# 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 $\Q$.

Contents

## 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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