Show that $M_2(\mathbb{R})$ has no non-trivial two-sided ideals

In addition to the title question, I also want to find a non-trivial right ideal and a non-trivial left ideal of $M_2(\mathbb{R})$ .

Attempt of title question:

Suppose $\exists I\subset M_2(\mathbb{R})$ s.t. $\forall x\in I, \forall r\in M_2(\mathbb{R})$, $rxr\in I$ $\Rightarrow$ $rx\in I, xr\in I$, but $M_2(\mathbb{R})$ is a non-commutative ring. So, w.l.o.g. if $rx\in I$ then $xr\not\in I$. Contradiction. I have an inkling this is mistaken.

I am also confused about the left and right ideals, but an idea that seems to work for a right ideal is to let $x=\begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}$. A left ideal would be similar.

• Your idea does not work. We could have $r_1x=xr_2$ with $r_1\ne r_2$ – Hagen von Eitzen Apr 4 '14 at 5:55

Your example for a left and right ideal is fine. Your argument for the other part is not right: noncommutative rings can have nontrivial ideals. (I can't think of a very simple example off the top of my head, but it's true.) To show that $M_2(\mathbb R)$ has no nontrivial ideals, assume a nonzero element $A=\big(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\big)$ is contained in a nontrivial ideal. You can multiply $A$ by elementary matrices (i.e. perform row and column operations) to reduce it to one of the three matrices $$\begin{bmatrix}0&0\\0&0\end{bmatrix},\, \begin{bmatrix}1&0\\0&0\end{bmatrix},\, \begin{bmatrix}1&0\\0&1\end{bmatrix}.$$ Now what can we conclude? (I omitted a lot of details. Can you fill them in?)

• It seems like the answer below you spoiled what you were getting at. Oh well... – user113525 Apr 4 '14 at 6:06
• Yeah, I'm relatively new to this site and that's one trend I've noticed. I try to not completely explain everything for some questions because I think you learn a lot by struggling a bit with problems. Anyway, I hope you understand now. – user134824 Apr 4 '14 at 6:08
• I agree with you, and I did see your process. I just didn't want to essentially copy the answer into a comment. – user113525 Apr 4 '14 at 6:09
• @user134824 Yeah, this spoiling happens frequently. Try not to get discouraged: hint-solutions like this are very valuable, especially in cases like this, where the OP is clearly putting thought into the problem. +1 – rschwieb Apr 4 '14 at 12:52

Assuming $M_2(\mathbb{R})$ refers to the set of all $2 \times 2$ real matrices:

If $A$ is a rank $2$ matrix, then $(A) = (I)$ is the whole ring. If $A$ is the zero matrix, then we get the trivial ideal. The only remaining possibility is that $A$ is a rank $1$ matrix.

By performing row operations (i.e. multiplying on the left by units) and column operations (i.e. multiplying on the right by units), we can produce any other rank $1$ matrix. So, we have $\pmatrix{1&0\\0&0} \in (A)$ and $\pmatrix{0&0\\0&1} \in (A)$. Because $(A)$ is closed under addition, $I \in (A)$, which means that $(A)$ is the entire ring.

Thus, the only two two-sided ideals are the trivial ideal and the ring itself.

As for nontrivial one-sided ideals, consider $$\left\{A\pmatrix{1&0\\0&0}: A \in M_2(\mathbb{R})\right\}\\ \left\{\pmatrix{1&0\\0&0}A: A \in M_2(\mathbb{R})\right\}$$

• Thanks for your answer! However, what does the parenthesis around $(I)$ and $(A)$ mean? – Frenzy Li Mar 24 '17 at 15:20
• For a ring element $r$, $(r)$ is the ideal generated by $r$. – Omnomnomnom Mar 24 '17 at 15:48

For your question on left/right ideals:

By "let $x= \begin{pmatrix}1 &1\\0 &0\end{pmatrix}$" do you mean the right ideal generated by $x$? (Obviously an idea is not equal to an element)? Of course this is true, given any element of a ring, the right ideal it generates is by definition a right ideal.

In more generality, simple submodules of $M_2(\mathbb R)$ (as a right module over itself) are isomorphic to $\mathbb R^2$ - corresponding to nonzero rows. Thus we have two simple submodules:

$$M_1 = \left\{\begin{pmatrix}\alpha &\beta \\0 &0\end{pmatrix} : \alpha , \beta \in \mathbb R \right\}$$ $$M_2 = \left\{\begin{pmatrix}0 &0\\\gamma &\delta\end{pmatrix} : \gamma , \delta \in \mathbb R \right\}$$

Can you show this? And moreover, now categorise all right ideals?

Similar arguments will show that all left ideals correspond to (direct sums of) columns in $\mathbb R^2$

• I never said $x$ was an ideal, I was simply describing the type of elements the ideal would contain. – user113525 Apr 4 '14 at 6:03
• Aha, sorry for confusion. You are of course right that this would be in a right ideal ($M_1$ to be specific). – ah11950 Apr 4 '14 at 6:09