Let $R= {\rm Mat}_2(\Bbb R)$ be the ring (with $1$) of $2\times2$-matrices with entries in $\Bbb R$. Let $$M = \left\{\begin{pmatrix}1&0 \\0&0\end{pmatrix},\begin{pmatrix}0&1\\0&0\end{pmatrix}\right\},$$ i.e. a set of two matrices. What are the left and right ideal generated by $M$ in $R$?

I tried to write my answer by it won't accept it. I don't know how to write Matrices (and their product) in the right format.

Is this correct: For left ideal: $$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}1&0\\0&0\end{pmatrix}=\begin{pmatrix}a&0\\c&0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&a\\0&c\end{pmatrix}$$

hence $L = \begin{pmatrix}a&a\\c&c\end{pmatrix}$.

For right ideal:

$$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}a&b\\0&0\end{pmatrix} \quad \text{ and } \quad \begin{pmatrix}0&1\\0&0\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}c&d\\0&0\end{pmatrix}$$

hence $R = \begin{pmatrix}a+c&b+d\\0&0\end{pmatrix}$.

  • $\begingroup$ Is this correct: For left ideal: (a,b;c,d)*(1,0;0,0)=(a,0;c,0) and (a,b;c,d)(0,1;0,0)=(0,a;0,c) => L=(a,a;c,c) For right ideal: (1,0;0,0)*(a,b;c,d)=(a,b;0,0) and (0,1;0,0)(a,b;c,d)=(c,d;0,0) => R=(a+c,b+d;0,0) $\endgroup$ – blondy Sep 4 '13 at 9:51
  • $\begingroup$ I tried my best to write the question. The things in (.) are supposed to be 2x2 matrices. How do I write matrices here? Can a mod please edit my question? $\endgroup$ – blondy Sep 4 '13 at 12:26
  • $\begingroup$ For some basic information about writing maths at this site see e.g. here, here, here and here. $\endgroup$ – Lord_Farin Sep 4 '13 at 12:27
  • $\begingroup$ I have edited the question for you. I hope to not have made any transcription errors. $\endgroup$ – Lord_Farin Sep 4 '13 at 12:35
  • $\begingroup$ Yeah everything is correct now. Thanks much. Do you know if my results are correct? $\endgroup$ – blondy Sep 4 '13 at 12:44

Sorry, there are some issues here caused by your choice of notation. By multiplying the same $a,b,c,d$ matrix with both generators, you've overlooked that there's nothing wrong with multiplying the generators with two different matrices and adding, which will produce many more possiblities.

In the left ideal generated by those two things, a general element will look like this:


Notice how we didn't recycle the $a,b,c,d$ matrix for both generators. Since you can pick $a,e,c,g$ to be whatever you want, you can see that the left ideal generated by these two things is the entire matrix ring.

Try to apply similar reasoning to the right ideal generated by these two things. You will get a different answer than the left ideal, and the answer you gave is not incorrect. It's just that you can express what the right ideal looks like more simply.

  • $\begingroup$ Isn't the right ideal $\begin{bmatrix}a+g&b+h\\0&0\end{bmatrix}$ then? How can I make it simpler? $\endgroup$ – blondy Sep 4 '13 at 13:20
  • $\begingroup$ @blondy Yes, that's all correct: everything in the right ideal can be written that way. But, after you've realized this, you can reexpress what the ideal looks like without $+$ signs, and even without talking about $a,b,c,d,e,f,g,h$. Can you see how? I just want you to focus on which matrices are in the right ideal rather than the detailed analysis we just did. $\endgroup$ – rschwieb Sep 4 '13 at 13:24
  • 1
    $\begingroup$ @blondy Your answer would fill in this blank to answer this question: "Q: Which matrices are in the right ideal generated by these?" A: "The ones with ___________." (No references to $a,b,c,d,e,f,g,h$ would be even needed, actually :) ). $\endgroup$ – rschwieb Sep 4 '13 at 13:26
  • 1
    $\begingroup$ \begin{bmatrix}i&j\\0&0\end{bmatrix} ??? $\endgroup$ – blondy Sep 4 '13 at 13:35
  • 1
    $\begingroup$ @blondy Yes, that is 99% what I was guiding you towards :) I was hoping for "matrices with zeros in the bottom row." Good job! $\endgroup$ – rschwieb Sep 4 '13 at 13:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.