find the number of ideals of R. let $R=\{$$  \left(
\begin{matrix}
 a & a & a \\
a & a & a \\
a & a & a
\end{matrix}
\right)$$ | a \in\mathbb{R}\}$ now find the number of ideals of R.
I think the ring $R$ has exactly trivial ideal because let $I$ be a pure  ideal of $R$  I claim that $I=R $ because $$  \left(
\begin{matrix}
 1& 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{matrix}
\right) \in I .$$
if we set $a\in \mathbb{Z_p}$ what is answer ?
 A: Hint:
$\phi:\mathbb R\to R$ given by $\phi(a)=\frac{a}{3}\begin{bmatrix}1&1&1 \\
1&1&1 \\
1& 1&1\end{bmatrix}$ is an isomorphism of rings.
Your conclusion is right, but your reasoning doesn't make any sense.  
The problems that stick out to me are


*

*Why should 
$\begin{bmatrix}
1&1&1 \\
1&1&1 \\
1& 1&1\end{bmatrix}\in I$? You didn't explain.

*Even if it is, why does it matter?  Presumably you thought it was the identity (it isn't) or a unit (it is, but you didn't explain.)

If you use $\mathbb Z/p\mathbb Z$ instead of $\mathbb R$, the conclusion that there are only two ideals still holds if $p\neq 3$. But if $p=3$ then you are looking at a ring in which every pair of things multiplies to zero. In such a ring, the ideals are exactly the additive subgroups. Since it is still isomorphic to $\mathbb Z/3\mathbb Z$, there are only two such subgroups, but what you are looking at is not a field, but a nilpotent ring.

One less important thing: the word you're looking for is proper ideal, not "pure" ideal. I imagine a lot of ring/module theorists would think of pure submodules, even though they would probably guess you meant this far simpler thing.
A: Note that $\mathbb R$ and $\mathbb Z/p\mathbb Z$ (where $p$ is prime) are both fields, so we can divide elements with no repercussions. Note that we require $p\neq3$ so that we can divide by $3$.
Let's define $M(a)$ as $$M(a):=\left(
\begin{matrix}
 a & a & a \\
a & a & a \\
a & a & a
\end{matrix}
\right)$$
Note that $M\big(\frac13\big)$ is the identity of the ring. 
For any non-zero $M(a)$, $M(a)\cdot M\big(\frac1{9a}\big)=M\big(\frac1{9a}\big)\cdot M(a)=M\big(\frac13\big)$. 
Hence, this ring is a field (since it's commutative), assuming the underlying constant structure is also a field where $0\neq3$.
A: With $A_a:=\left(
\begin{matrix}
 a & a & a \\
a & a & a \\
a & a & a
\end{matrix}
\right)$, observe that $A_aA_b=A_{3ab}$, so that $A_a\mapsto 3a$ is a ring isomorphism $R\to\Bbb R$. Thus $R$ is a field and has only $0$ and $R$ as ideals
