$A\otimes\bar{k}\simeq\bar{k}\times\cdots\times\bar{k}$, then the number of $k$-algebra maps $A\to\bar{k}$ is the dimension of $A$? I'm struggling to understand an equivalence of two definitions of a separable algebra. Let $\bar{k}$ be the algebraic closure of a field $k$, and $A$ a finite dimensional $k$-algebra. Theorem 6.3 of Waterhouse's Affine Group Schemes states that if $A\otimes\bar{k}\simeq\bar{k}\times\cdots\times\bar{k}$, then the number of $k$-algebra homomorphisms $A\to\bar{k}$ equals the dimension of $A$. (Over $k$ I think? The author doesn't specify.)
My thinking is
$$
\operatorname{hom}_{\bar{k}}(A\otimes\bar{k},\bar{k})\simeq\operatorname{hom}_k(A,\operatorname{hom}_{\bar{k}}(\bar{k},\bar{k}))\simeq\operatorname{hom}_k(A,\bar{k}).
$$
If $A\otimes\bar{k}\simeq\bar{k}\times\cdots\times\bar{k}$, say $n$ factors, Isn't
$$
\operatorname{hom}_{\bar{k}}(A\otimes\bar{k},\bar{k})\simeq\prod_{i=1}^n\hom_{\bar{k}}(\bar{k},\bar{k})\simeq\prod_{i=1}^n\bar{k}?
$$
Then the number of $k$-algebra maps $A\to\bar{k}$ is the cardinality of this space, but it seems like this is infinite, since $\bar{k}$ is necessarily infinite, and can't possbily be the dimension of $A$. Have I misapplied something?
 A: I'm assuming that algebra homomorphisms are required to preserve units, and that $\hom_k(A,B)$ denotes the set of $k$-algebra homomorphisms from $A$ to $B$.
In your first equation, $\hom_k(A,\hom_{\bar k}(\bar k,\bar k))$ doesn't make sense since $\hom_{\bar k}(\bar k,\bar k)$ is not a $k$-algebra, but the conclusion
$$
  \hom_{\bar k}(A\otimes_k \bar k,\bar k)\cong\hom_k(A,\bar k)
$$
is correct. Secondly, in general
$$
  \hom_k(A\times B,C)\neq\hom_k(A,C)\times\hom_k(B,C).
$$
Indeed given $f:A\to C$ and $g:B\to C$, if we try to define $h:A\times B\to C$ by $h(a,b)=f(a)+g(b)$, we'll run into trouble because
$$
  h((a_1,b_1)(a_2,b_2))=h(a_1a_2,b_1b_2)=f(a_1)f(a_2)+g(b_1)g(b_2)
$$
might not equal
$$
  h(a_1,b_1)h(a_2,b_2)=(f(a_1)+g(b_1))(f(a_2)+g(b_2)).
$$
In fact for the case $C=k$, we have
$$
  \hom_k(A\times B,k)=\hom_k(A,k)\sqcup\hom_k(B,k).
$$
Indeed given $f:A\times B\to k$, we have
$$
  f(1,0)f(0,1)=f((1,0)(0,1))=f((0,0))=0,
$$
$$
  f(1,0)+f(0,1)=f((1,1))=1.
$$
The first equation gives $f(1,0)=0$ or $f(0,1)=0$, and the second gives $f(0,1)=1$ or $f(1,0)=1$ respectively. In the first case define $g:B\to k$ by $g(b)=f(0,b)$. Then
$$
  f(a,b)=f((a,0)(1,0)+(0,b)(0,1))=f(a,0)\cdot0+g(b)\cdot1=g(b),
$$
so these homomorphisms correspond to $\hom_k(B,k)$, and similarly for the second case.
Now it is easy to see that $\hom_{\bar k}(A\otimes_k\bar k,\bar k)$ is a union of $n$ copies of $\hom_{\bar k}(\bar k,\bar k)$, where $A\otimes_k\bar k$ is a product of $n$ copies of $\bar k$.
