Describe $Spec( \mathbb C[x,y]/x(x-a))$ where $a$ is some complex number.

Question

I'm solving some exercises from my class notes on Commutative Algebra,On the following exercise I got stuck:

Describe $Spec( \mathbb C[x,y]/x(x-a))$ where $a$ is some complex number.

As far as I know the maximal ideals of $\mathbb C[x,y]$ are of the form $(x-a,y-b)$ where $a,b \in \mathbb C$.But I don't know any result regarding prime ideals of $\mathbb C[x,y]$,So I am facing difficulty in solving this exercise.I'm pretty sure that we've to use the Correspondence theorem of ideals for solving this particular problem,but I'm unable to solve this problem.Please help!

Answer 1

(i) If $a\neq0$ the prime ideals you want are the maximal ideals $(x-a,y-b)\; (b\in \mathbb C)$ and $(x,y-b)\; (b\in \mathbb C)$ about which you already know PLUS the two prime but not maximal ideals $(x)$ and $(x-a)$.

(ii) If $a=0$ the prime ideals you want are the maximal ideals $(x,y-b)\; (b\in \mathbb C)$ PLUS the prime ideal $(x)$

Thinking geometrically (which one should!) this is evident:
In (i) we obtain the closed (=traditional) points of the union of the two lines $x=0, x=a\:$ PLUS one generic (=Grothendieck) point for each line.
And in (ii) we obtain the closed points of the non reduced line $x^2=0$ PLUS its generic point.

Note carefully
Quite generally, given a ring $A$, its spectrum (as a topological space) is homeomorphic to that of its reduction $A_{red}=A/Nil(A)$.
In particular the spectrum of a quotient $A=R/I$ is homeomorphic to that of $A_{red}=R/\sqrt I$ and in our case we have (in accordance with (ii)) homeomorphisms $$Spec( \mathbb C[x,y]/(x^2))\equiv Spec( \mathbb C[x,y]/(x))\equiv Spec(\mathbb C[y])$$

Answer 2

Let's observe that if $a \neq 0$ the ideals $(x)$ and $(x-a)$ are coprime, by CRT you obtain:

$$\frac{\mathbb{C}[x,y]}{x(x-a)} \simeq \frac{\mathbb{C}[x,y]}{(x)} \oplus \frac{\mathbb{C}[x,y]}{(x-a)} \simeq \mathbb{C}[y]\oplus \mathbb{C}[y]$$

Now is easy to look at the prime ideals of $\mathbb{C}[x]\oplus \mathbb{C}[x]$ and I think you can easely understand their correspondence in the ring $\mathbb{C}[x,y]$

Otherwise, if $a=0$, by isomorphism theorem you obtain $$\frac{\mathbb{C}[x,y]}{x^2} \simeq \frac{\mathbb{C}[x]}{(x^2)} \left[y \right]=A[y]$$ Now if $P$ is a prime ideal of $A[y]$ think to $P \cap A$ : it is a prime ideal of $A$ and there are not ideals of this kind that contains $x^2$...

Answer 3

Note that $(x(x-a))$ decomposes as $(x) \cap (x-a)$. Thus $\textrm {Spec}$ of this corresponds to the disjoint union of two lines in $\mathbb C^2$.

However, if $a=0$, geometrically, this is only one line, but it is a so-called "fat line", in that the ideal is representend by $x^2$.