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Given a ring $R = k[x,y]$, what is the difference between $R$ as a $R[x]$ - module and $R$ as a $k[x]$ module. I just can't seem to comprehend the difference so any help would be appreciated.

user26857
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user901823
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    Do you mean $R$ as an $R$-module vs $R$ as a $k[x]$-module? It's not clear what you mean by $R[x]$. Taken literally, $R[x]=R$ since $x\in R$, but that may not be what you had in mind... – Eric Wofsey Feb 18 '18 at 19:02

2 Answers2

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If you think of the elements of a modules as "vectors", then the only difference is where the "scalars" come from.

  • If $R$ is an $R$-module (which is the same as an $R[x]$-module, since $x\in R$), then your "scalars" come from $R=k[x,y]$. In particular, this module is finitely generated $$R=\left <1\right >_R$$ that is, every element in $R$ can be written as $r\cdot 1$, with $r\in R$
  • Likewise, if $R$ is an $k[x]$-module, your scalars come from $k[x]$. In this case $R$ is not finitely generated, since $$R=\left < 1,y,y^2,\dots\right >_{k[x]}$$ that is, every element of $R$ can be written as either $r\cdot 1$, $r\cdot y$, $r\cdot y^2$, and so on, where $r\in k[x]$
cansomeonehelpmeout
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  • This is super helpful thank you. So when considering if R is Noetherian as a $k[x]$ module I need to consider submodules of R in terms of $y$? Which I think makes it neither? – user901823 Feb 19 '18 at 17:28
  • There may be submodules that contain more than $y$, e.g. $\left <x^2,y \right >_{k[x]}$, here every element is on the form $f\cdot x^2+g\cdot y$ with $f,g\in k[x]$, if I understood your question correct?

    Also $R$ is not finitely generated, as it has the basis (not just generating set) ${1,y,y^2,\dots}$, so $R$ cannot be Noetherian.

    This is a basis since $R\cong k[x]\times k[x]\times \dots$

     – cansomeonehelpmeout Feb 19 '18 at 17:42
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    You did understand it correctly, but I'm a little confused. I was going to suggest using the submodules that contain polynomials of degree at most $n$ in $y$, and since there are infinite powers of $y$ in the basis then this can be used to form an ascending chain that does not stabilise, is this not the case? – user901823 Feb 19 '18 at 17:49
  • That is very clever! And I think it will work! – cansomeonehelpmeout Feb 19 '18 at 17:54
  • Fab, thank you! Just one last thing, how would this change if I changed the ring to $k[x, y]/y^n$ for some $n$, as a $k[x]$-module this is now finitely generated so does this now mean it is Noetherian? – user901823 Feb 19 '18 at 17:58
  • Yes, I would think so. $M=k[x,y]/y^n=\left <1,y,y^2,\dots, y^{n-1} \right >_{k[x]}$ is finitely generated, and $k[x]$ is a Noetherian ring. So by this https://math.stackexchange.com/questions/533873/finitely-generated-modules-over-a-noetherian-ring-are-noetherian theorem $M$ is Noetherian. – cansomeonehelpmeout Feb 19 '18 at 18:04
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Well, what is the difference between $k[x]$ as a $k$-module (vector space) and as a $k[x]$-module? Answer: in the first case it is (free) of infinite countable dimension with basis $\{1,x,x^2,\ldots\}$, and in the second free of rank one with basis $\{1\}$.

Pedro
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