How to determine if a ring is a free module over a subring?

[Jeff Carlson]

I have a situation where I have a quotient ring B of a polynomial ring A. I want to generate a subring C of B using a certain finite list of variables and determine if B is a free module over C.

How might I do this? I've found there is a method that generates a subring of a polynomial ring, but this doesn't seem to help me with C, and I know there is an isFreeModule routine, but it seems to determine whether the submodule is free (?) rather than vice versa.

[Frank Moore]

Jeff,

As long as C is finitely generated over B, you should be able to accomplish this with a combination of isFreeModule and pushForward.

Here is some example code that I think has all that you may need:

A = QQ[x,y]
B = A/ideal{y^2}
C = QQ[z,Degrees => {2}]
phi = map(B,C,{x^2})
M = pushForward(phi,B^1)
isFreeModule M

One can do the same thing even when the ring homomorphism is not injective, but you have to pushForward along the induced map on the quotient (which I think you must build 'by hand').

Cheers,

Frank

On Sat, Jun 24, 2017 at 12:39 AM, Jeff Carlson <jdkca...@gmail.com> wrote:

I have a situation where I have a quotient ring B of a polynomial ring A. I want to generate a subring C of B using a certain finite list of variables and determine if B is a free module over C.

How might I do this? I've found there is a method that generates a subring of a polynomial ring, but this doesn't seem to help me with C, and I know there is an isFreeModule routine, but it seems to determine whether the submodule is free (?) rather than vice versa.

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Dr. W. Frank Moore
Associate Professor
Department of Mathematics, Wake Forest University

email: moo...@wfu.edu

[Daniel R. Grayson]

But please be aware that "isFreeModule" does no computation -- it merely reports whether the module

is presented as a free module, not whether it's isomorphic to one.