Localization at prime ideals of the infinite direct product of $\mathbb Z/2\mathbb Z$

Question

How to find the localization at a prime ideal of infinite direct product of $\mathbb Z/2\mathbb Z$, i.e. $ \mathbb Z/2\mathbb Z \times \mathbb Z/2 \mathbb Z \times\mathbb Z/2\mathbb Z \times \cdots$ ?

I know there is a question related to this but that question only prove the localization of this ring is a field, but did not specifically tell what the ring is after localization.

Some hints and explanations would be really helpful!

Thank you!

Answer 1

Let $R$ be the ring given by $$\prod_{n=1}^\infty F_2$$ and let $K = R_P$, for some prime ideal $P$ of $R$.

Let $S = R\setminus P$.

Noting that $r^2 = r$, for all $r\in R$, it follows that one of $r,\;1-r$ is in $P$, and the other is in $S$.

Suppose $x$ is a nonunit of $K$.

Write $x = r/s$, where $r \in R$, and $s \in S$.

Since $x$ is a nonunit, we must have $r \notin S$, hence $1-r \in S$.

Then $x = {\large{\frac{r}{s}}} = {\large{\frac{0}{1-r}}}$, hence $x=0$.

Thus, zero is the only nonunit of $K$, so $K$ is a field.

Let $x$ in $K$.

Writing $x = {\large{\frac{r}{s}}}$, where $r\in R$, and $s \in S$, we get $$x^2 = \left({\small{\frac{r}{s}}}\right)^2 = {\small{\frac{r^2}{s^2}}} = {\small{\frac{r}{s}}} = x $$ but then, since $K$ is a field, $x^2=x$ implies $x=0$ or $x=1$.

Thus, $K$ is isomorphic to $F_2$.

Answer 2

If $A$ is a Boolean ring, meaning every element is idempotent, then $A_P$ is a local ring where every element is idempotent. Local rings have no non-trivial idempotents, so $A_P$ is the field of two elements.