Posted by Jason Polak on 26. November 2016 · Write a comment · Categories: commutative-algebra · Tags:

Let $A\to B\to C$ be ring homomorphisms. Consider the following statements:

  1. $B$ is a projective $A$-algebra,
  2. $C$ is a projective $B$-algebra,
  3. $C$ is a projective $A$-algebra.
Question. Which pairs of these statements imply the third?

1 and 2 imply 3. This follows because a module is projective if and only if it is the direct summand of a free module. Alternatively, it is a special case of the following formula: if $C$ is a $B$ module and $A\to B$ is a ring homomorphism then ${\rm pd}_A(C)\leq {\rm pd}_B(C) + {\rm pd}_A(B)$, where ${\rm pd}_X(Y)$ denotes the projective dimension of $Y$ as an $X$-module.

1 and 3 do not imply 2. Indeed, consider the ring homomorphisms $k\to k[x]\to k[x]/x^2$ where $k$ is a field. Here, $k\to k[x]$ is the inclusion of $k$ into the polynomial ring $k[x]$, and $k[x]\to k[x]/x^2$ is the quotient map. Since $k$ is a field, $k[x]$ and $k[x]/x^2$ are both $k$-free and hence $k$-projective. Yet, $k[x]/x^2$ is not a projective $k[x]$-module because projective modules over integral domains are torsion-free, and $k[x]/x^2$ is not a torsion-free $k[x]$-module.

2 and 3 does not imply 1. Take any ring homomorphism $A\to B$ such that $B$ is not a projective $A$-module. For example $k[x]\to k[x]/x^2$ as in the previous paragraph. Consider $A\to B\to 0$, where $0$ is the zero ring, the final object in the category of rings. The zero module is projective over $A$ and $B$ because it is projective over any ring. Yet $B$ is not projective over $A$.

Cheap?! What if we stipulate $C\not= 0$. Sadly, 2 and 3 still do not imply 1. For example, $\Z\times\Z\to \Z\times\Q\to\Z$. The first homomorphism is given by $(a,b)\mapsto (a,b)$ and the second is given by $(a,b)\mapsto a$.

Under this homomorphism, $\Z$ is by construction a direct summand of $\Z\times\Z$ and $\Z\times\Q$, and hence projective as modules over both of these rings. Yet, $\Z\times\Q$ is not a projective $\Z\times\Z$-module, since $\Z\times\Q$ contains infinitely divisible elements as an abelian group, yet $\Z\times\Z$ does not.

Fun Fact. At the beginning, if we replaced the property “projective” by “finitely generated projective and separable”, then any two of the statements would imply the third.

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