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I am trying to prove

Every prime ideal in $\mathbb{Z}[x]$ can be generated by at most two elements.

Since I have not seen any prime ideal generated by 3 elements, this statement seems true to me. But I cannot find a way to prove it. Thanks for any help in advance!

user26857
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Lev Bahn
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    Hint: for any prime $p \in \mathbb{Z}$, $\mathbb{Z}/p\mathbb{Z}$ is a field; hence, $(\mathbb{Z}/p\mathbb{Z})[X]$ is a PID. Likewise, $\mathbb{Q}[X]$ is also a PID. Take any prime ideal $P \subset \mathbb{Z}[X]$, and consider $Q := P \cap \mathbb{Z}$. Since $Q$ is a prime ideal of $\mathbb{Z}$, it is generated by either $0$ or a prime number $p$. If $Q = p\mathbb{Z}$, try to use the cited fact about about $(\mathbb{Z}/p\mathbb{Z})[X]$. If $Q = {0}$, try the same with $\mathbb{Q}[X]$. – Alex Wertheim Aug 08 '19 at 20:49

1 Answers1

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(This is more "things that work in $\mathbb{Z}[x]$" more than "general algebraic methods", but it works.)

Suppose you have three (or more) generators. Then either two of them are constants or two of them are polynomials (of positive degree). With what can you replace these two generators? (Hint: Same answer for both cases.)

