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I have a question from a book which I am trying to attempt.

Let $k$ be a field not of characteristic 2 and let $a\in k$ be not a square (i.e. for all $b\in k$, $b^{2}\neq a$). I want to show that $X=\mbox{Spec}(k[U,T]/(T^{2}-aU^{2}))$ is geometrically reduced and geometrically connected.

My attempt is as follows: I am hoping to show that if $\overline{k}$ denotes the algebraic closure of $k$, then $X_{\overline{k}}$ is reduced. Since we have $X_{\overline{k}}=\mbox{Spec}(k[U,T]/(T^{2}-aU^{2})\otimes_{k}\overline{k})=\mbox{Spec}(\overline{k}[U,T]/(T^{2}-aU^{2}))$, and I need to show that this is reduced. To do this, I will have to show that localisation of $A=\overline{k}[U,T]/(T^{2}-aU^{2})$ at any prime ideal is a reduced ring. I was actually hoping to show that $A$ is a reduced ring (since then localisation of a reduced ring is also reduced), but algebraically I am not sure how to do it.

I have no idea how to start on proving geometrically connectedness. Glad if someone can give me some hints.

Thanks! (Will update this page if I have ideas on proving geometrically connectedness)

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up vote 1 down vote accepted

a) Yes, the ring $A=\overline{k}[U,T]/(T^{2}-aU^{2})$ is reduced (and hence the scheme $X_\bar k$ is reduced) since the ideal $(T^{2}-aU^{2})\subset \overline{k}[U,T]$ is equal to its nilradical.
This is because in the prime factorization $T^{2}-aU^{2}=(T-\sqrt a U)(T+\sqrt a U)$ there are no repeated factors.

b) Geometric connectedness is clear geometrically (!) because $\bar X=\operatorname {Spec} A$ is the union of the two intersecting (connected !) lines $V(T-\sqrt a U), V(T+\sqrt a U) \subset \mathbb A^2_\bar k\operatorname ={Spec} (\overline{k}[U,T])$ .

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Hi @Georges, thanks for answering! Just want to confirm: for (a) is it because $\overline{k}[U,T]$ is UFD, so whenever $(T^{2}−aU^{2})$ divides say $(P(U,T))^{n}$, then $(T^{2}−aU^{2})$ divides $P(U,T)$ (both $(T−\sqrt{a}U)$ and $(T+\sqrt{a}U)$ are distinct irreducible polynomials)... for b, it seems like a play of words that the definition of "geometric connectedness" has to do with "drawing lines in a suitable field extension and see geometrically if they are connected". is this also true for reduced, irreducible etc as well heuristically? – enoughsaid05 Nov 9 '13 at 22:42
Dear enoughsaid05: a) In a UFD a principal ideal $(f)$ is equal to its nilradical if and only if $f$ has no repeated factors in its decomposition into irreducibles. b) There is no play of words but a precise (albeit elementary) result in general topology: if a topological space is the union of two intersecting connected subsets, then it is itself connected. – Georges Elencwajg Nov 9 '13 at 22:49

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