If $X=Spec(R)$ and $Y=Spec(S)$ are affine schemes, then the disjoint union $X \sqcup Y$ is an affime scheme with $X \sqcup Y = Spec(R \times S)$

Question

Let $R,S$ be commutative rings with identity.

Proving that $X \sqcup Y$ is an affine scheme is the same as proving that $Spec(R) \sqcup Spec(S) = Spec(R \times S)$.

I proved that if $R,S$ are rings, then the ideals of $R \times S$ are exactly of the form $P \times Q$, where $P$ is an ideal of $R$ and $Q$ is an ideal of $S$.

However, for prime ideals this is not true in general.

If $I$ is a prime ideal of $R \times S$, then $I = \mathfrak{p} \times \mathfrak{q}$, where $\mathfrak{p}$ is a prime ideal of $R$ and $\mathfrak{q}$ is a prime ideal of $S$.

But if $\mathfrak{p}$ is a prime ideal of $R$ and $\mathfrak{q}$ is a prime ideal of $S$, it is not true in general that $\mathfrak{p} \times \mathfrak{q}$ is a prime ideal of $R \times S$.

Then, $Spec(R \times S) \subseteq Spec(R) \times Spec(S)$ and the reverse inclusion is false in general.

My question is, what is $Spec(R) \sqcup Spec(S)$ set-theoretically, in order to use what I proved above?

Answer 1

Bearing in mind that product of two nontrivial rings can never be an integral domain, we have that any prime ideal of $R\times S$ must be of the form $\mathfrak{p}\times S$ or $R\times\mathfrak{q}$. That explains why spec of a product of two rings is the disjoint Union of the spec of the respective rlngs.

Answer 2

let $I\subset R\times S$ be a prime ideal. Then $R\times S/I$ is an integral domain.

Note the idempotents $(1,0),(0,1) \in R\times S$, and consider their images.

Since $\overline{(0,1)}\cdot\overline{(1,0)} = \overline{(0,0)}$ in the quotient, $R\times S/I$ being an intgral domain forces one of the two $(1,0),(0,1)$ to be in $I$.

If $(1,0)\in I$, then $R\times {0}\subset I$, so tha, by your observation of ideals in general, $I = R\times P$ with $P\subset S$ prime.

If $(0,1)\in I$, then we get $I = Q\times S$ for $Q\subset R$ prime.

Thus, $Spec(R\times S) = Spec(R)\sqcup Spec(S).$