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Let $R$ be a commutative ring. If $I\leq R$ is an ideal s.t. $\sqrt I=m$ is maximal, then $I$ is primary.

Proof

Since $m$ maximal, $Nil(R/I)=m/I=:n$ is maximal.

Q1) Why $m/I$ is maximal in $R/I$ when $m$ is maximal in $R$ ? I have problem to prove this fact.

Hence, every prime ideal contain $n$.

Q2) Why is $p$ is prime in $R/I$, then $p\supset n$ ?

Watson
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user386627
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  • For Q1, there should be a theorem in your book, probably right when they first defined quotient rings, that there is a one-to-one correspondence between ideals in a ring $R$ that contains an ideal $I$, and ideals of $R/I$. – Arthur Jan 29 '17 at 10:58
  • See also http://math.stackexchange.com/questions/649146 – Watson Jan 29 '17 at 11:07

1 Answers1

2

1)

The quotient of $R/I$ by $m/I$ is $R/m$ by the third isomorphism theorem. It is a field, so that $m/I$ is a maximal ideal of $R/I$.

Alternatively, any ideal of $R/I$ is of the form $J/I$, where $J$ is an ideal of $R$ containing $I$. So if $J/I$ contains $m/I$, can you prove that $J/I=m/I$ or $J/I=R/I$?

2)

You know that $$\mathrm{Nil}(R/I) = \{x \in R/I \mid \exists r>0, x^r=0\}.$$ So proving that any prime of $R/I$ contains $n$ is equivalent to proving that $$n \subset \bigcap_{p \; \text{prime}} p = \mathrm{Nil}(R/I).$$

Assume that $x \in n = \sqrt I / I$. Then $x^r = 0$ for some $r>0$, since $x=[y]_{I}$ for some $y \in \sqrt I$. This shows that $x \in \mathrm{Nil}(R/I)$, hence $n \subset p$ for any prime ideal $p$ of $R/I$.

Therefore $R/I$ has only $n=m/I$ as prime ideal. If $x \in R/I$ is a zero divisor, then it is not a unit, then it belongs to $n$ (because any proper ideal, for instance $(x) \neq R$, is contained in a maximal ideal — but here $n$ is the only maximal ideal of $R/I$). Then $x=[y]_I$ where $y \in \sqrt I$, which implies that $x$ is nilpotent.

Hence, $I$ is $m$-primary.

Watson
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