The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to $\Q$.

Problem 525

Let
$R=\left\{\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \quad \middle | \quad a, b\in \Q \,\right\}.$ Then the usual matrix addition and multiplication make $R$ an ring.

Let
$J=\left\{\, \begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix} \quad \middle | \quad b \in \Q \,\right\}$ be a subset of the ring $R$.

(a) Prove that the subset $J$ is an ideal of the ring $R$.

(b) Prove that the quotient ring $R/J$ is isomorphic to $\Q$.

Proof.

(a) Prove that the subset $J$ is an ideal of the ring $R$.

Let
$\alpha=\begin{bmatrix} 0 & a\\ 0& 0 \end{bmatrix} \text{ and } \beta=\begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix}$ be arbitrary elements in $J$ with $a, b\in \Q$.
Then since we have
\begin{align*}
\alpha-\beta=\begin{bmatrix}
0 & a-b\\
0& 0
\end{bmatrix}\in J,
\end{align*}
the subset $J$ is an additive group.

Now consider any elements
$\rho=\begin{bmatrix} a & b\\ 0& a \end{bmatrix} \in R \text{ and } \gamma=\begin{bmatrix} 0 & c\\ 0& 0 \end{bmatrix}\in J.$ Then we have
\begin{align*}
\rho \gamma&=\begin{bmatrix}
0 & ac\\
0& 0
\end{bmatrix}\in J \text{ and }\6pt] \gamma \rho &=\begin{bmatrix} 0 & ca\\ 0& 0 \end{bmatrix}\in J. \end{align*} Thus, each element of J multiplied by an element of R is still in J. Hence J is an ideal of the ring R. (b) Prove that the quotient ring R/J is isomorphic to \Q. Consider the map \phi:R\to \Q defined by \[\phi\left(\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \,\right)=a, for $\begin{bmatrix} a & b\\ 0& a \end{bmatrix}\in R$.

We first show that the map $\phi$ is a ring homomorphism.
First of all, we have
\begin{align*}
\phi\left(\, \begin{bmatrix}
1 & 0\\
0& 1
\end{bmatrix} \,\right)=1.
\end{align*}
Thus $\phi$ maps the unity element of $R$ to the unity element of $\Q$.

Take
$\begin{bmatrix} a & b\\ 0& a \end{bmatrix}, \begin{bmatrix} c & d\\ 0& c \end{bmatrix}\in R.$ Then we have
\begin{align*}
\phi\left(\, \begin{bmatrix}
a & b\\
0& a
\end{bmatrix}+\begin{bmatrix}
c & d\\
0& c
\end{bmatrix} \,\right)&=\phi\left(\, \begin{bmatrix}
a+c & b+d\\
0& a+c
\end{bmatrix} \,\right)=a+c\6pt] &=\phi\left(\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \,\right)+\phi\left(\, \begin{bmatrix} c & d\\ 0& c \end{bmatrix} \,\right) \end{align*} and \begin{align*} \phi\left(\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix}\begin{bmatrix} c & d\\ 0& c \end{bmatrix} \,\right)&=\phi\left(\, \begin{bmatrix} ac & ad+bc\\ 0& ac \end{bmatrix} \,\right)=ac\\[6pt] &=\phi\left(\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \,\right)\phi\left(\, \begin{bmatrix} c & d\\ 0& c \end{bmatrix} \,\right). \end{align*} It follows from these computations that \phi:R\to \Q is a ring homomorphism. Next, we determine the kernel of \phi. We claim that \ker(\phi)=J. If \rho=\begin{bmatrix} a & b\\ 0& a \end{bmatrix}\in \ker(\phi), then we have \[0=\phi(\rho)=\phi\left(\, \begin{bmatrix} a & b\\ 0& a \end{bmatrix} \,\right)=a.

So $\rho=\begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix}\in J$, and hence $\ker(\phi)\subset J$.

On the other hand, if $\rho=\begin{bmatrix} 0 & b\\ 0& 0 \end{bmatrix}\in J$, then it follows from the definition of $\phi$ that $\phi(\rho)=0$.
Thus, $J \subset \ker(\phi)$.
Putting these two inclusions together yields $J=\ker(\phi)$.

Observe that the homomorphism $\phi$ is surjective.
In fact, for any $a\in \Q$, we take $\begin{bmatrix} a & 0\\ 0& a \end{bmatrix}\in R$. Then we have
$\phi\left(\, \begin{bmatrix} a & 0\\ 0& a \end{bmatrix} \,\right)=a.$
In summary, $\phi:R\to \Q$ is a surjective homomorphism with kernel $J$.

It follows from the isomorphism theorem that
$R/J\cong \Q,$ as required.

Remark.

Recall that the kernel of a ring homomorphism $\phi:R\to S$ is always an ideal of $R$.

Thus, the proof of (b) shows that $J$ is an ideal of $R$. This gives an alternative proof of part (a).

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