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Let $R$ be a ring, $m$ is a maximal ideal of $R$. Thereofre, $R\oplus m$ is a maximal ideal of $R\oplus R$. What is the local ring $(R\oplus R)_{(R\oplus m)}$? I think this involves a component of $R_R$ which makes no sense to me.

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Answer and intuition

The ring $A = (R\oplus R)_{(R\oplus\mathfrak{m})}$ is isomorphic to the ring $B = R_\mathfrak{m}$ via the map $f:A\to B$ given by $$ f\left(\frac{(a,b)}{(r,x)}\right) = \frac{b}x. $$ My intuition for obtaining this came from the observation that the first coordinate of either the numerator or denominator of an element of $A$ is insignificant. Indeed, we have for any $(a,b)/(r,x) \in A$ that $$ \frac{(a,b)}{(r,x)} = \frac{(a,b)(0,1)}{(r,x)(0,1)} = \frac{(0,b)}{(0,x)}. $$ because $(0,1)$ is a valid denominator in $A$.

To prove $f$ is a ring isomorphism, we need to show it is well-defined and bijective. I'll leave showing $f$ is a homomorphism to you--it's a menial exercise using the "first coordinate observation" from above.

Well-definition

Suppose $(a,b)/(r,x) = (c,d)/(s,y)$ in $A$. We need to show $b/x = d/y$ in $B$. By our above observation, we have that $$ \frac{(0,b)}{(0,x)} = \frac{(a,b)}{(r,x)} = \frac{(c,d)}{(s,y)} = \frac{(0,d)}{(0,y)}, $$ and so there exists a $(u,v) \in (R\oplus R) \setminus (R\oplus \mathfrak{m})$ such that $$ (0,0) = (u,v)\cdot\left[(0,b)(0,y) - (0,d)(0,x)\right] = (u,v)(0,by-dx) = (0,v\cdot[by-dx]). $$ Comparing second corrdinates reveals $0 = v\cdot[by-dx]$. Since $u \in R$ automatically, we must have that $v \notin \mathfrak{m}$, hence $v \in R\setminus \mathfrak{m}$ and we conclude $b/x = d/y$ in $B$ as desired. So $f$ is well-defined.

Bijectivity

Surjectivity is clear. For injectivity, suppose $(a,b)/(r,x) \in \ker(f)$. Then $b/x = 0/1$ in $B$, meaning there exists a $v \in R\setminus \mathfrak{m}$ with $0 = v\cdot[b\cdot1 - 0\cdot x] = vb$. Now $(0,v) \in (R\oplus R) \setminus (R\oplus \mathfrak{m})$ and $$ \frac{(a,b)}{(r,x)} = \frac{(0,b)}{(0,x)} = \frac{(0,b)(0,v)}{(0,x)(0,v)} = \frac{(0,vb)}{(0,vx)} = \frac{(0,0)}{(1,1)}. $$ Hence, the kernel of $f$ is trivial; $f$ is injective.

Thus, $(R\oplus R)_{(R\oplus\mathfrak{m})} \cong R_\mathfrak{m}$ via $f$.

Nick F
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