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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.

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    Can we prove it *with* the axiom of choice?2012-11-24
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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.2012-11-24
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    This question has had 5 downvotes. This is shockingly unjustified.2013-10-01

2 Answers 2

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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 $1 \in (f^{n_1}_1,\dots, f^{n_r}_r)$.

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    I prefer this one.2012-11-24
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    @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.2012-11-24
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    @froggie I agree.2012-11-24
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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 $(f_1^n,\ldots, f_r^n) = (1) = A$.

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    As simple as that, without AC or whatever. +12012-11-24
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    The expansion may be done via http://en.wikipedia.org/wiki/Multinomial_theorem2012-11-24
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    @DonAntonio Yeah, but nobody had answered it until after 2 hours it was asked.2012-11-24
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    @DonAntonio But apparently you paid attention to the question and wrote "That is wrong".2012-11-24
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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.2012-11-24
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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.2012-11-24