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I need a hint for this exercise in Lang's Algebra, Chapter 5, #28.

Let $f$ be a homogeneous degree 2 polynomial in $n$ variables over a field $k$. If $f$ has a non-trivial zero in an extension of odd degree then $f$ has a non-trivial zero in $k$.

I guess my trouble is that I am unsure how to deal with the roots of multi-variable polynomials. Most of Lang's Chapter 5 deals with single variable polynomials over a field.

Mykie
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2 Answers2

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Lang has been known to cloak theorems in the form of exercises, and this is an instance of this. What you are asking about is an elementary but important result in quadratic form theory, called Springer's Theorem.

(Beware: there is at least one other important result in this area called "Springer's Theorem". The other Springer's Theorem is Theorem 7 in these notes: it's a bit more technical.)

A quadratic form $q(x_1,\ldots,x_n)$ over a field $K$ is called anisotropic if for any vector $v \in K^n$ other than the zero vector, $q(v) \neq 0$. Then the contrapositive, hence equivalent, form of Lang's exercise is:

Theorem (Springer): Let $q$ be an anisotropic quadratic form over $K$, and let $L/K$ be an odd degree field extension. Then viewed as a quadratic form over $L$, $q$ remains anisotropic.

A proof of Springer's Theorem appears (e.g.) in $\S 3$ of these notes on quadratic forms. Despite the fact that this is almost 40 pages into a systematic study of the algebraic theory of quadratic forms, the proof should be self-contained provided the terminology is explained, which I hope I have done above.

Remark: A purely field theoretic result which can be viewed as a special case of Springer's Theorem is: if $K$ is a formally real field -- i.e., $-1$ is not a sum of squares -- and $L/K$ is a finite extension of odd degre, then $L$ is also formally real. (Indeed, a field is formally real iff for all $n$, the quadratic form $x_1^2 + \ldots + x_n^2$ is anisotropic.) This result appears in $\S 15.3$ of my field theory notes, and indeed the proof is essentially the same as the proof of Springer's Theorem.

Pete L. Clark
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  • Unfortunately, all the links are defunct now. I find a nice proof of the theorem in section 1.5 of B. Kahn's Formes Quadratiques Sur Un Corps, together with a little historical note acknowledging it had been conjectured from special cases by Witt, and apparently Artin had communicated a proof orally some years before Springer. (If one would start calling it "Artin-Springer theorem", one would avoid confusion with the other Springer's Theorem.) – Torsten Schoeneberg Jan 14 '21 at 18:42
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    @Torsten: Every few years my university level IT makes some change that unwittingly breaks a bunch of my links. This time around, all links that include math.uga.edu can be fixed by putting alpha.math.uga.edu instead. – Pete L. Clark Jan 22 '21 at 01:31
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    @PeteL.Clark: When I change trains at a train station, is that a "dévissage argument"? :) Nice proof (of Theorem 19)! I haven't seen this "lift to the polynomial ring and jump to the other factor" tactic before; it sounds like it could be useful elsewhere. – darij grinberg Mar 11 '22 at 10:57
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Here is a proof for two variables. So let $\mathbb L$ be an odd-degree extension of $\mathbb K$, and $f(x,y)=Ax^2+Bxy+Cy^2$. If $f$ has a non-trivial zero in ${\mathbb L}^2$, then by homogeneity either $g(x)=f(x,1)$ or $h(y)=f(1,y)$ has a zero in $\mathbb L$. Say it’s $g$. Then $g$ has a root $\alpha\in {\mathbb L}$. The degree $[{\mathbb K}(\alpha):{\mathbb K}]$ divides the odd number $[{\mathbb L}:{\mathbb K}]$ and is at most $2$, so $\alpha \in \mathbb K$ ; then $(\alpha,1)$ is a non-trivial zero for $f$ in $\mathbb K$.

Ewan Delanoy
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  • Could you elaborate on "then by homogeneity either $g(x)=f(x,1)$ or $h(y)=f(1,y)$ have a non-trivial zero in $\mathbb L$"? – JSchlather Feb 19 '13 at 06:10
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    @JacobSchlather Let $(x_0,y_0)$ be a non-trivial zero of $f$ : $f(x_0,y_0)=0$. Then by hypothesis, one of $x_0$ and $y_0$ is nonzero. If, say, $x_0 \neq 0$, then $f(1,\frac{y_0}{x_0})=\frac{1}{x_0^2}f(x_0,y_0)$ shows that $\frac{y_0}{x_0}$ is a zero for $g$. – Ewan Delanoy Feb 19 '13 at 06:26
  • @JacobSchlather : by the way, the scrap you just quoted was badly worded, I edited it. – Ewan Delanoy Feb 19 '13 at 06:32
  • Thanks I was thinking along the same lines but I didn't notice that you can divide by $x_{0}$ or $y_{0}$. – Mykie Feb 19 '13 at 21:17
  • I don't see how induction can be applied. How would you use the result for two variables to prove the result for three variables? –  Sep 23 '18 at 09:23
  • @Brahadeesh I wrote this a long time ago and I can't remember what my initial argument was. So I removed the claim "this follows by induction" from my answer. – Ewan Delanoy Sep 24 '18 at 14:46