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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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Do you mean, 'localization at every maximal ideal'? Because localizing at the units gives the ring R again. –  Greg Muller Sep 6 '10 at 19:54
I mean localization at every prime ideal, though it is probably equivalent to consider localization at every maximal ideal. –  John Voight Sep 6 '10 at 20:11
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 $(0)$. –  David Speyer Sep 6 '10 at 20:37
David: you mean it would be an example! :) –  KConrad Sep 6 '10 at 20:41
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. –  David Speyer Sep 6 '10 at 20:49

2 Answers 2

up vote 17 down vote accepted

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 $(t^2, t^2, t^2)$.

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This looks good, very nice! –  Hailong Dao Sep 6 '10 at 22:16
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. –  David Speyer Sep 7 '10 at 11:49
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. –  John Voight Sep 7 '10 at 15:05

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: $x=4x^2$!)

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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...) –  David Speyer Sep 6 '10 at 21:24
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 $y$. –  David Speyer Sep 6 '10 at 21:27
I like this idea though. I'll see if I can modify it. –  David Speyer Sep 6 '10 at 21:28
@David: looks like you are right. I still wonder if one can find an example with 2 charts (yours has 3)? –  Hailong Dao Sep 6 '10 at 22:42
Hmmm. Maybe pairs of polynomials with $f(1)=g(1)$ and $f(-1)=g(-1)$, and look for a square root of $(t^2, t^4)$? –  David Speyer Sep 6 '10 at 23:11

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