Going down theorem

Question

Here $A$ is a commutative ring with unity.

How to show that going down theorem holds for $A$ contained in $A[x]$, the polynomial ring.

Lying over is ok. I cannot do the other part.

What have you tried? Have you looked at a commutative algebra text like Atiyah Macdonald (where this is either proved or an exercise, I bet)? — davidlowryduda, Apr 15 '13 at 13:21

score 1 · Answer 1 · answered Apr 15 '13 at 13:51

1

Going down holds for arbitrary flat extensions, a proof can be found in every book on commutative algebra, or in the Stacks project Lemma 9.36.17, or at Wikipedia.

But for $A \to A[x]$ it's almost trivial: We have $\mathfrak{p} \subseteq \mathfrak{p}' \subseteq A$ and $\mathfrak{q}' \subseteq A[x]$ lying over $\mathfrak{p}'$. Then $\mathfrak{q}:=\mathfrak{p}[x]$ is a prime ideal lying over $\mathfrak{p}$ satisfying $\mathfrak{q} \subseteq \mathfrak{q}'$

answered Apr 15 '13 at 13:51

Martin Brandenburg

163,620

Why is q contained in q(prime)? – kushal Apr 15 '13 at 14:07
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Try to prove it yourself. – Martin Brandenburg Apr 15 '13 at 14:41

Going down theorem

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