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From Fraleigh's "A First course in Algebra"-book: Theorem 27.9 says

Let R be a commutative ring with unity. Then M is a maximal ideal of R if and only if $R/M$ is a field.

And there's this (probably silly and elementary) detail in the proof I don't understand, Fraleigh says:

Suppose M is a maximal ideal in R. Observe that if R is a commutative ring with unity, then $R/M$ is also a nonzero commutative ring with unity if $M\neq R$...

How do we immediately know that $R/M$ is a nonzero commutative ring with unity here?

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    $\begingroup$ Nonzero comes from $M\neq R$. Ring with unity comes from standard properties of the quotient that the author probably explained earlier in the book $\endgroup$ – Max May 11 '18 at 13:35
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by definition of any maximal ideal $M \subseteq R$, you have $M \neq R$, hence the quotient is not zero. Now observe that for an ideal $M$ in a commutative ring $R$ with unit the quotient $R/M =\{x+M:x\in M \} $ is itself a commutative ring with unit $1+M$.
That the quotient $R/M$ is, in fact, a field can be also seen by the following observation: There is a bijective correspondence between ideals $I$ of $R$ containing an ideal $J \subseteq R$ and the ideals of $R/J$. Hence if you choose $J$ to be maximal, $R/J$ has only trivial ideals (the factor ring itself and the zero ideal). A commutative nonzero ring with unit is a field if and only if it contains only these two ideals.

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