Let $A$ be a ring, $B$ a finite projective $A$-algebra, and $C$ a finite projective $B$-algebra. We can show that $C$ is also finite and projective when regarded as an $A$-algebra (by, for instance, taking direct sum with appropriate $A$-module to make it free in $(A\text{-Mod})$.)

The following are what I would like to prove / figure out:

(1) The map $$\hom_A(B,A)\otimes_B \hom_B(C,B)\to\hom_A(C,A),\;\;\; f\otimes g\mapsto f\circ g$$ is a surjection. Note that the $\hom_A$ stands for $A$-linear maps (instead of $A$-algebra maps). So the statement makes sense if we only require $C$ to be a $B$-module (instead of a $B$-algebra).

(2) Express $[C:A]$ in terms of $[C:B]$ and $[B:A]$. Here $[C:A]$ is the map $$[C:A]:\text{Spec }A\to\mathbb Z,\;\;\;\mathfrak p\mapsto\text{rank}_{A_\mathfrak p}C_\mathfrak p.$$ ($[C:B]$ and $[B:A]$ are defined in the same way.) Note that $B_\mathfrak p$ is free over $A_\mathfrak p$, the rank is just the "free rank".

For (1), I localize the map at a prime ideal $\mathfrak p$ to see whether it is surjective. Since both $B_\mathfrak p$ and $C_\mathfrak p$ are free over $A_\mathfrak p$, we can write $C_\mathfrak p \simeq A_\mathfrak p ^n$ and $B_\mathfrak p\simeq A_\mathfrak p ^m$ as $A_\mathfrak p$-modules. On the other hand, since $B_\mathfrak p=B\otimes_A A_\mathfrak p$ and $C_\mathfrak p=C\otimes_A A_\mathfrak p$, we know $C_\mathfrak p$ is projective (but not necessarily free) $B_\mathfrak p$-module. Suppose in $(B_\mathfrak p\text{-Mod})$that $C_\mathfrak p \oplus D \simeq B_\mathfrak p^k$. Then in $(A_\mathfrak p\text{-Mod})$ we have $C_\mathfrak p\oplus D\simeq A_\mathfrak p ^{mk}$. Now, given any $A_\mathfrak p$-linear map $C_\mathfrak p\to A_\mathfrak p$, I tried to decompose it into a sequence of maps $$C_\mathfrak p\to C_\mathfrak p \oplus D \simeq B_\mathfrak p^k \to B_\mathfrak p \simeq A_\mathfrak p^ m\to A_\mathfrak p$$ so that $C_\mathfrak p\to B_\mathfrak p^k\to B_\mathfrak p$ is $B_\mathfrak p$-linear. But the isomorphism $C_\mathfrak p \oplus D \simeq B_\mathfrak p^k$ is so unclear that I do not know how to write down the maps in the sequence.

For (2), I have even less idea to get started. I have just drawn a diagram about the maps, and to guess from the free case that we should have something like $$\text{(??) }\text{rank}_{A_\mathfrak p}C_\mathfrak p = \text{rank}_{B_\mathfrak p} C_\mathfrak p\cdot \text{rank}_{A_\mathfrak p} B_\mathfrak p$$ or $$\text{(??) }[C:A](\phi^{-1}(\mathfrak q))=[B:A](\phi^{-1}(\mathfrak q))\cdot [C:B](\mathfrak q), $$ where $\mathfrak q$ is a prime ideal in $B$ and $\phi: A\to B$ is the natural map which determent the $A$-algebra structure on $B$. But both of the two identities seem to be problematic.

  • 1
    $\begingroup$ What do you mean by projective algebra? A retract of a polynomial algebra? $\endgroup$ Dec 27, 2013 at 23:00
  • 1
    $\begingroup$ A projective algebra means an algebra that is projective as a module. $\endgroup$
    – Michael
    Dec 27, 2013 at 23:02
  • $\begingroup$ in the title, do you mean Compound? You do not use the word again, so I cannot tell whether it is a real word borrowed from some other language. $\endgroup$
    – Will Jagy
    Dec 27, 2013 at 23:07
  • $\begingroup$ Well, I just couldn't think of an appropriate word to describe the relation among A, B and C. Perhaps it is better to remove the word "compound". $\endgroup$
    – Michael
    Dec 27, 2013 at 23:12
  • $\begingroup$ Makes sense. You do use the word decomposition, I bet composition, compositum, and compound all come from the same ancient word. en.wikipedia.org/wiki/… $\endgroup$
    – Will Jagy
    Dec 27, 2013 at 23:34

1 Answer 1


I have tried out the first one. I realized that localization is not necessary.

Define a map $$\theta:\hom_A(B,A)\otimes_B \hom_B(P,B)\to\hom_B(P,\hom_A(B,A))$$ that is given by $$\theta(f\otimes g)(p)(b)=f(g(bp)),$$ where $f\in\hom_A(B,A)$, $g\in\hom_B(P,B)$, $p\in P$ and $b\in B$. One can easily check that it is a well-defined $B$-linear map. We claim that $\theta$ is an isomorphism. Indeed, if $P=B$, then both sides of $\theta$ are isomorphic to $\hom_A(B,A)$, and $\theta$ is clearly induced from the identity map. It can be generalized to the case in which $P=B^{\oplus n}$ for some finite $n$ since finite direct sums commute with both tensor products and Hom functors. And so $\theta$ is an isomorphism for any finitely generated projective $B$-module $P$ since $P$ is a direct summand for some $B^{\oplus n}$ of finite rank.

Now let $$\varphi:\hom_B(P,\hom_A(B,A))\to \hom_A(P,A)$$ be the map defined by $$\varphi(k)(p)=k(p)(1_B),$$ where $k\in\hom_B(P,\hom_A(B,A))$ and $p\in P$. Then the map in the statement is equal to $\varphi\circ\theta$, in which $\varphi$ is a surjection since each $h\in\hom_A(P,A)$ is the image of the $B$-linear map $k:P\to\hom_A(B,A)$ defined by $k(p)(b)=h(bp)$. So we are done.


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