The left adjoint of the restriction functor in field extension

Question

Let $k$ be a field and $A$ a $k$-algebra. Let $k \subseteq F$ be a field extension. Then naturally we have a restriction functor $\mathsf{res}: \operatorname{\mathit F-\mathbf{Alg}} \rightarrow \operatorname{\mathit k\,-\mathbf{Alg}}$. ($\operatorname{\mathit F-\mathbf{Alg}}$ means the algebras over field $F$).

I have seen that the functor $\mathsf{res}$ has the left adjoint $F \otimes_k-: \operatorname{\mathit k\,-\mathbf{Alg}} \rightarrow \operatorname{\mathit F-\mathbf{Alg}}$. The unit of this adjunction on $A$ is the $k$-algebra homomorphism $$\varphi: A \rightarrow F\otimes_kA$$ $$a\mapsto 1\otimes a$$ Also, $\varphi$ is injective.

I am not familiar with those things. So how to get that $\text{Hom}(F\otimes_k A_1,A_2) \cong \text{Hom}(A_1,\mathsf{res}(A_2))$ for any $k$-algebra $A_1$ and $F$-algebra $A_2$? Another question is that I think the unit of this adjunction is of the form $\varphi: A \rightarrow \mathsf{res}(F \otimes_k A)$, why it writes as $A \rightarrow F\otimes_k A$? Thank you for any help.

Answer 1

Let $R \to S$ be any homomorphism of commutative rings. We have a forgetful functor from $S$-algebras to $R$-algebras, denoted by $B \mapsto B|_R$ and called "restriction of scalars". It has a left adjoint which maps an $R$-algebra $A$ to the $S$-algebra $S \otimes_R A$, called "extension of scalars". The unit of the adjunction is $A \to (S \otimes_R A)|_R$, $a \mapsto 1 \otimes a$. (In this generality, it does not have to be injective.) The counit of the adjunction is $S \otimes_R B|_R \to B$, $s \otimes b \mapsto s \cdot b$. The triangle identities are satisfied, therefore we have an adjunction. Notice that the same adjunction holds for modules instead of algebras. The notation issue was explained by Derek Elkins in the comments. It is a common bad practice to ignore forgetful functors.