Hasse principle for quadratic forms over finitely generated fields Does the Hasse principle hold for quadratic forms over finitely generated fields (e.g. for the Henselisations/completions at height-$1$-primes or all places)?
 A: The examples that I wrote in the comments are wrong.  Thanks to Felipe Voloch for catching my mistake.  One can, indeed, form the symbol algebras that I indicated.  However, the Brauer-Hasse-Noether(-Albert) theorem from class field theory states that these symbol algebras are actually matrix algebras.  In these terms, the Brauer-Hasse-Noether theorem together with Hasse's Global Structure Theorem (injectivity on the left for the "reciprocity short exact sequence") implies that a Severi-Brauer variety over the function field $\mathbb{F}_q(B)$ of a curve has a point if and only if it has local points in the completions for all places.  So my examples in the comments were maximally wrong.  Sorry for the slip!
Nonetheless, there are examples of dimension $2$ or $3$ that are everywhere unramified at height one primes.  For instance, the Artin-Mumford threefolds are smooth, projective $3$-dimensional varieties that admit a conic fibration over a rational surface and such that there exists an everywhere unramified conic bundle over the threefold whose corresponding Brauer class is nonzero.  The nontriviality of the Brauer class is "witnessed" by geometric computations on codimension $2$ subvarieties of the threefold, so testing with prime ideals of height $>1$ should detect this failure of the Hasse principle.  
The original construction of the Artin-Mumford examples was in characteristic $0$ (if memory serves).  As usual, the smooth, projective $3$-fold $B$, the everywhere smooth conic bundle $C\to B$, and the "witness" subvarieties of $B$ can all be defined over a finitely generated integral domain $R$ that contains $\mathbb{Z}$.  Thus, there exists an integer $p_0$ such that for all primes $p$ that are greater than $p_0$, for every field $k$ of characteristic $p$, for every ring homomorphism $R\to k$, the base change $B_k$ will be smooth, the conic bundle $C_k\to B_k$ will be everywhere smooth, and the witnessing subvarieties behave well and prove that there is no rational section of $C_k\to B_k$.
Does this conic bundle $C_k\to B_k$ satisfy the Hasse principle?  If the restriction of $C_k$ over every surface in $B_k$ has a rational point, then by Hensel's Lemma, there is a local point in the corresponding completion of $k(B_k)$.  In that case $C_k\to B_k$ violates the Hasse principle (for height one primes).  
On the other hand, assume that there exists a surface $B'_k\subset B_k$ over which $C'_k = B'_k\times_{B_k} C_k$ has no rational point. Replace $B_k$ by $B'_k$, replace $C_k$ by $C'_k$, and start over (if you like, you can use resolution of singularities of surfaces to make $B'_k$ smooth).  Now the Hasse principle is definitely violated, since the Brauer-Hasse-Noether theorem tells us that the restriction of $C_k$ over every curve in $B_k$ has a rational section.  
So, one way or another, we get violations of the Hasse principle for conic bundles over a threefold or over a surface over a finite field $k$.
