ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 04 Apr 2022 18:23:17 +0200Determining whether module is free for polynomial/power series ringshttps://ask.sagemath.org/question/61796/determining-whether-module-is-free-for-polynomialpower-series-rings/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?Sat, 02 Apr 2022 06:50:20 +0200https://ask.sagemath.org/question/61796/determining-whether-module-is-free-for-polynomialpower-series-rings/Comment by whatupmatt for <p>Say I have two modules</p>
<pre><code>R = QQ[[x, y, z]]
S = QQ[[x, y^2, z]]
</code></pre>
<p>(we can also make <code>R</code> and <code>S</code> as polynomial rings).</p>
<p><code>R</code> is naturally an <code>S</code> module as <code>S</code> can act on <code>R</code> by multiplication.</p>
<p>Is there a command which determines whether <code>R</code> is free over <code>S</code> and the rank?</p>
https://ask.sagemath.org/question/61796/determining-whether-module-is-free-for-polynomialpower-series-rings/?comment=61824#post-id-61824Hello, 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
a*x^i*y^j*z^k
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
a*x^i*y^(2j)*z^k
So
QQ[ [ x,y^2, z] ]
will be power series ring in variables x, y^2, z.Mon, 04 Apr 2022 18:23:17 +0200https://ask.sagemath.org/question/61796/determining-whether-module-is-free-for-polynomialpower-series-rings/?comment=61824#post-id-61824Comment by Emmanuel Charpentier for <p>Say I have two modules</p>
<pre><code>R = QQ[[x, y, z]]
S = QQ[[x, y^2, z]]
</code></pre>
<p>(we can also make <code>R</code> and <code>S</code> as polynomial rings).</p>
<p><code>R</code> is naturally an <code>S</code> module as <code>S</code> can act on <code>R</code> by multiplication.</p>
<p>Is there a command which determines whether <code>R</code> is free over <code>S</code> and the rank?</p>
https://ask.sagemath.org/question/61796/determining-whether-module-is-free-for-polynomialpower-series-rings/?comment=61808#post-id-61808Shouldn'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 ?Mon, 04 Apr 2022 00:17:27 +0200https://ask.sagemath.org/question/61796/determining-whether-module-is-free-for-polynomialpower-series-rings/?comment=61808#post-id-61808