Adding an inverse of $(2, 0)$ to $\Bbb{R\times R}$

Question

This is a problem from Artin:

Describe the ring obtained by adding an inverse of $(2, 0)$ to $\Bbb{R\times R}$.

I can see that this also adds inverses for all elements of the form $(a, 0)$, but how do I proceed from here?

I'd like a small hint.

Answer 1

Actually the questions says "by inverting the element $(2,0)$".

In terms of rings of fractions this means to look for $R[a^{-1}]$, where $R=\mathbb R\times\mathbb R$, and $a=(2,0)$. (The multiplicative set is $S=\{1,a,a^2,\dots\}$).

In general, if $R_1,R_2$ are two commutative rings and $S_1,S_2$ are multiplicative sets, $S_i\subset R_i$, $i=1,2$, then $(S_1\times S_2)^{-1}(R_1\times R_2)\simeq S_1^{-1}R_1\times S_2^{-1}R_2$. (The correspondence is easily seen: $\frac{(r_1,r_2)}{(s_1,s_2)}\mapsto(\frac{r_1}{s_1},\frac{r_2}{s_2})$.)

In your case set $S_1=\{1,2,2^2,\dots,\}$, and $S_2=\{0,1\}$. Then $R[a^{-1}]\simeq\mathbb R\times\{0\}\simeq\mathbb R$.

Edit. The objection that $S=\{(1,1), (2,0), (2^2,0),\dots\}\subsetneq S_1\times S_2$ is right, but only of pedantic nature. We obviously have $S^{-1}(\mathbb R\times\mathbb R)=(S_1\times S_2)^{-1}(\mathbb R\times\mathbb R)$.

Answer 2

The multiplicative set is $S = \{(1,1), (2,0), (4,0), \dots\}$.

Since $(2,0) \cdot (0,a) = 0$ for all $a \in \mathbb{R}$, when we add an inverse to $(2,0)$, we make $(0,a)/1 = 0$ for all $a \in \mathbb{R}$.

Let's now show that $S^{-1}(\mathbb{R} \times \mathbb{R}) \cong \mathbb{R}$. Define $f \colon \mathbb{R} \to S^{-1}(\mathbb{R} \times \mathbb{R})$ by $f(x) = (x,0)/1$. Since $1 = (1,1)/1 = (1,0)/1 + (0,1)/1 = (1,0)/1$ this is clearly a ring homomorphism. This is also injective, since if $(x,0)/1 = (y,0)/1$, then $2^nx = 2^ny$ for some $n$, but then $x = y$.

All that is left is to show that this mapping is surjective. Let's take $(a,b)/(2,0)^n \in S^{-1}(\mathbb{R} \times \mathbb{R})$. We already know that $(a,b)/(2,0)^n = (a,0)/(2,0)^n$. Now, let $x = 2^{-n}a$. Then we get $(2,0)^n \cdot (x,0) = (2^nx,0) = (a,0)$ so $(x,0)/1 = (a,0)/(2,0)^n$. Hence $f$ is surjective, and so $\mathbb{R} \cong S^{-1}(\mathbb{R} \times \mathbb{R})$.

Answer 3

Let's look at the problem from a more general point of view. Let $R$ be a commutative ring and $f\in R$. If $f$ is a zero-divisor, you can't find an overring $S$ of $R$ such that $f$ is invertible in $S$, because $f$ would remain a zero-divisor also in $S$ and no zero-divisor is invertible, unless we're in the trivial ring.

So the best we can hope to find is another ring $S$ and a ring homomorphism $\lambda\colon R\to S$ such that $\lambda(f)$ is invertible. However we need to keep $\ker\lambda$ as small as possible, because choosing $S$ as the trivial ring would solve the problem, but it would be a very uninteresting solution.

Thus we try to add as little as possible. Suppose we found a solution, that is, a pair $(S,\lambda)$. We shall assume that $S$ is generated by the elements of the form $\lambda(a)$, for $a\in R$, and $\lambda(f)^{-1}$ (such elements are those “reached” by $\lambda$, together with the inverse of $\lambda(f)$). In $S$ we'll have elements of the form $\lambda(f)^{-m}\lambda(a)$, with $m\ge0$ and this is indeed a subring of $S$, because \begin{align} \lambda(f)^{-m}\lambda(a)+\lambda(f)^{-n}\lambda(b) &=\lambda(f)^{-m}\lambda(f)^{-n}\lambda(f^na)+ \lambda(f)^{-m}\lambda(f)^{-n}\lambda(f^mb)\\ &= \lambda(f)^{-(m+n)}\lambda(f^na+f^mb)\\[6px] \bigl(\lambda(f)^{-m}\lambda(a)\bigr)\bigl(\lambda(f)^{-n}\lambda(b)\bigr) &=\lambda(f)^{-(m+n)}\lambda(ab) \end{align} In order to save space, we set $$ \lambda(f)^{-m}\lambda(a)=\frac{a}{f^m} $$ that's only a formal notation. The above relations become $$ \frac{a}{f^m}+\frac{b}{f^n}=\frac{f^na+f^mb}{f^{m+n}}, \qquad \frac{a}{f^m}\frac{b}{f^n}=\frac{ab}{f^{m+n}} \tag{*} $$ Notice that $\frac{0}{f^m}=0$ (the zero of $S$), because $\lambda(0)=0$ by assumption. Note also that we must have $$ \frac{a}{f^m}=\frac{f^na}{f^{m+n}} $$ because $\frac{f}{1}=\lambda(f^{-0})\lambda(f)=\lambda(f)$ is invertible in $S$ by assumption.

We now ask when $$ \frac{a}{1}=\frac{0}{f^n} $$ should hold. If we multiply both sides by $f^n/1$, we get $$ \frac{f^na}{1}=\lambda(f^n)\lambda(f^{-n})\lambda(0)=\lambda(0) $$ Thus $$ \ker\lambda\supseteq\{a\in R: f^na=0\text{ for some }n\ge0\} $$ Note that the right-hand side set is an ideal of $R$, which is $\{0\}$ if and only if $f$ is not a zero divisor. (On the contrary, the set coincides with $R$ if and only if $f$ is nilpotent.)

In particular, by the homomorphism theorem, we get an embedding $$ R/\ker\lambda\to S $$

Now that we know a bit of what $S$ and $\lambda$ should look like, the trick is to define $S$ as the set of equivalence classes of elements of the form $(a,f^m)$ under the equivalence relation $$ (a,f^m)\sim(b,f^n) \quad\text{if and only if}\quad f^p(f^na-f^mb)=0 $$ for some $p\ge0$. The equivalence class of $(a,f^m)$ is denoted by $\frac{a}{f^m}$ as before.

The verification that we get a ring by defining operations like in (*) is just tedious. The map $a\mapsto \frac{a}{1}$ is a ring homomorphism and $$ \ker f=\{a\in R: f^na=0\text{ for some }n\ge0\} $$ so the kernel is as small as possible.

In your specific case, the kernel is $$ \{(a,b)\in\mathbb{R}\times\mathbb{R}:(2^n,0)(a,b)=0\text{ for some }n\ge0\} =\{0\}\times\mathbb{R} $$ so the ring we obtain is (isomorphic to) a subring of $$ \mathbb{R}\times\mathbb{R}/\ker\lambda\cong\mathbb{R} $$ and it's easy to see $\lambda$ is surjective, in this case.