Talk:Pseudo algebraically closed field

I'm afraid the new lead does not make sense. The Nullstellensatz does not hold for any non-algebraically closed field. In fact, already the weak Nullstellensatz for univariate polynomials implies that the field is algebraically closed: if f is a non-constant polynomial, then I = f K[x] is a proper ideal in K[x], hence V(I) is nonempty, i.e., f has a root. — Emil J. 13:35, 26 November 2009 (UTC)

Examples

The comment (consequence of Riemann Hypothesis for function fields) (twice) seems to be nonsense. I have removed it. Lichfielder (talk) 08:49, 20 September 2012 (UTC)

It appears to be correct, at least as far as the ultraproduct example is concerned. See Fried & Jarden (2004) p.217. Deltahedron (talk) 18:32, 21 September 2012 (UTC)

Formulation

I don't quite follow the statement If $R$ is a finitely generated integral domain over $K$ with quotient field which is regular over $K$, then there exist a homomorphism $h:R\to K$ such that $h(a)=a$ for each $a\in K$. The field K = C is algebraically closed and so PAC. The field C(X) is a regular extension of C (any extension of an AC field is regular), and hence R = C(X) is a finitely generated integral domain with itself as field of fractions. But there is no C-homomorphism of C(X) to C. Is there a reference for this claim? Deltahedron (talk) 07:30, 13 November 2012 (UTC)

While I know nothing about the claim or any references, C(X) is not finitely generated over C as a ring. C[X] is, but then there are plenty of evaluation homomorphisms.—Emil J. 11:39, 13 November 2012 (UTC)

Yes, my mistake — finitely generated as a field but not as a ring. But a reference would still be good. Deltahedron (talk) 18:14, 13 November 2012 (UTC)