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Show that $M$ and $\{0_M\}$ are the only ideals of the ring $M=\left\{ \begin{pmatrix}a&b\\ c&d\end{pmatrix} \mid a,b,c,d \in \Bbb R\right\}$. (Hint: Show if $I \ne \{0_M\}$ show that $1_M \in I.)$

Let $I$ be an ideal of $M$ such that $I \ne \{0_M\}$. Then $I$ is closed under addition and for all $r \in M, x \in I$ we have that $rx \in I$.

I think I need to use these properties to show that for any $\begin{pmatrix}a&b\\ c&d\end{pmatrix} \in I$ I can sum and multiply this by some elements to get $\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$.

If I multiply for example $$\begin{pmatrix}1&0\\ 0&0\end{pmatrix}\begin{pmatrix}a&b\\ c&d\end{pmatrix}-\begin{pmatrix}a&b\\ c&d\end{pmatrix}\begin{pmatrix}0&0\\ 0&1\end{pmatrix}=\begin{pmatrix}a&0\\ 0&-d\end{pmatrix}$$ but I don't have control over $a$ and $-b$ in the ideal?

geojaffa
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  • I use this as a guided multistep exercise in early courses on abstract algebra. The first step is to show that a non-trivial ideal contains a matrix $A$ such that $A$ has exactly one non-zero entry. Multiplying a matrix you start with from both left and right with matrices $$\pmatrix{1&0\cr0&0\cr}$$ and $$\pmatrix{0&0\cr0&1\cr}$$ will get you there. Do you see how? – Jyrki Lahtonen Aug 17 '22 at 07:02
  • In the next step you show that A) you can move that single non-zero entry wherever you want. Multiply (left or right) by $$P=\pmatrix{0&1\cr1&0\cr}$$ to achieve that, and B) show that you can scale the single non-zero entry to $1$ - all while staying inside the ideal. Then only in the last step you show that you can get the identity matrix in the ideal as well. At that point you are, of course, done. – Jyrki Lahtonen Aug 17 '22 at 07:05
  • Mind you, we have probably done this on this site already. Do search, also! – Jyrki Lahtonen Aug 17 '22 at 07:06
  • Check out this. Or this. – Jyrki Lahtonen Aug 17 '22 at 07:11

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Let $I\subset M_2(\mathbb{R})$ be an ideal s.t $I\neq \{0\}$. Since $I\neq \{0\}$, there exists $A\neq 0\in I$. WLOG, we can assume that $a_{1,1}\neq 0$ (otherwise, if $a_{i,j}\neq 0)$, we can multiply be elementary matrices that act as row and column operations, and since $I$ is closed under multiplications, we have $A'$ with $a'_{1,1}\neq 0$). Now, multiplying $A$ with $$\begin{pmatrix}1&&0\\0&&0\end{pmatrix}$$ And since $I$ is closed under multiplication, we have that $$\begin{pmatrix}a_{1,1}&&0\\0&&0\end{pmatrix}\in I$$By multiplying by elementary matrices that act as row and column operations, we also have: $$\begin{pmatrix}0&&0\\0&&a_{1,1}\end{pmatrix}\in I$$Since $I$ is closed under sum, we have that:$$\begin{pmatrix}a_{1,1}&&0\\0&&a_{1,1}\end{pmatrix}\in I$$Multiplying by $\frac{1}{a_{1,1}}$, we finally see that:$$\begin{pmatrix}1&&0\\0&&1\end{pmatrix}\in I$$ Now you need to show that if $I\subset R$ is an ideal that contains the identity, then $I=R$. Therefore, $I=M_2(\mathbb{R})$.

Math101
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