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Is it possible that $R/I$ is a field when $R$ is non-commutative ring with unit and $I$ is a maximal left ideal of $R$? If it is not, can anyone give an example of such $R$ and $I$? Thanks.

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Let $F$ be a field, let $R=F\langle x,y\rangle$ be the free (non-commutative) $F$-algebra on two generators, and let $I\subset R$ be the ideal $I=\langle x,y\rangle$ consisting of all elements with no constant term. Then obviously $I$ is maximal and $R/I\cong F$ is a field.

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You can do this artificially. Take

$$R = \mathbb{Z} \times M_{n \times n}(\mathbb{R})$$

Consider the ideal

$$I = (n \mathbb{Z}) \times M_{n \times n}(\mathbb{R})$$

Then $R/I \cong \mathbb{Z}/n\mathbb{Z}$. If $n$ was prime, we would have a field.

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  • $\begingroup$ On a less serious note: do you believe in the field with one element...? $\endgroup$ Mar 3 '14 at 13:47

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