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Prove that if $I$ is an ideal of the ring $M_2(\mathbb{R})$ and $X= \begin{bmatrix}0&1\\0&0\end{bmatrix}\in I$, then $I=M_2(\mathbb{R})$.

I think the question is wrong because there is no matrix $A\in M_2(\mathbb{R})$ such that $AX=I_2$ or $XA=I_2$, so $I_2$ is not in this ideal hence $I\neq M_2(\mathbb{R}).$

Am I correct?

user26857
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Eryczek
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2 Answers2

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Hint: Consider $AX$ and $XA$ where $A$ is a permutation matrix.

lhf
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Your proof is wrong because the ideal generated by one element in a non-commutative setting does not only include its left multiples and right multiples.

Kenny Lau
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    As I understand correctly, there is also sum of elements in ideal. Connecting those 2 conditions, we looking for matrix $A$ such that $A.X+X.A$ is identity matrix. Considering answer below, for matrix $A=\begin{bmatrix}0&1\1&0\end{bmatrix}$ we got identity matrix in result. – Eryczek Mar 05 '21 at 11:11