Why is it that $(\mathbb{C}[x]/(x^2-x))_{x-1} \cong \mathbb{C}?$

Question

Consider $\mathbb{C}[x]/(x^2-x)$. Why do we have for the localization at $x-1$ that $$(\mathbb{C}[x]/(x^2-x))_{x-1} \cong \mathbb{C}?$$

We are essentially inverting at $\{1, x-1, (x-1)^2, (x-1)^3, \dots \}$, but I don't see how we end up with $\mathbb{C}$ in the end?

Answer 1

I don't think it's a bad idea to work through this explicitly, to get comfortable with the definitions.

First, let's understand what $(\mathbb C[x] / (x^2 - x))_{x-1}$ actually is. This ring is a little tricky to understand, since we're dealing with a localisation of a quotient.

Each element in $(\mathbb C[x] / (x^2 - x))_{x-1}$ takes the form $$ \frac{[f(x)]_{\text{mod }(x^2 - x)}}{[(x-1)^n]_{\text{mod }(x^2 - x)}},$$ where $f(x)$ is some polynomial in $\mathbb C[x]$ and $n \in \mathbb N$.

And how do we determine if two elements in $(\mathbb C[x] / (x^2 - x))_{x-1}$ are equal? By the definition of localisation, we have $$ \frac{[f(x)]_{\text{mod }(x^2 - x)}}{[(x-1)^n]_{\text{mod }(x^2 - x)}} = \frac{[g(x)]_{\text{mod }(x^2 - x)}}{[(x-1)^m]_{\text{mod }(x^2 - x)}}$$ if and only if $$[(x-1)^{m + l}f(x) - (x-1)^{n + l} g(x)]_{\text{mod }(x^2 - x)} = 0$$ for some $l \in \mathbb N.$

Now let's prove that $(\mathbb C[x] / (x^2 - x))_{x-1}$ is isomorphic to $\mathbb C$.

Let $p : (\mathbb C[x] / (x^2 - x))_{x-1} \to \mathbb C$ be the homomorphism that evaluates elements in $(\mathbb C[x] / (x^2 - x))_{x-1}$ at $x = 0$. In other words, $p$ is the homomorphism that sends $$ \frac{[f(x)]_{\text{mod }(x^2 - x)}}{[(x-1)^n]_{\text{mod }(x^2 - x)}} \mapsto \frac{f(0)}{(0 - 1)^n} = (-1)^{-n} f(0).$$

This $p$ is well-defined: the answer that our definition gives us is independent of how the element of $(\mathbb C[x] / (x^2 - x))_{x-1}$ is represented. Indeed, if $$ \frac{[f(x)]_{\text{mod }(x^2 - x)}}{[(x-1)^n]_{\text{mod }(x^2 - x)}} = \frac{[g(x)]_{\text{mod }(x^2 - x)}}{[(x-1)^m]_{\text{mod }(x^2 - x)}},$$ then $$(x-1)^{m+l}f(x) - (x-1)^{n+l} g(x) = (x^2 - x)h(x)$$ for some $l \in \mathbb N$ and for some $h(x) \in \mathbb C[x]$. Since $x^2 - x$ evaluates to zero at $x = 0$, this implies that $$(-1)^{-n} f(0) = (-1)^{-m} g(0),$$ which shows that the result of acting on our element with $p$ is independent of how the element is represented.

It is easy to see that $p$ really is a ring homomorphism, like I claimed.

Clearly, $p$ is surjective. Indeed, for any $c \in \mathbb C$, the element $[c]_{\text{mod }(x^2 - x)} / [1]_{\text{mod }(x^2 - x)}$ evaluates to $c$ at $x = 0$.

The kernel of $p$ consists of elements of the form $$\frac{[x r(x)]_{\text{mod }(x^2 - x)} }{[(x - 1)^n]_{\text{mod }(x^2 - x)}}$$ where $r(x) \in \mathbb C[x]$. But such an element can be written as $$ \frac{[x(x-1) r(x)]_{\text{mod }(x^2 - x)} }{[(x - 1)^{n+1}]_{\text{mod }(x^2 - x)}},$$ which is manifestly zero, since $[x(x-1) r(x)]_{\text{mod }(x^2 - x)} = 0$. Therefore, the kernel of $p$ is trivial.

Thus $p$ is an isomorphism from $(\mathbb C[x] / (x^2 - x))_{x-1}$ to $\mathbb C$.

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By the way, there is an algebraic geometry interpretation of this.

• $\mathbb C[x]/(x^2 - x)$ is the coordinate ring for the closed subvariety $V_1$ of $\mathbb A_{\mathbb C}^1$ consisting of two points, at $x = 0$ and $x = 1$.
• $(\mathbb C[x]/(x^2 - x))_{x - 1}$ is the coordinate ring for the open subvariety $V_2$ of $V_1$, obtained by deleting the point at $x = 1$ from $V_1$. This $V_2$ consists of a single point at $x = 0$. The coordinate ring of $V_2$ is of course $\mathbb C$.

So that's how I got the idea for the homomorphism $p: (\mathbb C[x]/(x^2 - x))_{x - 1} \to \mathbb C$ that evaluates elements of $(\mathbb C[x]/(x^2 - x))_{x - 1}$ at $x = 0$.

Answer 2

This is just a way to get the intuition for why it is true, not a formal proof:

Firstly, the ring $R:=\mathbb C[x]/(x^2-x)$ is essentially $\mathbb C$, but with an additional element $x$, satisfying $x^2-x=0$, or equivalently, $x^2=x$. So its elements look like $a+bx$ with $a,b$ in $\mathbb C$, and when multiplying two such elements we follow the rule that $x^2=x$ and otherwise treat $x$ like a variable.

Then, the ring $R_{x-1}$ is trying to make all powers of $x-1$ invertible if possible, or kill them if it's not possible. Since $x(x-1)=0$, these are all zero-divisors (except for $(x-1)^0$), they are going to be killed, that is, sent to $0$. So then $x-1=0$, or $x=1$. But then the elements of our ring look like $a+b$, where $a,b\in\mathbb C$. But that's just $\mathbb C$ again.