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In a commutative ring with an identity, every maximal ideal is a prime ideal. However, if a commutative ring does not have an identity, I'm not sure this is true. I would like to know the counterexamples, if any. The more examples, the better.

EDIT I would like to know the counterexamples other than $2\mathbb{Z}$. The more examples, the better.

EDIT I also would like to know the counterexamples that are not given in the Arturo Magidin's answer if any, namely an example of a non-prime maximal ideal which does not contain $R^2$.

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    $\begingroup$ Let $R$ be the ring and $I$ be such a maximal ideal. Write $S=R/I$. I claim that if $S$ is finite, then $S$ has a trivial multiplication table, i.e. $ab=0$ for all $a,b\in S$. There exists an element $x\in S\setminus\{0\}$ such that multiplication by $x$ is not injective; then, $xS=0$ the only proper ideal. However, the set $N=\{x\in S:xS=0\}$ is an ideal, whence $N=S$. $\endgroup$ Jul 25, 2013 at 23:14
  • $\begingroup$ @YACP I'm asking more examples. What's wrong with this question? $\endgroup$ Jul 26, 2013 at 10:43
  • $\begingroup$ @MakotoKato As YACP points out, the other question has a general theorem, not only the $2\mathbb Z$ example. $\endgroup$
    – martini
    Jul 26, 2013 at 10:50
  • $\begingroup$ @YACP I added new edit. $\endgroup$ Jul 26, 2013 at 11:06
  • $\begingroup$ @MakotoKato What do you mean by Arturo's condition? The only time he uses the word "condition" explicitly is when he's talking about prime ideals, and I really doubt you are asking us to ignore those. Are you talking about something else? Please just spell it out in the post. Thanks! $\endgroup$
    – rschwieb
    Jul 26, 2013 at 13:41

6 Answers 6

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You won't find any examples of maximal, non-prime ideals other than those given in Arturo Magidin's lovely answer. I won't even assume commutativity. And I freely admit that this is basically the same argument as in Arturo's answer!

Claim: If $R$ is a rng with a maximal ideal $M$ that is not prime, then $R^2 \subseteq M$.

Proof: Let $M$ be such an ideal, and suppose that $A,B$ are ideals of $R$ not contained in $M$ such that $AB \subseteq M$. By maximality of $M$ we have $M + A = R = M + B$. It follows that $$\begin{align*} R^2 &= (M+A)(M+B) \\ &= M^2 + AM + MB + AB \\ &\subseteq M, \end{align*}$$ since $M^2,AM,MB,AB \subseteq M$. QED

Combining this with Arturo's theorem, we have:

Corollary: Let $R$ be a rng with maximal ideal $M$. Then $M$ is not prime if and only if $R^2 \subseteq M$.

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  • $\begingroup$ yes, very clear $\endgroup$ Dec 22, 2013 at 14:17
  • $\begingroup$ @MakotoKato why not accept this answer? $\endgroup$
    – Arrow
    Aug 3, 2017 at 13:47
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Take $R=2\mathbb{Z}$, the ring of even integers. The ideal $4\mathbb{Z}\subset R$ is maximal (the only larger ideal is $R$ itself), but not prime, as $2\cdot 2\in 4\mathbb{Z}$, but $2\not\in 4\mathbb{Z}$.

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There are lots of parodies on the $2\Bbb Z$ example.

You can look at $R=2\Bbb Z[x]$: the ideal $(2x,4)=I$ is maximal, and again $2^2\in I$ and $2\notin I$.

Here's another: let $M$ be any $2\Bbb Z$ module. You could, for example, let $M=2\Bbb Z$, or any number of copies of $2\Bbb Z$. Look at the following ring of matrices and ideal:

$$ R=\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\mid a\in 2\Bbb Z, b\in M\right\} $$

$$ I=\left\{\begin{bmatrix}a&b\\0&a\end{bmatrix}\mid a\in 4\Bbb Z, b\in M\right\} $$

Again, $\begin{bmatrix}2&0\\0&2\end{bmatrix}^2$ shows $I$ isn't prime.

Finally to get a noncommutative example, try

$$ R=\begin{bmatrix}2\Bbb Z&2\Bbb Z\\0&2\Bbb Z\end{bmatrix} $$

$$ I=\begin{bmatrix}2\Bbb Z&2\Bbb Z\\0&4\Bbb Z\end{bmatrix} $$ $$ J=\begin{bmatrix}4\Bbb Z&2\Bbb Z\\0&2\Bbb Z\end{bmatrix} $$

$I$ is a maximal ideal of $R$, $J\nsubseteq I$, but $J^2\subseteq I$, so $I$ isn't prime.

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I will prove the following fact. Let $R$ be a commutative ring without an identity. Let $M$ be a non-prime maximal ideal. Then $R/M$ has a prime order.

Proof Let $S = R/M$. Since $R^2 \subset M$ by the Manny Reyes' answer, $S^2 = 0$. Hence every subgroup of the additive group $S$ is an ideal of $S$. Since $S$ has no non-trivial ideals, the order of $S$ must be a prime number. QED

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The following example is a generalization of the example $2\mathbb{Z}$. Let $p$ be a prime number. Let $R = p\mathbb{Z}$. Then $R^2 = p^2\mathbb{Z}$ is a maximal ideal but not a prime ideal of $R$.

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Let $R$ be an abelian group of prime order. We define multiplication on $R$ by $ab = 0$ for all $a, b \in R$. Then $0$ is a maximal ideal but not a prime ideal.

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