If $R$ is a commutative ring with unit and $p$ is a prime number, then is $pR$ a maximal ideal of $R$? If not what conditions should I impose on $R$?
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$\begingroup$ Clearly not if $R$ is a field for instance. $\endgroup$– lhfCommented Aug 25, 2015 at 13:21
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$\begingroup$ @lhf Why is this true? $\endgroup$– HAHACommented Aug 25, 2015 at 13:22
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$\begingroup$ Zero is the only maximal ideal in fields. $\endgroup$– lhfCommented Aug 25, 2015 at 13:22
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$\begingroup$ @lhf Ah and zero is not a prime, correct? $\endgroup$– HAHACommented Aug 25, 2015 at 13:23
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$\begingroup$ @HAHA zero is prime if R is a domain. $\endgroup$– LolmanCommented Aug 25, 2015 at 14:29
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3 Answers
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A silly example : Consider the field of real numbers, $\mathbb{R}$. Clearly $p\mathbb{R}=\mathbb{R}$, so $\mathbb{R}/p\mathbb{R} \cong \lbrace 0\rbrace$. More interesting: Let $R=\mathbb{Z}[x]$. $\mathbb{Z}[x]/p\mathbb{Z}[x] \cong (\mathbb{Z}/p\mathbb{Z})[x]$ which is not a field!
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$\begingroup$ I fixed a typo. Hope you don't mind! $\endgroup$ Commented Aug 25, 2015 at 14:42
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There are also plenty of counterexamples apart from fields. For example $p$ could be a unit even if $R$ is not a field, e.g. in $\mathbf Z [\frac{1}{p}]$, or $p$ could fail to be irreducible, e.g in $\mathbf Z[i]$ we have $(1+i)(1−i)=2$.
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A prime $p$ is not prime in $\mathbb Z[\sqrt p]$ because $(\sqrt p)^2 \in p\mathbb Z[\sqrt p]$ but $\sqrt p \notin p\mathbb Z[\sqrt p]$. Thus $p\mathbb Z[\sqrt p]$ is not a prime ideal of $\mathbb Z[\sqrt p]$.
Therefore, no prime remains prime in the ring of all algebraic integers.
