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I'm trying to teach myself linear algebra from an abstract algebra perspective with this book: http://www.math.miami.edu/~ec/book/. I'm having some difficulty with this problem:

Recall that if $A$ is a ring and $a \in A$, then $aA$ is right ideal of $A$. Let $A = R_2$ (the ring of all matrices of size $2\times 2$) and $a = (a_{ij})$ where $a_{11}= 1$ and the other entries are $0$. Find $aR_2$ and $R_2a$. Show that the only ideal of $R_2$ containing $a$ is $R_2$ itself.

My approach to the problem is to try and demonstrate that the ideal must contain the identity element, but I'm not too sure how to go about that.

So far, this is what I have. Let $I$ be some ideal in $R_2$. Since $a = \left( \begin{array}{ccc}1 & 0 \\0 & 0\end{array} \right) $, $\exists$ some $i \in I$ and $r \in R_2$ such that $ ir = a $.

I think that since $a$ is singular, either $i$ or $r$ reduces to $a$, and $i$ or $r$ reduces to the identity matrix. And if the ideal contains the identity, then it must also be the entire ring. I'm not sure how to demonstrate that the ideal contains the identity though.

Thanks a lot!

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The real upshot of what is going on here is that the ring $M_2(F)$, with $F$ a field has exactly two ideals: the zero ideal and the whole ring. These types of rings are called simple rings. –  rschwieb Dec 26 '12 at 14:55
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If $a\in I$, then $$ \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \cdot a= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$ is also in $I$.

Right multiplying this by $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ shows $\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$ is also in $I$. As an ideal, $I$ is also an additive subgroup, so adding this element to $a$ shows $I_2\in I$.

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Note that: $$ \left( \begin{array}{cc}1 & 0 \\0 & 1\end{array} \right) = \left( \begin{array}{cc}1 & 0 \\0 & 0\end{array} \right) + \left( \begin{array}{cc}0 & 1 \\1 & 0\end{array} \right) \left( \begin{array}{cc}1 & 0 \\0 & 0\end{array} \right) \left( \begin{array}{cc}0 & 1 \\1 & 0\end{array} \right) $$ So the identity must be in the ideal.

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