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Let K be a field, I an infinite indexed set and A be the product ring $K^I$. For every $p\in Spec A$, prove that the localization $A_p$ is a field. In particular, $p \in Spm A$.

This question was asked in my end term exam on module theory and I couldn't solve it, I tried it again but in vain.

$A_p = ({A/p})^{-1} A$, Clearly , all element of A are invertible. All elements of A/p are invertible in A but how do I prove that it is invertible in A/p? Let $x \in A/p $ is not invertible in A/p . So, it means that there exists $b \in p$ such that ab=1. How exactly can I find a contradiction here?

I also have to prove that $p\in Spm A$, ie p is a maximal ideal. Let it not be a maximal ideal , there exists an ideal Q such that $P\subset Q \subset A$. But I am not able to proceed foreward and I need help.

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    It's not true that every element in $A^{I}$ is invertible. For instance, if $i_0 \in I$ and $x$ is the element $x = (a_i){i \in I}$ where $a_i = 0$ if $i \ne i{0}$ but $a_{i_0} = 1$, then $x$ has no inverse in $A^I$ – Geoff Dec 20 '21 at 16:26
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    (1) Where you wrote $A/p$, you presumably meant $A\setminus p$ (the set-theoretic difference, not the quotient ring). (2) You wrote "there exists $b\in p$ such that $ab=1$. How exactly can I find a contradiction here?" You have an invertible element ($b$) in a prime ideal; surely that contradicts a theorem (or remark) that you know. – Andreas Blass Dec 20 '21 at 22:29
  • @AndreasBlass Thank you very much for your comment! –  Jun 07 '22 at 11:00
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    Apart from superficial wording in the title, this question has little in common with the question with which it is associated. Please look beyond the title before voting to close. e.g. If someone asks for a proof that $2^p-1$ prime implies $p$ prime, then it is not helpful to direct them to a proof that prime is equivalent to irreducible in the integers. – tkf Jul 10 '22 at 11:59
  • @tkf Not true. It is trivial that fields are (von Neumann) regular, and that regular rings are closed under products, so the sought result is a special case of the simple proof given in this answer in the linked dupe - which characterizes such rings. – Bill Dubuque Jul 10 '22 at 15:11
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    @BillDubuque I think the term 'trivial' is at least subjective. Someone asking this question need not know what a (von Neumann) regular ring is let alone the two 'trivial' results needed to solve this question from the associated one. In any case that is just one solution to this question, which also invites others. Referring to my earlier example, once you know the definition of prime number, it is fairly easy to show $2^p-1$ prime implies $p$ prime. However I would not consider the two questions to be duplicates. – tkf Jul 10 '22 at 17:05
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    @BillDubuque By the way, if you expand your comment to an answer, I for one would certainly upvote it. I do not think the OP fully follows my answer yet, but they may follow yours if you explain it. – tkf Jul 10 '22 at 17:06
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    @tkf I follow your answer. Your answer is great. –  Jul 10 '22 at 18:36
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    @Bill I have done only 1 course course in commutative algebra and you expect me to learn Von Neumann (regular) and regular rings. I am sorry these terms had nothing to do with my course. –  Jul 10 '22 at 18:40
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    @Avenger The only thing you need to know about them is the simple definition (and said trivial consequences). You can get this from the first few paragraphs of the Wikipedia page. This is easily accessible to anyone studying a first course in commutative algebra. If perchance anything remains unclear then simply pose another question. – Bill Dubuque Jul 10 '22 at 18:50
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    @Bill This answer of OP is a great tool for this type of questions and was very informative for me and I am thankful to OP for it. I am sorry that I cannot talk to you much on this because I remain chronically ill. I am sorry. –  Jul 10 '22 at 18:52

1 Answers1

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Denote $R=k^I$ (I used the letter $A$ a lot in my answer for subsets of $I$, so should not use it to denote the ring).

Consider the elements $e_A\in k^I$ for $A\subseteq I$, which have $i$'th co-ordinate $1$ if $i\in A$ and all other co-ordinates $0$. Let $\mathcal{V}_p$ denote the set of subsets $A$, with $e_A\in p$, where $p$ is a prime ideal of $R$.

Clearly $\mathcal{V}_p$ determines $p$ - the elements of $p$ are precisely those whose support is an element of $\mathcal{V}_p$.

