Pseudo algebraically closed fields

Question

i am reading Fried-Jarden's book Field Arithmetic and I have a question on Pseudo Algebraically closed (PAC) fields.

We say that a field $K$ is PAC if every variety absolutely irreducible over $K$ has a $K$-rational point. Recall that a variety is absolutely irreducible if it is irreducible over $\overline{K}$, and a $K$-rational point is an element on the variety with all its coordinates in $K$.

My question is the following: Is $\mathbb{Q}$ a PAC field? And more generally, is any number field PAC?

My attempt: I would like to find a polynomial in two variables with coefficients in $\mathbb{Q}$ which is irreducible over $\mathbb{C}$ and that does not have any root in $\mathbb{Q}^2$. This would show that $\mathbb{Q}$ is not PAC, which is my guess, but so far i've been unable to find this example.

I am also guessing that, more generally, every number field is not PAC, although this question is much more difficult...

Thanks in advance!

Related: https://math.stackexchange.com/questions/1885165/fx-x2y21-is-irreducible-in-mathbbcx-y — Watson, May 19 '18 at 14:54

score 3 · Answer 1 · answered May 19 '18 at 14:44

3

I think that the projective conic with equation $$X^2+Y^2+Z^2=0$$ is absolutely irreducible over $\Bbb Q$.

answered May 19 '18 at 14:44

Angina Seng

158,341

But it has a root in $\mathbb{Q}^3$, the $(0,0,0)$. I found $x^2+y^2+1$. This works to prove that $\mathbb{Q}$ is not PAC. – El.Gon.Zalo May 21 '18 at 13:02
It has no point in $\mathbb P^2(\Bbb Q)$. @El.Gon.Zalo – Angina Seng May 21 '18 at 14:51

Pseudo algebraically closed fields

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