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So I know that the "intersection of two prime ideals being a prime ideal" is false. There are some simple examples to disprove that. But does that mean that the intersection of two prime ideals has no possibility of being a plain ideal? Can it result in a different type of ideal?

user338113
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3 Answers3

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For the ring $\mathbb{Z}$, consider the prime ideals $(2)$ and $(3)$. Their intersection is $(6)$, which is a non-prime ideal.

The intuition is that intersection of ideals corresponds roughly to multiplication of elements. That is why the intersection of prime ideals is typically not prime.

vadim123
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  • oh okay, so even though the intersection of (2) and (3) is (6) which is non-prime, its still an ideal – user338113 May 09 '16 at 16:29
  • Not to nitpick, but: the intersection of two prime ideals can be prime, e.g. with $(2)$ and $(0)$ in $\mathbb{Z}$. – Alex Wertheim May 09 '16 at 16:30
  • But wouldn't (2) and (0) result in (0)? Which isn't a prime? or would it be (2), because then I understand that it's prime – user338113 May 09 '16 at 16:33
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    @user338113: $(0)$ is a prime ideal of $\mathbb{Z}$, and indeed of any domain, since if $xy = 0$ then one of $x, y$ must be zero. Another way of putting this is: an ideal $I$ of a commutative ring $R$ is prime if and only if the quotient $R/I$ is an integral domain, and $R/(0) \cong R$. – Alex Wertheim May 09 '16 at 16:35
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The intersection of two ideals $I$ and $J$ is always an ideal, regardless of whether $I$ and $J$ are prime or not. If $x,y\in I\cap J$, then $x+y\in I$ and $x+y\in J$ since $I$ and $J$ are ideals, hence $x+y\in I\cap J$.

Similarly, if $x\in I\cap J$ then $rx\in I$ and $rx\in J$ since $I$ and $J$ are ideals, hence $rx\in I\cap J$. If $R$ is commutative then this is enough to show that $I\cap J$ is an ideal, and if $R$ isn't commutative then use the same argument for right multiplication. (For non-commutative rings I'm assuming that ideal means two-sided ideal).

carmichael561
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The intersection of two ideals $I$ and $J$ is the kernel of the homomorphism $R \to R/I \times R/J$ given by $x \mapsto (x \bmod I, x \bmod J)$ and so is an ideal.

If $I$ and $J$ are both prime ideals, then $R/I \times R/J$ is a product of two domains, which is never a domain except in trivial cases. So, it is unlikely that $R/(I \cap J)$ is a domain and $I \cap J$ is a prime ideal.

lhf
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