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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.

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    $\begingroup$ It's common to notationally identify a mathematical structure with its underlying set. This is an abuse of notation. In this case, $F\otimes_k A$ and $\mathsf{res}(F\otimes_k A)$ have the same underlying set, they are just equipped with slightly different scaling operations. This leads to the same notation being used for them and the intended scaling operation being determined by context. Personally, I prefer being much more explicit and unambiguous, but that's probably due to my background as a programmer. $\endgroup$ – Derek Elkins left SE Apr 18 '17 at 3:27
  • $\begingroup$ I get your meaning, thank you for your explaination $\endgroup$ – Xiaosong Peng Apr 18 '17 at 7:03
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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.

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  • $\begingroup$ Thank you for your help. The unit of the adjunction in my question is injecitve, but you say that the unit does not have to be injective in general. I am confused at here: can't we get $a=0$ from $1 \otimes a=0$? $\endgroup$ – Xiaosong Peng Apr 19 '17 at 2:02
  • $\begingroup$ The unit is injective if $A$ is flat over $R$ (for instance when $R$ is a field). Notice what I have written in the first sentence of my answer. It is more general than your situation. $\endgroup$ – HeinrichD Apr 19 '17 at 6:57
  • $\begingroup$ I have noticed your answer is more general. Could you tell me why the unit is injecitve if $A$ is flat over $R$ for the general case?(If $R \rightarrow S$ is injecitve, then I can get the unit is injecitve by the flat of $A$, but I can't get the conclusion if $R \rightarrow S$ is not inejctive) $\endgroup$ – Xiaosong Peng Apr 19 '17 at 7:45
  • $\begingroup$ Sorry, I was assuming that $R \to S$ is injective here, because this is the special case $A=R$ and therefore we have to assume it anyway (if we want to have a condition for injectivity). Notice that injectivity is absolutely irrelevant when it comes to the construction of the adjunction. $\endgroup$ – HeinrichD Apr 19 '17 at 7:51
  • $\begingroup$ Ok, thank you again for your help. $\endgroup$ – Xiaosong Peng Apr 19 '17 at 7:52

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