Eric Towers
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  • Thank you for the answer! Could I ask why either two of them are constant or two of them are polynomials, – Lev Bahn Aug 08 '19 at 21:39
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    @LevBan Another way to look at a constant is as a polynomial of degree $0$. The degree of a generator is a non-negative integer. So either two generators have degree $0$ or two generators have positive degree. – Robert Shore Aug 08 '19 at 22:28
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    @LevBan : Pigeonhole principle. There are two pigeonholes: constants and polynomials. Every element of $\mathbb{Z}[x]$ is one or the other. – Eric Towers Aug 08 '19 at 22:28
  • @EricTowers Thank you so much! So in the case when there are two constants, we can replace it with the gcd of them. As for when we have two nonconstant polynomials we mod the ideal out by appropriate prime ideal as the comment of Alex then we get the PID so we can reduce the ideal generated by two polynomials into the ideal generated by one polynomial. Is it correct so far? – Lev Bahn Aug 09 '19 at 03:18
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    @LevBan : In the two polynomial case, we can also replace them with their (polynomial) gcd. (Which is what you said, but perhaps obscuring that this parallels the two constants case.) – Eric Towers Aug 09 '19 at 05:55
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    @EricTowers My concern is if there is $gcd$ of two polynomials. Since $\mathbb{Z}[x]$ is not a Euclidean domain, it is not necessary to have gcd of two arbitrary polynomials over $\mathbb{Z}$. So in my solution in the comment, I actually don't know how to bring the gcd in $\mathbb{Z}/p\mathbb{Z}[x]$ to $\mathbb{Z}[x]$. :( Could you give me more clarification? – Lev Bahn Aug 09 '19 at 13:55
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    @LevBan : Being a ED is too strong. Every ideal in an ED is principal. However, you should know how to compute polynomial GCDs in $\mathbb{Z}[x]$, since this application of Euclid's algorithm is something taught in high school. It is enough that $\mathbb{Z}[x]$ is a UFD and we have a GCD algorithm in the coefficients. – Eric Towers Aug 09 '19 at 16:27
  • @EricTowers Could you help me out to remove my confusion? So let's consider $(3x+1,x+1,3)$. As you suggested, I choose $3x+1$ and $x+1$. Their gcd should be $1$ by the property of UFD. By the Euclidean Algorithm, we have $3x+1=3(x+1)-2$ which terminates the algorithm immediately... So using Euclidean algorithm here does not give good way to compute the gcd. – Lev Bahn Aug 09 '19 at 19:44
  • @EricTowers Actually, $(3x+1,x+1,3)=\mathbb{Z}[x]$ because $2=3(x+1)-(3x+1)$ so $\mathbb{Z}[x]=(1)=(2,3)\subseteq (3x+1,x+1,3)$. But if we conclude that the gcd of $x+1$ and $3x+1$ is $1$, without using $3$, we immediately get that the ideal is $\mathbb{Z}[x]$. So it seems we need more argument – Lev Bahn Aug 09 '19 at 19:48
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    @EricTowers Also, we have not used the property of prime ideal here... – Lev Bahn Aug 09 '19 at 19:48
  • @LevBan : You realize the polynomial GCD algorithm stops as soon as you get a non-polynomial (i.e., not when you get the remainder zero, but when you get a remainder of degree zero -- because you've just shown that the two have no positive degree common divisor). – Eric Towers Aug 09 '19 at 23:11
  • This answer doesn't look right: it seems that it works for any ideal of $\mathbb Z[x]$, and the ideals of $\mathbb Z[x]$ can have arbitrary numbers of generators. – user26857 Dec 30 '21 at 04:55
  • Btw, there is no polynomial GCD algorithm in $\mathbb Z[x]$. – user26857 Dec 30 '21 at 05:05
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    Last but not least, I have no idea why the OP, who raised some serious doubts, decided to accept this answer. – user26857 Dec 30 '21 at 05:24
  • @user26857 : There may be a GCD algorithm in any GCD domain. $\Bbb{Z}$ is a UFD, so $\Bbb{Z}[x]$ is a UFD. All UFDs are GCD domains. This shows there is a GCD and LCM in $\Bbb{Z}[x]$. The algorithm generating primitive pseudo-remainder sequences computes such GCDs. It is not the only such algorithm. – Eric Towers Dec 30 '21 at 20:16
  • This is not the main question here (and I still believe that the "algorithm" you invoke is useless in this case; more useful could be to consider the polynomials in $\mathbb Q[x]$ and clear the denominators then). The problem is that your answer fails to produce two generators as in the trivial case of the ideal $(2,x,x+2)$. Moreover, even if your process would be correct (but once again, it is not!) how do you know that every prime ideal has a finite number of generators? – user26857 Dec 30 '21 at 20:25
  • @user26857 : It is not required that it produce two generators for the ideal you write. It is only required that it produce at most two generators for every prime ideal. Further, this Answer is responsive to the Question, not, as you seem to expect, to the quotation in the Question -- it constitutes the key insight I had in solving this problem largely the same way as in the Answer you reference in your comment to the Question. – Eric Towers Dec 30 '21 at 20:39
  • You can't produce two or at most two generators for every prime ideal in the way you suggested. In the example I mentioned I introduced a fake generator and your process fails to produce two or less generators. That's why I said that your suggestion doesn't work and consequently the answer is wrong. I understand that you want to produce a constructive solution, but if there is one then it must take care of all situations. – user26857 Dec 30 '21 at 21:02
  • @user26857 : Step 1 of Euclidean reduction of the pair $(x+2,x)$ is $(x+2-x, x) = (2,x)$, and the algorithm halts as reminded in comments above. Replacing that pair in your ideal yields $\langle 2, x, 2 \rangle = \langle 2,x \rangle$. After denying the existence of the polynomial GCD algorithm, your refusal to understand it is wasting both our times. Based on your history at MSE, you might benefit from this answer re-phrased in terms of S-polynomial reduction as used in Buchberger's algorithm, but such a rephrase does not improve the answer given OP's apparent context. – Eric Towers Dec 30 '21 at 21:13
  • If everything is so clear to you, then why don't try to elaborate in such a way that every aspect is covered and finally give a full proof? (For instance, I can't see the Euclidean reduction mentioned so far.) Posting a truncated answer, based on question marks and vague hints, is actually of waste of everyone's time. – user26857 Dec 30 '21 at 22:08
  • "Why [I] don't" is: (1) OP agrees this Answer resolved his Question, (2) the Question requests "Help", not a full proof, and (3) because the full proof exists elsewhere on MSE. A Question requesting a full proof would have been a duplicate of the Question you reference in your comments to this Question, so this Question would have been closed as a duplicate. However, this Question does not request a full proof, so your comments do not constitute an improvement to this Answer. – Eric Towers Dec 30 '21 at 22:15
  • This recent thread is the living proof that your answer needs an improvement. – user26857 Dec 30 '21 at 22:15
  • @user26857 : That thread asks for a generalization outside the scope of the Question here. This answer warns that it is responsive the Question that is actually asked here, not to a more general Question. Your thread shows that this Answer would not answer that Question, but it is not offered as an answer to that Question and is clearly marked to not be a response to the generalization in that Question. – Eric Towers Dec 30 '21 at 22:20
  • The other full proof is quite different than yours. Your approach is taking a different route and for some reasons you still prefer to write tones of comments instead of clarifying your answer. – user26857 Dec 30 '21 at 22:21
  • Come one! What generalization? Everything works similarly for PIDs. You simply refuse to improve your answer. – user26857 Dec 30 '21 at 22:23
  • You seem to wish that this Answer responded to a different Question or that the Question that was asked were different. I have patiently explained that both wishes are wrong. Your comments do not constitute an improvement to this Answer to this Question. – Eric Towers Dec 30 '21 at 22:23
  • I have the feeling that we are living in parallel worlds. Let's read what this question asks: "I am trying to prove Every prime ideal in $\mathbb{Z}[x]$ can be generated by at most two elements.". Did your answer prove this? I don't think so! – user26857 Dec 30 '21 at 22:28