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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

user26857
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Vatsa Srinivas
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    Why do you use the direct summand symbol $\bigoplus$ here? Do you mean to denote the product $\prod_α B_α$ here? – k.stm Jul 15 '19 at 10:52
  • If I'm not mistaken, the direct product over a finite family of modules is the same as the direct sum. I used the direct summand symbol because there is a theorem in the book saying that finite direct sums "distribute" over a tensor product. – Vatsa Srinivas Jul 15 '19 at 13:16
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    Stricty speaking, there is no such thing as a direct sum of rings, see Martin’s answer to Is $A × B$ the same as $A \oplus B$?, but I get what you mean. – k.stm Jul 15 '19 at 13:23
  • Thanks. I unknowingly swapped the categories I was working in. Ill keep this in mind in the future – Vatsa Srinivas Jul 15 '19 at 14:03

1 Answers1

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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.

k.stm
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