In Attiyah commutative algebra page 71, it is given some equivalent definitions of Jacobson ring. One of the definitions are that every prime ideal which is not maximal is equal to the intersection of prime ideals which contains it strictly. How is this compatible with the prime avoidance lemma? I.e if a prime ideal is equal to the intersection of a family of ideals then one of the ideals is the given prime ideal.

is every prime ideal in a Jacobson ring maximal?

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    $\begingroup$ Don’t have a counterexample right now, but the prime avoidance lemma affects finite intersections only. $\endgroup$ – k.stm Apr 3 '15 at 20:09
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    $\begingroup$ @k.stm, harajm: the prime avoidance lemma has absolutely nothing to do with intersections, finite or not. Also, there is no book by "Attiyah"but Atiyah and Macdonald have written a splendid introduction to commutative algebra. $\endgroup$ – Georges Elencwajg Apr 3 '15 at 20:13
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    $\begingroup$ @GeorgesElencwajg Prp. 1.11 in Atiyah–MacDonald (on p. 8) covers both the prime avoidance lemma and a similar statement roughly saying: »$\mathfrak p \supseteq \mathfrak a_1 ∩ … ∩ \mathfrak a_n ⇒ \mathfrak p \supseteq \mathfrak a_i$ for some $i$« with equality implying equality. I think this is what harajm is refering to. $\endgroup$ – k.stm Apr 3 '15 at 20:17

In a Jacobson ring $R$ it is not true that every prime ideal is maximal.

For example, a field $k$ is Jacobson. By a general version of the Nullstellensatz (see Eisenbud, Commutative algebra, Thm. 4.19), $k[x,y]$ is also Jacobson. But we know there are many non-maximal prime ideals in $k[x,y]$, for example $(x)$.

With respect to the question about the "prime avoidance lemma" (Prop. 1.11(ii) in Atiyah-Macdonald), k.stm is correct in saying that the prime avoidance lemma is only about finite intersections. In the example above, $(x) = \bigcap (x,p(y))$ where $p(y)$ run over polynomials in $y$. If $k$ is algebraically closed, you can even say the $p(y)$ run over linear polynomials. But regardless, you will not be able to write $(x) = (x,p(y))$ for $p(y)$ as above.


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