Z[x]\to S$ that sends $p(x)$ to $p(B)$ contains $x^2-5x-2$, so this homo factors through $R$ by the universal property of ring homomorphism, and there is a unique ring homomorphism $f:R\to S$ with $f(x+I)=B$ defined by $p(x)+I\mapsto p(B)$.

I want to show that the image of the ideal $J=((x-1)+I)\subset R$ under $f$ is not an ideal in $S$.

My thoughts are: the matrix with rows $(0,2),\ (3,3)$ lies in the image. If the image were an ideal, then multiplication by an elementary matrix would result in a matrix that lies in the image of $f$. E.g. the matrix with rows $(3,3),(0,2)$ would be in the image of $f$. But I'm not sure how to show that it isn't in the image of $f$. The image of $f$ is all elements of the form $f(\overline {p(x)}\overline{(x-1)})=p(B)(B-I)$ if I understand correctly.

Answer 1

Let $T:=\text{im}(f)$. Clearly, $T$ is not the whole $S$, as elements of $T$ commute with $B$. Observe that $$f\big((x-1)+I\big)=f(x)-1_S=B-1_S=\begin{bmatrix}1&2\\3&4\end{bmatrix}-\begin{bmatrix}1&0\\0&1\end{bmatrix}=\begin{bmatrix}0&2\\3&3\end{bmatrix}=:C\,.$$ Thus, the image of the ideal $J$ of $R$ under $f$ is $TC$. It is strictly contained in the left ideal $SC$ of $S$. This is because everything in $TC$ commutes with $B$, whereas the element $$\begin{bmatrix}0&1\\1&0\end{bmatrix}\,C=\begin{bmatrix}3&3\\0&2\end{bmatrix}$$ of $SC$ does not commute with $B$, as we can readily check $$\begin{bmatrix}3&3\\0&2\end{bmatrix}\,B=\begin{bmatrix}3&3\\0&2\end{bmatrix}\,\begin{bmatrix}1&2\\3&4\end{bmatrix}=\begin{bmatrix}12&18\\6&8\end{bmatrix}\neq\begin{bmatrix}3&7\\9&17\end{bmatrix}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\,\begin{bmatrix}3&3\\0&2\end{bmatrix}=B\,\begin{bmatrix}3&3\\0&2\end{bmatrix}\,.$$ Thus, $TC$ is not even a left ideal of $S$.

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Note that the left ideal $SC$ of $S$ consists of matrices $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, where $(a,b)$ and $(c,d)$ are in the $\mathbb{Z}$-span of $(0,2)$ and $(3,3)$. The right ideal $CS$ of $S$ is composed by matrices $\begin{bmatrix}a&b\\c&d\end{bmatrix}$, where $(a,c)$ and $(b,d)$ lie within the $\mathbb{Z}$-span of $(0,3)$ and $(2,0)$. Finally, the two-sided ideal $SCS$ of $S$ equals $S$ itself.

Answer 2

If $f(R/I)$ were an ideal, $C(B-I)$ should be an element of $f(R/I)$ for any $C \in M_2(\mathbb{Z})$ and thus there should exist $p \in \mathbb{Z}[X]$ such that $p(B)(B-I) = C(B-I)$, that is,

$$ p(B) = C $$

since $B-I$ is invertible. In particular, the eigenvalues of $C$ have to be those of $p(B)$ which in turn are the eigenvalues of $B$ via $p$. However, $B$ has real eigenvalues and therefore this would imply that any matrix with integer coefficients has real eigenvalues, which is not the case: take for example

$$ \left[ {\begin{array}{cc} 0 & 1 \\ -1 & 0 \\ \end{array} } \right] $$

with characteristic polynomial $X^2 + 1$.