# For $k$-algebras $B_1, \dots, B_n$, $\# \operatorname{Hom}_k( \prod_{i=1}^nB_i, \Omega) = \Sigma_{i=1}^n \# \operatorname{Hom}_k(B_i, \Omega)$?

Let $$k$$ be a field with $$k \subseteq \Omega$$ a algebraically closed field. Let $$B_1 , \dots, B_n$$ be ( possibly finite local ) $$k$$-algebras. Then next equality of cardinals holds

$$\# \operatorname{Hom}_k( \prod_{i=1}^nB_i, \Omega) = \Sigma_{i=1}^n \# \operatorname{Hom}_k(B_i, \Omega)$$ ?

We can refer to the Rotman's An introduction to Homological Algebra, Theorem 2.31 so that

$$\operatorname{Hom}_k( \prod_{i=1}^nB_i, \Omega) \cong \prod_{i=1}^n \operatorname{Hom}_k(B_i , \Omega)$$

Then, $$\# \prod_{i=1}^n \operatorname{Hom}_k(B_i , \Omega) = \Sigma_{i=1}^n \# \operatorname{Hom}_k(B_i, \Omega)$$ ?

This question originates from second question of next linked question : Understanding the Proposition 12.18 in Gortz's Algebraic Geometry. In the linked question I don't know whether the equality marked by question symbol - in procedure of deriving the second equality - is true ; i.e., $$\# \operatorname{Hom}_{k'} (\prod_{i=1}^{n}B_{\mathfrak{p}_{x_i'}} , \Omega) \overset{\mathrm{?}}{=} \Sigma_{i=1}^{n}\#\operatorname{Hom}_{k'}(B_{\mathfrak{p}_{x_i'}}, \Omega)$$. An issue that makes me stuck is, the cardinal of finite product of sets is the product of cardinals of each sets, not the sum. Am I missing something?

• Your context is not an additive category, so the result from Rotman doesn't apply Commented Jul 18 at 9:42
• But $\#\prod_{i=1}^nX_i=\prod_{i=1}^n\#X_i$ for sure, so that's also a mistake Commented Jul 18 at 9:43
• You can use \sum to produce the correct sized symbol for summation. Commented Jul 18 at 9:52
• @Asaf Karagila : Ah~ O.K. Thanks ~ Commented Jul 18 at 11:13

Your first statement is correct: in fact, it is true that: $$\hom\left(\prod_{i=1}^nB_i,\Omega\right)=\coprod_{i=1}^n\hom(B_i,\Omega).$$

You can see this as follows: pick any morphism $$\phi \in \hom\left(\prod_{i=1}^nB_i,\Omega\right)$$, and let $$e_i = (0,...,1,...,0) \in \prod_{i=1}^nB_i$$ (a 1 on the $$i$$-th component). The $$(e_i)$$ for a complete set of nontrivial orthogonal idempotents of $$\prod_{i=1}^nB_i$$, i.e. they satisfy:

• $$e_i^2 = e_i$$
• $$e_i e_j = 0$$ if $$i \neq j$$
• $$\sum_i e_i = 1$$
• $$e_i$$ is neither 0 nor 1 in $$\prod_{i=1}^nB_i$$

Now notice that the images $$\phi(e_i)$$ do also satisfy the three first conditions, but the only way for this to happen is that one of the $$\phi(e_i)$$ is one, and the other $$\phi(e_j)$$ are all zero (there are no nontrivial idempotents in $$\Omega$$). In particular, this shows that $$\phi$$ factors as: $$\prod_{i=1}^nB_i \xrightarrow{\phi} B_i \rightarrow \Omega$$ This defines a map $$\hom\left(\prod_{i=1}^nB_i,\Omega\right) \rightarrow \coprod_{i=1}^n\hom(B_i,\Omega)$$, which is easily seen to be a bijection.

You are actually not quoting Rotman's statement quite right: the actual isomorphism is $$\hom\left(\bigoplus_{i=1}^nB_i,\Omega\right)=\prod_{i=1}^n\hom(B_i,\Omega)$$ and in the context of $$k$$-algebras (or more generally, commutative rings), direct sums (tensor products) are very different from products.

As mentionned by FShrike below, if $$k$$-algebras were an abelian category, finite products and coproducts would be canonically isomorphic, and the original (wrong) statement would be correct. This is what happens for groups, or modules over a ring.

• It help for op to mention (i) my own mistake of forgetting that of course $k$-algebras don't form an additive category, so the result doesn't apply and (ii) explicitly, if $\phi(e_i)=1=\phi(e_j)$ for $i\neq j$ then $\phi(e_ie_j)=0=1\cdot1$ is a contradiction, since they seemed confused in comments. Thanks for catching that; I've been doing too much hom. alg., "everything is additive" Commented Jul 18 at 9:46
• I don't understand your construction $\Psi : \hom\left(\prod_{i=1}^nB_i,\Omega\right) \rightarrow \coprod_{i=1}^n\hom(B_i,\Omega)$ at all until now. Let me follow your construction. Fix $\phi : \prod_{=1}^n B_i \to \Omega$. As you said, $\phi(e_i)=1$, $\phi(e_j)=0$ for all $j \neq i$. Assume that $\phi(e_1)=1$, $\phi(e_j)=0$ for all $j \neq 1$. Then does there exists $f_1 : B_1 \to \Omega$ such that for $(b_i)= (b_1 , \dots b_n) \in B := \prod_{i=1}^n B_i$, $\phi((b_i))= f_1 \circ pr_1 ((b_i))$, where $pr_1 : B \to B_1$ is the projection ? Commented Jul 18 at 11:10
• That's right, $f_1$ is simply defined by the fact that $\phi$ is equal to $\phi \circ (\operatorname{id} \times 0 \times ... \times 0)$ (i.e. $\phi$ only reads one coordinate), so $\phi$ factors through $B_1 \times \{0\} \times ... \times \{0\}$ which is obviously isomorphic to $B_1$. And in that case, the corresponding element of $\coprod_{i=1}^n\hom(B_i,\Omega)$ (which is the disjoint union of all $\hom(B_i,\Omega)$) is the corresponding map in $\hom(B_1,\Omega)$ Commented Jul 18 at 11:32
• If you denote by $\iota_1 : B_1 \rightarrow B_1 \times \{0\} \times ... \times \{0\}$ the isomorphism, then $\Psi : \hom\left(\prod_{i=1}^nB_i,\Omega\right) \rightarrow \coprod_{i=1}^n\hom(B_i,\Omega)$ is defined as: $\Psi (\phi) = \phi \circ \iota_1$. But it is better to "feel" the situation rather than to make every single correspondance explicit (although I agree that it is reassuring). Commented Jul 18 at 11:36
• All you need is the observation that for any $b \in \prod_{i=1}^nB_i$, $b = b \times 1= \sum_i b e_i$ (because the $e_i$ for a complete set of nontrivial indempotents). Then apply $\phi$. Then write explicitely who $b e_i$ is in terms of coordinates. Commented Jul 18 at 12:48