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).

The post The Quotient Ring by an Ideal of a Ring of Some Matrices is Isomorphic to $\Q$. first appeared on Problems in Mathematics.

Prove that the rings $2\Z$ and $3\Z$ are not isomorphic.
 

Definition of a ring homomorphism.

Let $R$ and $S$ be rings.

• A homomorphism is a map $f:R\to S$ satisfying
  1. $f(a+b)=f(a)+f(b)$ for all $a, b \in R$, and
  2. $f(ab)=f(a)f(b)$ for all $a, b \in R$.
• A bijective ring homomorphism is called an isomorphism.
• If there is an isomorphism from $R$ to $S$, then we say that rings $R$ and $S$ are isomorphic (as rings).

Proof.

Suppose that the rings are isomorphic. Then we have a ring isomorphism
\[f:2\Z \to 3\Z.\] Let us put $f(2)=3a$ for some integer $a$. Then we compute $f(4)$ in two ways.
First we have
\[f(4)=f(2+2)=f(2)+f(2)=3a+3a=6a.\]

Next we have
\[f(4)=f(2\cdot 2)=f(2)\cdot f(2)=3a\cdot 3a=9a^2.\] These are equal and hence we have
\[6a=9a^2.\] The only integer solution is $a=0$.

But then we have $f(0)=0=f(2)$, which contradicts that $f$ is an isomorphism (hence in particular injective).
Therefore, there is no such isomorphism $f$, thus the rings $2\Z$ and $3\Z$ are not isomorphic.

The post Rings \Z$ and \Z$ are Not Isomorphic first appeared on Problems in Mathematics.