# The going-up theorem

I am reading Introduction to Commutative Algebra / Atiyah & Macdonald, Theorem 5.11 ("Going-up theorem").

The statement is:

Let $A \subset B$ be rings, $B$ integral over $A$; let $p_1 \subset \dotsm \subset p_n$ be a chain of prime ideals of $A$ and $q_1 \subset \dotsm \subset q_m$ (m < n) a chain of prime ideals of $B$ such that $q_i \cap A = p_i$ ($1 \leq i \leq m$). Then the chain $q_1 \subset \dotsm \subset q_m$ can be extended to a chain $q_1 \subset \dotsm \subset q_n$ such that $q_i \cap A = p_i$ for $1 \leq i \leq n$.

I read the proof and I don't think the fact that $q_1$ (and hence $p_1$) is prime is used.

My question:

Does the conclusion still hold if we omit the assumption that $q_1$ (and $p_1$) is prime?

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A related question: math.stackexchange.com/questions/150295/… –  KCd May 27 '12 at 9:10

You are right, we donot need that $\mathfrak{q}_1,\ldots,\mathfrak{q}_m$ are prime. In the proof, we need $\mathfrak{p}_{i+1}$ where $i+1\geq m+1$ is prime.

For example, $m=1$, we need $\mathfrak{p}_2$ is prime, and then it follows $B/\mathfrak{q}_1$ is integral over $A/\mathfrak{p}_1$. By lying over theorem, we can find a prime ideal $\mathfrak{q}_2\subset B$ such that $\mathfrak{q}_2\supset \mathfrak{q}_1$ and lie over $\mathfrak{p}_2$.

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That the proof relies on a hypothesis doesn't really mean the conclusion is false without the hypothesis. It would be more effective to give a counterexample without the hypothesis. Take $A = {\mathbf Z}[2i] = {\mathbf Z} + {\mathbf Z}2i$ and $B = {\mathbf Z}[i] = {\mathbf Z} + {\mathbf Z}i$. For the ideal $I = 2A = {\mathbf Z}2 + {\mathbf Z}4i$ in $A$, there is no ideal $J$ in $B$ such that $J \cap A = I$. If $J$ existed, write $J = \alpha{B}$ (since $B$ is a PID). If $\alpha{B} \cap A = 2A$ then $\alpha$ is a factor of $2$ in ${\mathbf Z}[i]$, so as far as knowledge of $J$ (contd). –  KCd May 26 '12 at 20:42
is concerned we could take $\alpha$ to be 1, $1+i$, or 2, so $J$ would be $B$, $(1+i)B$, or $2B$. Check that $B \cap A = A$ and $(1+i)B \cap A = 2B \cap A = {\mathbf Z}2 + {\mathbf Z}2i$. These intersections have index 1 and 2 in $A$, but $I$ has index 4 in $A$. So $J$ does not exist. Thus the ideal $I$ in $A$ is not the contraction of any ideal in $B$. Note $I$ is not prime. No chain of ideals in $A$ that involves $I$ can be lifted to $B$ because $I$ doesn't lift by itself at all. –  KCd May 26 '12 at 20:50
The ideal $I = 2A$ in $A$ isn't prime since, for instance, $2+2i$ is in $A$, $(2+2i)^2 = 8i \in 2A$, but $2+2i \not\in 2A$. In fact, the quotient ring $A/2A$ is isomorphic to ${\mathbf F}_2[x]/(x^2)$ (think of $x$ as $2+2i$ in $A/2A$), which is not a domain. –  KCd May 26 '12 at 20:58
@KCd: I'm not sure if the example you gave really is a counter-example to dropping the single assumption that $q_1$, the first ideal in the chain of ideals in $B$, is not prime. –  user3533 May 26 '12 at 21:55
@DonAntonio: The proof makes direct use of (5.10) and (5.6) and it seems that $p_1$ and $q_1$ (the first elements in the chains) are only mentioned when (5.6) is used, which does not require primality. –  user3533 May 26 '12 at 22:09