Regarding Atiyah-Macdonald Chapter 3 Exercise 27 ii)

Question

Let $B = \Pi_{\alpha} B_{\alpha}$ be the direct product of finitely many $A$-algebras $f_{\alpha}: A \to B_{\alpha}$. Then if $f: A \to B$ is given by $f(x) = (f_{\alpha}(x))_{\alpha}$ show that $$ f^*(\text{Spec}(B)) = \bigcup_{\alpha}f_{\alpha}^*(\text{Spec}(B)) $$

My Proof: With all tensors being over $A$ and letting $k$ be the residue field at a particular prime ideal $p \in \text{Spec}(A)$ we have that

\begin{align*} f^{*-1}(p) &= \text{Spec}(B\otimes k) \\ &= \text{Spec}\left(\left(\bigoplus_{\alpha}B_{\alpha}\right)\otimes k\right) \\ &= \text{Spec}\left(\bigoplus_{\alpha}B_{\alpha}\otimes k\right) \\ &= \bigsqcup_{\alpha} \text{Spec} \left( B_{\alpha} \otimes k\right) \\ &= \bigsqcup_{\alpha} f_{\alpha}^{*-1}(p) \quad \quad (1) \end{align*}

Thus we have that $f^{*-1}(p) \neq \varnothing \iff \exists \alpha \, \,, \, \, f_{\alpha}^{*-1}(p) \neq \varnothing $. This gives us that $$ f^*(\text{Spec}(B)) = \bigcup_{\alpha}f_{\alpha}^*(\text{Spec}(B)) $$

My only concern is that there is a disjoint union in eq (1). I was wondering whether this disjoint union is an external one rather than an internal one.

Thanks,
Vatsa

Answer 1

So, apart from using “$\bigoplus$” instead of “$\prod$”, your proof is fine. The disjoint union is an external one.

However, your proof is unnecessarily abstract. Instead

• consider how primes in $\prod_α B_α$ exactly look like, and
• compare their preimages under $f$ to preimages of primes under $f_α$.

You will quickly find that every preimage of a prime under $f$ can be realised as a preimage of a prime under $f_α$ for some $α$ and vice versa.