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I've been stuck on this for a few days.

I'm supposed to find an example of a continuous function $f$ (with values in the field) defined on an affine variety $V=V_1\cup V_2$ with two irreducible components, such that the restrictions to each component is regular, but $f^n$ is not regular for all $n\geq1$.

This doesn't exist in one dimension, so I tried finding such an example by playing with plane curves that are tangent, and considered various functions on each component like the restrictions of projections on different coordinate axes. but I had no success, other than finding a function that is not regular, but it's square is regular, so it's not enough (like the one described by Georges).

A follow-up question that interests me, if those two components are disjoint, would it then have to be regular? (Answer: yes, see Georges comment)

what if they intersect in sufficiently nice ways, would the function have to be regular? I just don't know where to start looking, so some criteria might help.

Feelix
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  • Yes, if the irreducible components are disjoint they are both open (since they are closed !) and you can glue arbitary regular functions on them to a global regular function on their union. – Georges Elencwajg Mar 06 '17 at 09:55
  • But it sounds to me it only guarantees continuity, I think some justification must be said about regularity – Feelix Mar 06 '17 at 12:40
  • Not at all: regularity is guaranteed without any further justification. This is just the fact that $\mathcal O$ is a sheaf, since $\mathcal O(U_1 \coprod U_2)=\mathcal O(U_1)\oplus \mathcal O(U_2)$ for disjoint open subsets $U_i$. By the way, your question would be much clearer if you introduced some mathematical notations/symbols in your text. – Georges Elencwajg Mar 06 '17 at 13:05
  • thanks, that is indeed reasonable. I tried adding some symbols now. – Feelix Mar 06 '17 at 13:10
  • "$f^n$ is not regular for all $n \geq 1$" <-- This is ambiguous. Do you mean "$\forall n$, $\lnot f^n$ is regular" or do you mean "$\lnot \forall n$, $f^n$ is regular"? – apt1002 Mar 14 '17 at 18:01
  • the initial option. though I should note that I don't know what to believe if there is such an example or not.

    if one can show it doesn't exist (i.e there always exists a power for which $f^n$ is regular) it is very welcome aswell.

     – Feelix Mar 14 '17 at 18:04

1 Answers1

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Let $V=V(Y(Y-X^2))\subset \mathbb A^2=\operatorname {Spec} k[X,Y]$ be the plane curve with irreducible components $V'=V(Y), V''=V(Y-X^2)$ and with ring of regular functions $k[x,y]=k[X,Y]/\langle Y(Y-X^2)\rangle$.
The function $x$ on $V'$ and $y$ on $V''$ glue to a continuous function $\phi:V\to k$ but
Claim:
The function $\phi$ is not regular on $V$.

Indeed, any regular function on $V$ can be written uniquely as $$f(x,y)=p(x)+yq(y)+xyr(x)\quad (\star)$$ for some polynomials in one variable $p,q,r$.
In order for $f(x,y)$ to have a restriction to $V'=V(y)$ equal to $x$ we must have $p(x)=x$ (set $y=0$ in $(\star)$) so that $$ f(x,y)=x+yq(y)+xyr(x)\quad (\star \star) $$
In order for $\phi$ moreover to restrict to $y$ on $V''=V(y-x^2)$ we must also have $x^2=x+x^2q(x^2)+x^3r(x)$ on $k[V'']=k[x]$ (set $y=x^2$ in $(\star \star) $).
But the equality $x^2=x+x^2q(x^2)+x^3r(x)$ is impossible in $k[x]$, so that $f$ does not exist and the claim is proved.

An explanation
If you know scheme theory, the reason for this phenomenon is that the scheme-theoretic intersection of $V'$ and $V''$ is not the origin but the origin with non reduced structure: $V'\cap V''=\operatorname {Spec} \frac {k[x]}{\langle x^2\rangle }$.
The functions $x$ on $V'$ and $y$ on $V''$ do not coincide on $S$: they are equal respectively to $x$ and $0$ and thus they cannot be glued to a regular function on $V$.

Georges Elencwajg
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  • About the example you gave, it is one I already found, I was specifically asking for a function that any of its powers aren't regular, and the function you provided squared is regular... I have mentioned it in the second paragraph – Feelix Mar 06 '17 at 12:38
  • Thank you for your elaboration thus far – Feelix Mar 06 '17 at 12:41
  • For clarity I suggest you rewrite your requirements by carefully giving names to the functions on the irreducible components and then writing what you require of the global function with respect to the two first ones. Also, your terminology "projection maps on each" doesn't make sense to me: do you mean restriction? – Georges Elencwajg Mar 06 '17 at 12:58
  • I meant the restriction on each irreducible component is the projection map. which is what you've done aswell. I will try to improve the phrasing. – Feelix Mar 06 '17 at 13:01
  • I wonder if you have any idea for the question after the clarifications? maybe there is no such function? – Feelix Mar 14 '17 at 15:43
  • I gave you the bounty as no one has answer, a shame that my lowly 50 reputation points would go to waste. – Feelix Mar 16 '17 at 07:10
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    Dear Feelix, unfortunately I don't know whether high enough powers of the given function might be regular. (And thanks for the bounty, but of course my wish to answer your question was quite independent of it...) – Georges Elencwajg Mar 16 '17 at 07:19