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If $Z$ is a ring and $p$ is a prime ideal of $Z$ then $p[X]$ is a prime ideal of $Z[X]$. Is it true or false?

I believe that it is true and I try to prove it like that:

Take $f(x)\in p[X]$ and suppose that $f(x)=g(x)h(x)$ then $n=\deg f=\deg g+\deg h=m+r$. Put $g(x)=a_{m}x^{m}+...+a_{1}x+a_{0}$ and $h(x)=b_{r}x^{r}+...+b_{1}x+b_{0}$ + $a_0b_0\in p$ then either $a_{0}\in p$ or $b_{0}\in p$ (choose $a_{0}\in p$) + $a_{1}b_{0}+b_{1}a_{0}\in p$ then either $a_{1}\in p$ or $b_{0}\in p $

-If $a_{1}\in p$ then $a_{2}\in p$ because $a_{2}b_{0}+b_{2}a_{0}+a_{1}b_{1}\in p$

-If $b_{0}\in p$ then either $a_{1}\in p$ or $b_{1}\in p$,go back to the first one case. But I think that my proof is quite impossible. Could someone help me?

  • One more thing, just a little bit question I want to ask that is about the Krull dimension. If we have a chain $P_{0} \subseteq P_{1}...$ do we have $P_{0}=0$ because $0$ is also a prime ideal and it is the smallest. In particular, in PID $\dim K=1$ mean that $0\subseteq P$ with $P$ being a prime ideal
user26857
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Thế Long Lê
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  • I think you want to organize this a little differently. It may be that $a_0 \in p$ but $g \notin p[X]$ (of course one would then hope that $h \in p[X]$ and in particular $b_0 \in p$). There are many ways to do this. You could assume that $g \notin p[X]$ and try to prove $h \in p[X]$. This would focus your attention, so you wouldn't have to keep bouncing back and forth between $g$ and $h$. – Hoot Jul 02 '16 at 05:03
  • The second question means that in PID, do we have that any prime chain of Krull dimension is $0=P_{0}\subseteq P_{1}\subseteq...$ – Thế Long Lê Jul 02 '16 at 07:31
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    Your proof becomes easier if you assume that $g(x),h(x) \not\in p[X]$ and then look at the lowest $i$ such that $a_i \not\in p$ and the lowest $j$ such that $b_j \not\in p$. Now look at the coefficient of $x^{i+j}$ in $g(x)h(x)$ and argue that that is not in $p$. A slicker version of exactly this argument is given in the answer by See-Woo Lee. – Magdiragdag Jul 02 '16 at 07:33

1 Answers1

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Since $R[x]/P[x]\simeq (R/P)[x]$, we only have to show that $A$ is a domain $\Rightarrow A[x]$ is a domain. If $f,g\in A[x]$ and both are nonzero, then consider their minimal term with nonzero coefficient.

CPM
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Seewoo Lee
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  • This approach is nice. You can use McCoy's theorem to prove the contrapositive of what you indicate there. If $f(x)$ is a non-zero, zero-divisor in $A[x]$, then actually there is a non-zero element $a\in A$ such that $a\cdot f(x)=0$. This is a non-zero zero-divisor in $A$, so it would fail to be a domain. http://math.stackexchange.com/questions/83121/zero-divisor-in-rx sketches the proof in the top answer. – CPM Jul 02 '16 at 06:50
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    @CPM If $f\in A[x]$, $f\ne0$ is a zero-divisor then there is $g\in A[x]$, $g\ne0$ such that $fg=0$. We may assume $f(0)\ne0$ and $g(0)\ne 0$ (why?). Then $f(0)g(0)=0$. Why need McCoy's theorem? – user26857 Jul 02 '16 at 07:27
  • Thank you. How about the second question can you help me explain it? – Thế Long Lê Jul 02 '16 at 07:49
  • @CPM As user26857 said, I think we don't need McCoy's theorem here. If $f(x)=a_{l}x^{l}+\dots$ and $g(x)=b_{k}x^{k}+\dots$ with $a_{l}, b_{k}\neq 0$, then $f(x)g(x)=a_{l}b_{k}x^{l+k}+\dots$ which is nonzero since $a_{l}b_{k}\neq 0$. – Seewoo Lee Jul 02 '16 at 08:36
  • @ThếLongLê For the second question, you are right. For example, If $D$ is a domain with $\dim(D)=1$ then it is equivalent to say that every nonzero prime ideal of $D$ is maximal ideal. If $\dim(D)=0$ and $D$ is domain, the only candidate is field. But $D=\mathbb{Z}/4\mathbb{Z}$ also has dimension 0. – Seewoo Lee Jul 02 '16 at 08:38
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    I know McCoy's theorem is overkill, but it is interesting that for any zero-divisor in the polynomial ring, the whole polynomial is actually killed by a single element from the ground ring. Of course this element kills the coefficients as well which is all you need for this question and can be proven more easily, but why not say when something stronger is actually true? I remember finding this result surprising and not as well known – CPM Jul 02 '16 at 12:28
  • @CPM Oh, I see. I understand your purpose. – Seewoo Lee Jul 02 '16 at 12:30