1. Subring Elements

Let $R$ be a ring and $a$ an element of $R$. Let $S = \{ x \in R: ax = 0_R\}$. $S$ is a subring of $R$.

Let $R = M_2(\mathbb{Z}_5)$ (the ring of 2×2 matrices with entries from the field $\mathbb{Z}_5$) and

$a= \begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix}$. Determine the elements of the subring S defined previously. (I) How many elements are in S? (II) Show that S is not an ideal of R.

Attempt:

(I) the elements of S are 2x2 matrices x such that $ax=0_R$.

let $x=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$

$ax=\begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix}4a+2c & 4b+2d \\ 3a+4c & 3b+4d \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$

If I solve the homogeneous system
4a+2c=0
4b+2d=0
3a+4c=0
3b+4d=0

I get a=b=c=d=0.

So does this mean the only element in the subring S is the zero matrix?

(II) I will use the "ideal test":

Since the zero matrix belongs to S, $S \neq \emptyset$.

Then $S \triangleleft R \iff (x-y) \in S$ and $rS, Sr \subseteq S \forall r \in R$. (x,y are in S).

The question says show that S is NOT an ideal. But the problem is that S passes the ideal test since (x-y) is equal to the zeo matrix and therefore $a(x-y)=0_R \in S$. The same is true with multipication. So, is something wrong with the question or did I make a mistake?!

2. Originally Posted by demode
Let $R$ be a ring and $a$ an element of $R$. Let $S = \{ x \in R: ax = 0_R\}$. $S$ is a subring of $R$.

This may depend on the definition. Many authors require that if R is unitary then in order to be considered a

candidate to be a subring a subset must contain the unit of R. In this case, though, it

must be not so since then a = 0.

Let $R = M_2(\mathbb{Z}_5)$ (the ring of 2×2 matrices with entries from the field $\mathbb{Z}_5$) and

$a= \begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix}$. Determine the elements of the subring S defined previously. (I) How many elements are in S? (II) Show that S is not an ideal of R.

Attempt:

(I) the elements of S are 2x2 matrices x such that $ax=0_R$.

let $x=\begin{pmatrix}a & b \\ c & d \end{pmatrix}$

$ax=\begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix}a & b \\ c & d \end{pmatrix} = \begin{pmatrix}4a+2c & 4b+2d \\ 3a+4c & 3b+4d \end{pmatrix}= \begin{pmatrix}0 & 0 \\ 0 & 0 \end{pmatrix}$

If I solve the homogeneous system
4a+2c=0
4b+2d=0
3a+4c=0
3b+4d=0

I get a=b=c=d=0.

Even without checking this cannot be right since the matrix x is singular, and indeed:

$3a+4c=2(5a+2c)\,,\,\,3b+4d=2(4b+2d)\Longrightarrow c=-2a=3a\,,\,\,d=-2b=3b\Longrightarrow$

any matrix of the form $\begin{pmatrix}a&b\\3a&3b\end{pmatrix}$ belongs to $S$

So does this mean the only element in the subring S is the zero matrix?

(II) I will use the "ideal test":

Since the zero matrix belongs to S, $S \neq \emptyset$.

Then $S \triangleleft R \iff (x-y) \in S$ and $rS, Sr \subseteq S \forall r \in R$. (x,y are in S).

The question says show that S is NOT an ideal. But the problem is that S passes the ideal test since (x-y) is equal to the zeo matrix and therefore $a(x-y)=0_R \in S$. The same is true with multipication.

No, it's not. Try to find a counterexample to $x\in S\,,\,r\in R\Longrightarrow rx\in S$ now that you know the general

form of elements in $S$

Tonio

So, is something wrong with the question or did I make a mistake?!
.

3. So if the elements are of the form $\begin{pmatrix}a&b\\3a&3b\end{pmatrix}$, then $a \in \{0,1,2,3,4\}$, $b \in \{0,1,2,3,4\}$ in $\mathbb{Z}_5$. 5x5=25 possible combinations of a and b. Therefore there are 25 elements in S?

Try to find a counterexample to $x\in S\,,\,r\in R\Longrightarrow rx\in S$ now that you know the general

form of elements in S
Let a=b=1 then $x= \begin{pmatrix}1&1\\3&3\end{pmatrix}$, and $r= \begin{pmatrix}1&1\\1&1\end{pmatrix}$.

rx= $\begin{pmatrix}4&4\\4&4\end{pmatrix}$, and

$a(rx)= \begin{pmatrix}4&2\\3&4\end{pmatrix} \begin{pmatrix}4&4\\4&4\end{pmatrix} = \begin{pmatrix}4&4\\3&3\end{pmatrix} \neq R_0$

Therefore $rx \notin S$. Is this okay?

4. Originally Posted by demode
So if the elements are of the form $\begin{pmatrix}a&b\\3a&3b\end{pmatrix}$, then $a \in \{0,1,2,3,4\}$, $b \in \{0,1,2,3,4\}$ in $\mathbb{Z}_5$. 5x5=25 possible combinations of a and b. Therefore there are 25 elements in S?

Let a=b=1 then $x= \begin{pmatrix}1&1\\3&3\end{pmatrix}$, and $r= \begin{pmatrix}1&1\\1&1\end{pmatrix}$.

rx= $\begin{pmatrix}4&4\\4&4\end{pmatrix}$, and

$a(rx)= \begin{pmatrix}4&2\\3&4\end{pmatrix} \begin{pmatrix}4&4\\4&4\end{pmatrix} = \begin{pmatrix}4&4\\3&3\end{pmatrix} \neq R_0$

Therefore $rx \notin S$. Is this okay?

Yes and yes, but it is S, not R_o.

Tonio

5. Originally Posted by tonio
Yes and yes, but it is S, not R_o.

Tonio
Why is that? S is defined as x in R such that ax=0_R. Since rx is in R, but a(rx) is not equal to 0_R, it follows that rx doesn't belong to S. Then what's wrong with what I did?

6. Originally Posted by demode
Why is that? S is defined as x in R such that ax=0_R. Since rx is in R, but a(rx) is not equal to 0_R, it follows that rx doesn't belong to S. Then what's wrong with what I did?

I think I do. You wrote $a(rx)\neq R_0$ , instead of $a(rx)\neq 0_R$

Tonio