Isomorphisms in Product Rings I'm working on a problem in Michael Artin's Algebra that asks:
Describe the ring obtained from the product ring $\mathbb{R} \times \mathbb{R}$ by inverting the element $(2,0)$.
In the last problem, I proved $$R[x]/(x^2-1) \cong \mathbb{R} \times \mathbb{R}$$
I'm note really sure what to do with this, however.
I considered creating $\varphi: \mathbb{R}[x] \to \mathbb{R} \times \mathbb{R}$ given by $\varphi(f(x))=(f(1),\ f(-1))$. Then $\varphi(x+1)=(2,0)$.
But I don't think that helps me because modding out $(x+1)$ wouldn't really invert the element $(2,0)$, so I'm not really sure where to proceed from here.
 A: Let $f:R\times R\to R$ be the first projection, which is a surjective ring morphism. The image of $(2,0)$ is invertible. Suppose that $g:R\times R\to S$ is a ring morphism such that $g(2,0)$ is invertible in S. Since $(0,0)=(2,0)(0,a)$, applying $g$ to both sides and using that $g(2,0$) is invertible you get that $g(0,a)=0$. It follows that $g$ does not depend on the second component of its argument, so there is a unique function $h:R\to S$ such that $g(x,y)=h(f(x,y))$. Indeed, $h(x)$ is $g(x,0)$ for all $x \in R$.
Show that $h$ is morphism of rings.
This implies that the map $f$ has the correct universal property to be a localization of $R\times R$ at $(2,0)$ and therefore the localization you are looking for is $R$.
A: If you're following along in Artin and are trying to solve the problem with only the tools provided up to that point in the book, here's another solution:
In the previous problem, it is shown that:
$$\mathbb{R}[\alpha]=\mathbb{R}[x]/(x^2-1) \cong e\mathbb{R} \times e'\mathbb{R} \cong \mathbb{R} \times \mathbb{R}$$ where $e=1/2-1/2\alpha$, $e'=1-e$ and $\alpha$ satisfies the relationship $\alpha^2=1$.
With this, inverting $(2,0)$ is akin to identifying the ring:
$$\mathbb{R}[\alpha][y]/(y(1-\alpha)-1)$$ since $e$ is the multiplicative identity in $e\mathbb{R}$.
Note:
$$\mathbb{R}[\alpha][y]/(y(1-\alpha)-1)=(\mathbb{R}[x]/(x^2-1))[y]/(y(1-x)-1)=\mathbb{R}[x,y]/(x^2-1,\ y(1-x)-1)$$
It is simple to show that $(x+1,\ y(1-x)-1)=(x^2-1,\ y(1-x)-1)$ by first noting: $$x+1=-y(x^2-1)-(x+1)(y(1-x)-1)$$
We thus identify the ring
$$\mathbb{R}[x,y]/(x+1,\ y(1-x)-1)=(\mathbb{R}[x,y]/(x+1))/(y(1-\bar{x})-1)$$
One can first show $\mathbb{R}[x,y]/(x+1) \cong \mathbb{R}[y]$ by sending $x\mapsto -1$ and then show $$\mathbb{R}[y]/(y(1-\bar{x})-1)=\mathbb{R}[y]/(2y-1) \cong \mathbb{R}[1/2]=\mathbb{R}$$
We thus have that inverting $(2,0) \in \mathbb{R} \times \mathbb{R}$ gives us back a ring isomorphic to $\mathbb{R}$.
