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Let $A$ be a commutative ring with $1$. Let $f_1,\dots,f_r$ be elements of $A$. Suppose $A = (f_1,\dots,f_r)$. Let $n > 1$ be an integer. Can we prove that $A = (f_1^n,\dots,f_r^n)$ without using axiom of choice?

EDIT It is easy to prove $A = (f_1^n,\dots,f_r^n)$ if we use axiom of choice. Suppose $A≠(f_1^n,…,f_r^n)$. By Zorn's lemma(which is equivalent to axiom of choice), there exists a prime ideal $P$ such that $(f_1^n,…,f_r^n) \subset P$. Since $f_i^n \in P$, $f_i \in P$ for all $i$. Hence $A = (f_1, \dots,f_n) \subset P$. This is a contradiction.

Makoto Kato
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    Can we prove it with the axiom of choice? – Chris Eagle Nov 24 '12 at 12:10
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    @ChrisEagle Suppose $A \neq (f_1^n,\dots,f_r^n)$. By Zorn's lemma(which is equivalent to axiom of choice), there exists a prime ideal $P$ such that $(f_1^n,\dots,f_r^n) \subset P$. Since $f_i^n \in P$, $f_i \in P$ for all $i$. This is a contradiction. – Makoto Kato Nov 24 '12 at 12:15
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    This question has had 5 downvotes. This is shockingly unjustified. – Georges Elencwajg Oct 01 '13 at 21:25

2 Answers2

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Since $(f_1,\ldots, f_r) = A$, there exist elements $a_i\in A$ such that $a_1f_1 + \cdots + a_rf_r = 1$. We then get that $$1 = (a_1f_1 + \cdots + a_rf_r)^{rn}.$$ Expanding the right hand side out explicitly shows that it is in the ideal $(f_1^n,\ldots, f_r^n)$. Therefore $(f_1^n,\ldots, f_r^n) = (1) = A$.

froggie
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    As simple as that, without AC or whatever. +1 – DonAntonio Nov 24 '12 at 14:46
  • The expansion may be done via http://en.wikipedia.org/wiki/Multinomial_theorem – Martin Brandenburg Nov 24 '12 at 14:52
  • @DonAntonio Yeah, but nobody had answered it until after 2 hours it was asked. – Makoto Kato Nov 24 '12 at 19:02
  • @DonAntonio But apparently you paid attention to the question and wrote "That is wrong". – Makoto Kato Nov 24 '12 at 20:26
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    I just don't see why MSE is a forum where one needs to bicker. If you have a problem with Makoto or his questions, then you should just not participate in his questions instead of telling him people have lives they need to live. – Rankeya Nov 24 '12 at 21:23
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    It's so easy to say that it's so simple(implying the question is stupid or unimportant) after one knows the answer someone else gave. Just because a proof is trivial, it does not necessarily mean that it is easy to find it, nor the proposition is unimportant. – Makoto Kato Nov 24 '12 at 22:45
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Alternatively, clearly $(f_1,\dots,f_r) \subset \sqrt{(f^{n_1}_1,\dots, f^{n_r}_r)}$. Thus, $\sqrt{(f^{n_1}_1,\dots, f^{n_r}_r)} = A$, and so $1 \in (f^{n_1}_1,\dots, f^{n_r}_r)$.

Rankeya
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  • I prefer this one. – Makoto Kato Nov 24 '12 at 18:17
  • @MakotoKato: This is certainly a much more elegant way of writing the proof, agreed. On the other hand, implicit in the "clearly" statement is that the radical of an ideal is an ideal, right? It is clear that $f_1,\ldots, f_r$ lie in the radical of $(f_1^{n_1},\ldots, f_r^{n_r})$, but how do you prove that this implies $(f_1,\ldots, f_r)$ is contained in the radical? You use an argument exactly like the one I gave (or else a prime ideal argument which uses AC). For this reason, the proofs are really no different: my argument is precisely the first inclusion. – froggie Nov 24 '12 at 19:34
  • @froggie I agree. – Makoto Kato Nov 24 '12 at 19:38