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

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

$\displaystyle a= \begin{pmatrix}4 & 2 \\ 3 & 4 \end{pmatrix}$. 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 $\displaystyle ax=0_R$.

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

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

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:

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

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

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, $\displaystyle S \neq \emptyset$.

Then $\displaystyle S \triangleleft R \iff (x-y) \in S$ and $\displaystyle rS, Sr \subseteq S \forall r \in R$. (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 $\displaystyle a(x-y)=0_R \in S$. The same is true with multipication.

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

form of elements in $\displaystyle S$

Tonio

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