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Assuming all rings are commutative, suppose $M$ is finite projective module over $R$ of rank $m$ which is itself an $R'$-algebra that is as a module finite projective of rank $n$. How can we show that $M$ is of rank $mn$ over $R'$?

Where the rank means localized at any prime, the module is free of that rank.

It is straightforward that $M$ is finite projective but I have difficulty because even though primes of $R'$ lie below primes of $R$, $M$ localized at the two primes is different.

This is question 2 in section 7.9 of Jacobson's Basic Algebra II.

user26857
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davik
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1 Answers1

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Let me use a slightly different notation: We have $R' \to R \to M$, where $R$ is $R'$-projective of rank $s$ and $M$ is $R$-projective of rank $r$.

The previous exercise states that

if $M$ is $R$-projective of rank $r$, then for any maximal ideal $m$ of $R$, $\dim_{R/m} M/mM = r$.

Since you already showed that $M$ is $R'$-projective, it suffices to show that for a maximal ideal $m$ of $R'$, $\dim _{R'/m} M/mM = sr$.

The extension $R'/m \to R/ mR$ is integral and $R'/m$ is a field. Hence $R / mR$ is a finite product of Artinian local rings, and $M /mM$ is a module over this product. Write $$R/ mR \cong R_1 \times R_2 \times \cdots \times R_q.$$ Here, each $R_i$ is a localization of $R / mR$ at a maximal ideal, and $R_i$ is finite over $R'$. We conclude that

$\dim_{R'/mR'} R/mR = \sum \dim_{R'/mR'} R_i = s$.

Now, we apply the exercise to $R$ and $M$. As $M$ is $R$-projective, for each maximal ideal $n$ of $R$, $\dim_{R/n} M/ nM = r$. Let $n_i$ be the maximal ideal which is the kernel of the morphism $$ R \to R/ mR \cong R_1 \times R_2 \times \cdots \times R_q \to R_i.$$ Then $(R/mR)/ n_i(R/mR) \cong R/n_i R$, and $\dim_{R/{n_i}} M/{n_i}M = r$.

Lastly, we these numbers together. We have that $M/mM \cong \oplus (R/{n_i})^r$, and $$\dim_{R'/m} M/mM = \sum r \dim_{R'/m} R_i = r \sum \dim_{R'/m} R_i = r \dim_{R'/m} R/mR = rs.$$

Youngsu
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  • I have a concern, how do we know M has a rank over R' at all. Sorry I haven't fully read the proof but I thought I couldn't apply the previous exercise because of this problem – davik May 31 '18 at 13:33
  • I also don't see why R/mR is artinian? – davik May 31 '18 at 13:47
  • After some thought I agree that you can use the exercise but I think it has to be more roundabout. Suppose $P'$ is a prime ideal of $R'$ we can localize everything at its complement. Then we get (localized) $M$ is $R$ finite projective rank $m$ and $R$ is $R'$ free of rank n and we need to show $M$ is rank $mn$ over $R'$ now a local ring. So now we can apply a dimension argument because there's no question about whether $M$ has a rank – davik May 31 '18 at 14:13
  • Ah I see $R/mR$ is finite dimensional. – davik May 31 '18 at 14:36
  • So following my comment above we can quotient by the maximal ideal of the now local ring to get to a situation where R is $s$ dimensional and M is rank $r$ and actually R is krull dimension zero as well but I don't think you use that. – davik May 31 '18 at 14:41
  • I can follow until the line $M/mM \cong \oplus (R/n_i)^r$ can you elaborate on why that's true? – davik May 31 '18 at 15:35
  • OK I see why. You map $M/mM$ to $\oplus M/n_iM$ and this is clearly injective and we can show $M/n_iM$ is $R_i$ free of rank $s$ (by usual arguments) then we need to show surjective but that follows by decomposing 1 in $R$ as the sum of the 1's in $R_i$ and using that to construct a preimage. – davik May 31 '18 at 15:51