find the number of ideals of R.

Question

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 ?

Answer 1

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

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

2. 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.)

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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.

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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.

Answer 2

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

Answer 3

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 $0\neq3$.