# Spectrum of a product of rings isomorphic to the product of the spectra

I've found in an exercise this statement:

If $A$ is a commutative ring with unit and $A = A_{1} \times \dots \times A_{n}$ then $$\def\Spec{\operatorname{Spec}} \Spec(A) \cong \Spec(A_{1})\times \dotsb \times \Spec(A_{n})$$ Why is this true ?

• I suppose they are direct product of prime ideals, but why ? – WLOG May 2 '14 at 14:52
• @WLOG Note that a product of rings is never an integral domain, so the product of prime ideals can't be prime in general. – Seth May 2 '14 at 15:04
• I hope my edit is useful. I took the liberty of rewriting the title of the question to make it more easy to find when searching the site. – kahen May 2 '14 at 15:08

This is not true. Direct products of rings correspond to disjoint unions of affine schemes. Tensor products, on the other hand, correspond to fibre products (in particular, products) of affine schemes.

Explicitly,

$$\operatorname{Spec}(A_1\times\cdots\times A_n) \cong \operatorname{Spec}(A_1)\sqcup\cdots\sqcup \operatorname{Spec}(A_n)$$ and

$$\operatorname{Spec}(A_1\otimes_R A_2) \cong \operatorname{Spec}(A_1)\times_{\operatorname{Spec} R}\operatorname{Spec}(A_2).$$

The reason for this is that the functor $\operatorname{Spec}$ is a contravariant equivalence from the category $\operatorname{Ring}$ of commutative unital rings to the category $\operatorname{Aff}$ of affine schemes. In particular, it sends (fibre) products to (fibre) coproducts, and vice versa. The product in $\operatorname{Ring}$ is the cartesian product $\times$ and the coproduct in $\operatorname{Aff}$ is disjoint union $\sqcup$, giving the first statement above. The fibre coproduct in $\operatorname{Ring}$ is given by the tensor product, giving the second statement above.

Without resorting to abstract nonsense, 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 $A_2$.

• Where can I find the proof ? – WLOG May 2 '14 at 15:25
• I've edited the answer to include some explanation. – bradhd May 2 '14 at 19:55

This isn't true. Consider the simple case where $A_1 = A_2 = \mathbb{C}[x]$. Then $Spec(A_i) = \mathbb{C}$.

However, $Spec(A_1 \times A_2) = \mathbb{C} \sqcup \mathbb{C}$, not $\mathbb{C}^2 = Spec(A_1 \otimes A_2)$.

• The max spec of $\mathbb{C}[x]$ is $\mathbb{C}$ but the prime spec has an extra point. – Seth May 2 '14 at 20:01