Determining whether module is free for polynomial/power series rings

asked 2022-04-02 06:50:20 +0200

whatupmatt gravatar image

updated 2024-05-04 21:21:13 +0200

FrédéricC gravatar image

Say I have two modules

R = QQ[[x, y, z]]
S = QQ[[x, y^2,  z]]

(we can also make R and S as polynomial rings).

R is naturally an S module as S can act on R by multiplication.

Is there a command which determines whether R is free over S and the rank?

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Shouldn't the firs line be

R = QQ[["x, y, z"]]

I can't make head or tail of

S = QQ[[x, y^2 + 1, z]]

What is this supposed to mean ?

Emmanuel Charpentier gravatar imageEmmanuel Charpentier ( 2022-04-04 00:17:27 +0200 )edit

Hello, I just made that as an example. So to make thing simpler, let's ignore the +1. If we think of polynomial ring first,

QQ[ x, y, z]

is the polynomial ring with variables (generators if you like) x,y,z. So terms here look like


where a is a rational number. So

QQ[ x, y^2, z]

will be polynomial ring with variables x,y^2, z. Terms here look like



QQ[ [ x,y^2, z] ]

will be power series ring in variables x, y^2, z.

whatupmatt gravatar imagewhatupmatt ( 2022-04-04 18:23:17 +0200 )edit