In this post I set out to prove an equation holding in a ring $R$ graded over $\mathbb{Z}$ or $\mathbb{N}$.

The equation was: $\sqrt{J^\ast}=(\sqrt{J})^\ast$, where $J$ is an ideal of $R$, $J^\ast$ is the subideal of $J$ generated by homogeneous elements, and $\sqrt{J}$ denotes the intersection of all prime ideals containing $J$.

The containment $\supseteq$ is obviously valid for a grading over any semigroup.

The other containment is what I want to ask about. Let $x=\sum x_g$ be the decomposition of $x$ into homogeneous elements. As you can see, my proof pivots on being able to say that at least one of the homogeneous terms of $x^n$ is just $(x_g)^n$. Using this, a contradiction can be produced. In the case of that post, it was possible to isolate such a term because the well behaved ordering of $\mathbb{Z}$ and $\mathbb{N}$.

So my question is:

Does the equation break down for gradings over general semigroups?

If anybody can see how the proof can be streamlined to avoid that problem, or if anybody has an example of the equation breaking down, I would love to see. Thanks!

  • $\begingroup$ Why does this question fall so flat?! :P $\endgroup$ – rschwieb Jul 3 '12 at 14:08
  • $\begingroup$ @YACP I would very much like to know what nonsense you are spouting now about answer-copying. If I ever knowingly transfer someone else's work, I always give full credit, as I did below. $\endgroup$ – rschwieb Nov 26 '12 at 16:57
  • $\begingroup$ @YACP People will see the length of my post and the six minute separation and correctly conclude that you edited your post before I finished editing mine. Even if I had been aware of the edit you made, it would hardly be worth copying. I would have simply stopped writing my own post and saved the time. Anyhow, I think we've about spent this thread. I'll do my best to forget it. I wish you would try too, but your comments make it sound like you intend to be sore forever. Happy editing! $\endgroup$ – rschwieb Nov 26 '12 at 17:58

If $\sqrt{J^\ast}=(\sqrt{J})^\ast$ for all ideals $J$ of $R$, in particular we get that the radicals of homogeneous ideals of $R$ are also homogeneous. But, in general, this is false and a very simple example can be found here.

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  • $\begingroup$ Thanks for the source. For the benenfit of future readers, I have a request, though. Since comments are a little more mutable than answers, and the example is so simple, would you mind sketching it here? I would edit it in myself but I'm not in the habit of making radical edits. $\endgroup$ – rschwieb Nov 25 '12 at 14:26
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    $\begingroup$ That's OK, I think I'll leave radical improvements to your answers up to you. $\endgroup$ – rschwieb Nov 26 '12 at 0:30
  • $\begingroup$ I've removed a couple of off-topic comments $\endgroup$ – robjohn Jan 3 '13 at 0:26

For the benefit of future readers, the examples hidden in the comments of the link provided by YACP are reproduced here.

The first example was given by darij grinberg:

$R=\Bbb{F}_2[x]/(x^2+1)$ can be graded with the additive group $\Bbb{Z_2}$ giving $x$ degree $\overline{1}$. It is clear that $x+1$ is in the radical of the homogeneous ideal $\{0\}$.

Later on in the comments a characteristic zero example is given by quim:

"Let $N$ be the monoid formed by the nonnegative integers and an element $i$ such that $i+n=n+1$ for all $n>0$. Give to $K[x,y]$ the grading in which elements of $K$ have degree 0, $x$ has degree 1 and $y$ has degree $i$ . Then $(x^2+2xy+y^2)$ is homogeneous of degree 2 and its radical $(x+y)$ is not homogeneous."


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