Localization and Field [duplicate]

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.

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 $e_A\in P$.

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 $e_A\in \mathcal{V}_p$ or $e_B\in \mathcal{V}_p$.

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.