Let $$ M=\left\{ \begin{pmatrix} a & b \\ 2b & a \end{pmatrix} : a,b \in \Bbb{Q} \right\}.$$ We can show that $M$ is a ring. The problem is to find all ideals of $M$.
Here is my work so far.
In order for $I \subset M$ to be an ideal, it must first be a subgroup of $(M,+)$. But, I don't know how to find all subgroups of $M$, so I am stuck here. Any help is appreciated.
On
It seems to me that the only ideal is $M$.
We denote as $(a,b)$ the elements of $M$. The properties of $M$ are that $(1,0)$ is the identity, and $(0,1)(a,b)=(2b,a)$, all elements of $M$ are linear combinations of these 2 elements. Also, $M$ is a commutative ring.
If $i \in I$ is invertible, then $\forall m \in M$, we have $ m . i^{-1} . i$ which implies that $I = M$ as soon as it contains an invertible element.
The only interesting elements of of $M$ that would participate to a non trivial ideal would necessarily be non invertible. Since $\Delta = a^2-2b^2$, this leaves 2 types of elements: $$I_+ = \{(\sqrt{2} a, a \in Q)\}$$ $$I_- = \{(-\sqrt{2} a, a), a \in Q\}$$ As those elements are stable by multiplication $(1,0)$ and $(0,1)$ and linear combination, those would be bona fide ideals.
The only problem is that $\sqrt{2}$ is a real number, not a rational, and therefore $I_-$ and $I_+$ are not part of $M$.
Hints:
Your ring is commutative and generated by $\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $\begin{bmatrix}0&1\\2&0\end{bmatrix}$, the latter of which behaves like $\sqrt{2}$. Therefore you have a ring epimorphism sending $1\mapsto \begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $x\mapsto\begin{bmatrix}0&1\\2&0\end{bmatrix}$ that makes a map $\mathbb Q[x]\to M$ with $x^2-2$ in the kernel.
This allows you to look at $M$ as an easily understood quotient of $\mathbb Q[x]$.