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This is regarding Example 1.21 in the textbook $\textit{Algorithmic Methods in Non-Commutative Algebra,J.Bueso, J. Gomez-Torrecillas, A.Verschoren}$. It says: Let us denote by $C = \Bbbk[x]$, where $\Bbbk$ is an algebraically closed field and by $\mathfrak{m}$ the maximal ideal $\langle x \rangle$ of $C$. Let $R = \left[\begin{array}{l}C&\mathfrak{m}\\\mathfrak{m}&C\end{array}\right]$, i.e the subring of $M_2(C)$ consisting of all matrices whose off-diagonal entries belong to $\mathfrak{m}$ . It is fairly easy to see that there are exactly two surjective $\Bbbk$-maps $R \rightarrow \Bbbk$ corresponding to taking the quotient of $R$ by the maximal ideals $M_+ = \left[\begin{array}{l}\mathfrak{m}&\mathfrak{m}\\\mathfrak{m}&C\end{array}\right]$ resp. $M_{-} = \left[\begin{array}{l}C&\mathfrak{m}\\\mathfrak{m}&\mathfrak{m}\end{array}\right]$.

I am not able to see how we can show that there are the only two surjective maps and how these ideals are obtained.

Zoey
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  • While I don't know how to show there are exactly two surjective $K$-map, I think the two ideals are the kernels of the two maps respectively and their maximality follows from $K$ being a field. – Alex Vong May 18 '17 at 15:02
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    I have a problem. Is $M_+$ really an ideal? Take $K = \mathbb{C}, \begin{pmatrix} x & x \ x & 1 \end{pmatrix} \in M_+, \begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix} \in C$. Then $\begin{pmatrix} x & x \ x & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \ 1 & 1 \end{pmatrix} = \begin{pmatrix} 2x & 2x \ x + 1 & x + 1 \end{pmatrix} \notin M_+$. – Alex Vong May 18 '17 at 15:39
  • But $ \left[\begin{array}{l}1&1\1&1\end{array}\right] \notin \left[\begin{array}{l}C&\mathfrak{m}\\mathfrak{m}&C\end{array}\right] (=R)$ – Zoey May 19 '17 at 07:13
  • Let $R$ be (noncommutative) ring and $I$ be ideal of $R$. Isn't it true that $ar \in I$ and $ra \in I$ for any $a \in I$ and $r \in R$? – Alex Vong May 19 '17 at 14:09
  • yes it is. but the general ring element should come from $R$ not $M_2(C)$. – Zoey May 19 '17 at 14:19
  • I understand it now, thank you. – Alex Vong May 19 '17 at 14:22

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