Does there exist a (noetherian) commutative ring $R$ and an element $a \in R$ such that $a$ is a square in every localization of $R$ but $a$ itself is not a square?
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4$\begingroup$ Do you mean, 'localization at every maximal ideal'? Because localizing at the units gives the ring R again. $\endgroup$– Greg MullerCommented Sep 6, 2010 at 19:54
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1$\begingroup$ I mean localization at every prime ideal, though it is probably equivalent to consider localization at every maximal ideal. $\endgroup$– John VoightCommented Sep 6, 2010 at 20:11
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1$\begingroup$ If you meant localization at every maximal ideal, then the element $u$ in the ring $\mathbb{C}[u,v]/(uv-1)$ would be a counter-example. Localization at primes will be harder to break since, if my ring is a domain, it injects into the localization at the prime $(0)$. $\endgroup$– David E SpeyerCommented Sep 6, 2010 at 20:37
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$\begingroup$ David: you mean it would be an example! :) $\endgroup$– KConradCommented Sep 6, 2010 at 20:41
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$\begingroup$ Oh, wait, I'm sorry. That doesn't work. It would be square in the completion of the local ring, but not in the local ring itself. $\endgroup$– David E SpeyerCommented Sep 6, 2010 at 20:49
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3 Answers
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OK, I've got it. There is no such local criterion for squareness.
Let $k$ be a field of characteristic not $2$. Take the ring of triples $(f,g,h) \in k[t]^3$, subject to the conditions that $f(1)=g(-1)$, $g(1)=h(-1)$ and $h(1)=f(-1)$. Consider the element $(t^2,t^2,t^2)$. If this were a square, its square root would have to be $(\pm t, \pm t, \pm t)$. But two of those $\pm$'s would be the same sign, and $t$ evaluated at $1$ and at $-1$ are not equal.
Now, to check that $(t^2, t^2, t^2)$ is everywhere locally a square. Geometrically, we are talking about three lines glued into a triangle. Any prime ideal has a neighborhood which is contained in the union of two neighboring lines, say the first two. On the first two lines, $(-t, t, 1)$ is a square root of $(t^2, t^2, t^2)$.
For the suspicious, an algebraic proof. Set $u_1=(0, (1+t)/2, (1-t)/2)$ and let $u_2$ and $u_3$ be the cyclic permutations thereof. We have $u_1+u_2+u_3=1$ so, in any local ring, one of the $u_i$ must be a unit. WLOG, suppose that $u_1$ is a unit. Notice that $u_1 (1, -t, t)^2 = u_1 (t^2, t^2, t^2)$. So, in a local ring where $u_1$ is a unit, $(1,t,-t)$ is a square root of $(t^2, t^2, t^2)$.
answered Sep 6, 2010 at 21:54
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1$\begingroup$ A conceptual comment, to explain the geometric perspective. If $X$ is a space with nontrivial $\pi_1$, then there can be nonvanishing functions on $X$ which have no square root, because their square root lives on a double cover of $X$. That's why I tried
$\mathbb{C}^*$at first. But$\mathbb{C}^*$is non-simply-connected in the analytic topology; it is simply connected in the Zariski topology. That's why there were square roots in the completed local rings, not the local rings themselves. So I switched to something which is not simply connected even in the Zariski topology. $\endgroup$ Commented Sep 7, 2010 at 11:49 -
$\begingroup$ Thanks Dave! It's funny because the (spectrum of the) ring $R=\{ (f,g,h) \in k[t]^3 : f(1)=g(-1), g(1)=h(-1), h(1)=f(-1)\}$ is connected (in other words, $R$ has no nontrivial idempotents). You could also note that $R$ is indeed noetherian, since this isn't completely obvious. $\endgroup$ Commented Sep 7, 2010 at 15:05
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I would try something like this: $R=\mathbb C[x,y,u,v]/(f,g)$ with $f=x(y-u^2)$ and $g=(1+x)(y-v^2)$
When you localize at any prime ideal, you have to invert either $x$ or $1+x$. Either way, $y$ becomes a square. The only way to make $y$ a square in $R$ is to find $a,b,h$ such that $af+bg=y-h^2$. This looks unlikely, but I am too lazy to do the work.
EDIT: (this does not work) here is probably an easier example using the same idea $R=\mathbb Z[x]/(f,g)$ with $f=x(x-4)$ and $g=(1+x)(x-1)$. Then by the same argument $x$ is a square in all localizations. In $R$, the best you can get is $4x=1$.(Oops: $x=4x^2$!)
answered Sep 6, 2010 at 20:44
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$\begingroup$ I think this doesn't work, although I also haven't checked everything. This variety has four components:
$(x=0,\ y=v^2)$,$(x-1,\ y=u^2)$,$(u=v,\ y=u^2)$and$(u=-v,\ y=u^2)$. Each of these is isomorphic to $\mathbb{A}^2$. I think that the components meet with normal crossings, but I haven't checked carefully. If so, anything quadruple of polynomials which agrees on the overlaps is an element of $R$. (continued...) $\endgroup$ Commented Sep 6, 2010 at 21:24 -
$\begingroup$ In particular, consider the element which is $v$, $u$, $u$ and $(2x-1)u$ on these four components respectively. If I am right about the normal crossings, this is in $R$. And its square is $y$. $\endgroup$ Commented Sep 6, 2010 at 21:27
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$\begingroup$ I like this idea though. I'll see if I can modify it. $\endgroup$ Commented Sep 6, 2010 at 21:28
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$\begingroup$ @David: looks like you are right. I still wonder if one can find an example with 2 charts (yours has 3)? $\endgroup$ Commented Sep 6, 2010 at 22:42
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$\begingroup$ Hmmm. Maybe pairs of polynomials with $f(1)=g(1)$ and $f(-1)=g(-1)$, and look for a square root of $(t^2, t^4)$? $\endgroup$ Commented Sep 6, 2010 at 23:11
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Asher Auel pointed out to me another example in a paper of Colliot-Thelene, Parimala, and Suresh: http://arxiv.org/pdf/0812.3099.pdf
It is the claim in the appendix. I don't think it's so far in spirit from David Speyer's example, but it is may be more convenient for certain geometric settings.
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$\begingroup$ Well... except that the reference you gave deals with elements that are squares in each completion (as opposed to each localization). I would say that that's a big difference. For example, David Speyer's example was not irreducible. It's unclear to me whether there exist irreducible counterexamples (to your original question, i.e. the one formulated in terms of localizations). $\endgroup$ Commented Oct 25, 2015 at 19:57
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$\begingroup$ Yes, it is a difference; the matter of whether or not this is a "big" difference depends on the context. Personally, I found it helpful to see the example and how it was used, YMMV. That being said, I also do not know if there is an irreducible counterexample to the original question. $\endgroup$ Commented Oct 26, 2015 at 1:10