Then $\mathcal{V}_p$ is closed under taking subsets (as $p$ closed under multiplication by elements of $k^I$), does not contain $I$ (as $p$ a proper ideal) and is closed under finite unions ($e_{A\cup B}=e_A+e_{B\backslash A}e_B)$.

Further if $I=A\sqcup B$ (with $A\cap B=\emptyset$) then $e_Ae_B=0\in p$ so $p$ prime implies $e_A\in \mathcal p$ or $e_B\in \mathcal p$.

It is this property that tells us that $p$ is maximal. If $p\subsetneq p'$, then some $x\in p'$ has support $B\notin \mathcal{V}_p$. Then $e_A\in p$ where $A$ is the complement of $B$, by the above property. But then $e_A+e_B=1\in p'$ so $p$ is maximal.

The properties we have listed for $\mathcal{V}_p$ make its complement (which also equals the set of complements of its elements) an ultrafilter on $I$. Conversely, given any ultrafilter on $I$, we obtain a prime $p$ by taking the set $\mathcal{V}$ of complements of the ultrafilter, and letting $p$ consist of all elements whose support lies in $\mathcal{V}$.

I first learnt all this in a course on Ramsey theory! It is amazing how interconnected mathematics is.

Anyway to answer your question: We will show that $R_p\cong R/p$. We have already shown that $R/p$ is a field, as $p$ is maximal.

For $x\in P$ we have $\frac x1=\frac01$, as $(1x-0\cdot1)e_B=0$, where $B$ is the complement of the support of $x$, so $e_B\notin p$.

Thus we have a well defined map $\theta\colon R/p\to R_p$ sending $$[x]\mapsto \frac x1.$$

Suppose $\theta([x])=\frac01$. Then $xy=0$ for some $y\notin p$. Thus $x\in p$, so $\theta$ is injective.

Now consider arbitrary $\frac xy\in R_p$. As $y\notin p$ we know the support of $y$ is not in $\mathcal{V}_p$, so the complement $A$ of the support of $Y$ does lie in $\mathcal{V}_p$. Thus $e_A\in p$ and $y+e_A$ is invertible in $R$. It remains to show: $$\theta([x(y+e_A)^{-1}])=\frac xy.$$

As $A$ is the complement of the support of $y$, we have: \begin{eqnarray*} &\,\,&ye_A=0\\ &\implies& xy^2=x(y+e_a)y\\ &\implies& xy^2(y+e_a)^{-1}=xy\\ &\implies& (xy(y+e_a)^{-1} -x)y=0\\ &\implies& \frac{x(y+e_a)^{-1}}1=\frac xy. \end{eqnarray*} Thus we can conclude that $\theta$ is surjective as well as injective, hence the desired isomorphism.

tkf
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  • Can you please tell in line 5th of your answer , what you mean by "support" ? –  Jun 07 '22 at 11:08
  • An element $x\in k^I$ assigns to each element of $a\in I$ a co-ordinate $x_a\in k$. e.g. a vector $v\in \mathbb{R}^3$ has co-ordinates $v_1,v_2,v_3\in \mathbb{R}$. The set of $a$ such that $x_a\neq 0$ is called the $support$ of $x$. – tkf Jun 07 '22 at 11:36
  • In the 6 th para of your answer, shouldn't $e_A \in V_p$ if property in 5th paragraph is to be used. Still , I am not able to deduce how you wrote : then $e_A +e_B=1\in p'$ and how does that implies that p is maximal? –  Jul 10 '22 at 09:42
  • You are right - in para 5 I wrote $\mathcal{V}_p$ when I meant $p$. This is now corrected. The statement in para 5 is just the definition of prime ideal: if $e_Ae_B\in p$ then $e_A\in p$ or $e_B\in p$. – tkf Jul 10 '22 at 12:28
  • Para 6 says that if $p$ was properly contained in some $p'$, then we get $x\in p'$ with support $B$ not in $\mathcal{V}_p$, so $e_B\notin p$. Then using para 5 we get that $e_A\in p$ which means that $e_A\in p'$, so both $e_A$ and $e_B$ are in $p'$. But if they are both in $p'$ then so is their sum, which is $1$. This means that $p'$ cannot be a proper ideal. Conclusion: $p$ is not properly contained in any proper ideal. e.g. $p$ is maximal. – tkf Jul 10 '22 at 12:35