# Find irreducible and connected components of $\operatorname{Spec}(\mathbb{C}[x] \times \mathbb{C}[y])$

Find irreducible and connected components of $$\operatorname{Spec}(\mathbb{C}[x] \times \mathbb{C}[y])$$

That's the problem 11.1 from commutative algebra course

As answered here

we can see the bijection between $$\operatorname{Spec}(A_1\times A_2)$$ and $$\operatorname{Spec}(A_1)\sqcup\operatorname{Spec}(A_2)$$ as follows: the prime ideals of $$A_1\times A_2$$ are of the form $$\mathfrak{p}\times A_2$$ where $$\mathfrak p$$ is a prime ideal of $$A_1$$, or $$A_1\times\mathfrak q$$ where $$\mathfrak q$$ is a prime ideal of $$A_2$$.

That means that we should find irreducible and connected components of $$\operatorname{Spec}(\mathbb{C}[x])$$

As discussed here: Let $$A$$ be a commutative ring with unit, $$X = \operatorname{Spec}(A)$$ with the Zariski topology. The irreducible components are $$\lbrace V(p) : p\subset A \ \text{minimal prime ideal} \rbrace$$ where $$V(P) =\lbrace q \ \text{prime ideal } \mid p\subset q\rbrace$$.

An ideal is called prime if the complement is a multiplicative set.

I know that prime ideals of $$\mathbb{C}[x]$$ are the principal ideals generated by $$\lbrace (x-a) | a \in \mathbb{C}\rbrace$$ and $$0$$. $$0$$ is contained in all the other ideals.

The prime ideals $$\mathfrak{p}$$ of $$\mathbb{C}[x]$$ except $$0$$ are all minimal, so the irreducible components of $$\operatorname{Spec}\mathbb{C}[x]$$ are all $$\lbrace V(p) \rbrace$$.

Is the final answer that the connected component are the same as irreducible components and are in the bijection with $$I_{x-a} \times \mathbb{C}[y]$$ and $$\mathbb{C}[x] \times I_{y-a}$$ ?

Could you explain how this topology can be seen "in a picture"?

## 1 Answer

Your proof is correct!

The topology of $$\displaystyle\mathrm{Spec}(\mathbb{C}[x])$$ is the following:

1. on $$\mathrm{Spec}(\mathbb{C}[x])\setminus\{(0)\}$$ one has the cofinite topology, I mean that all and only closed sets are the whole of the space, the empty set, and the finite subsets;
2. $$\{(0)\}$$ is a fat point, I mean that $$\overline{\{(0)\}}=\mathrm{Spec}(\mathbb{C}[x])$$.

As a picture: $$\mathrm{Spec}(\mathbb{C}[x])\setminus\{(0)\}$$ is the affine (complex) line $$\mathbb{A}^1_{\mathbb{C}}$$ plus a dense point, that is a point arbitrally closed to any other point.