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I am reading about pseudo algebraically closed fields. The definition states that a field $K$ is PAC if for every absolutely irreducible polynomial $f(x,y)\in K[x,y]$ there are infinitely many points $(a,b)\in K^{2}$ s.t. $f(a,b)=0$.

I saw an example stating $\mathbb{R}$ is not PAC since $$ f(x,y)=y^{2}+x^{2}+1\in\mathbb{R}[x,y] $$ is absolutely irreducible but there is a finite (actually $0$) elements $(a,b)\in\mathbb{R}^{2}$ s.t. $f(a,b)=0$.

While it is easy to see that $f(a,b)\neq0$ for $(a,b)\in\mathbb{R}^{2}$ I don't understand why it is absolutely irreducible.

Wikipedia gives two examples: $$ x^{2}+y^{2}-1\in\mathbb{R}[x,y] $$ is also claimed to be absolutely irreducible, but $$ x^{2}+y^{2}\in\mathbb{R}[x,y] $$ admits the factorization $$ x^{2}+y^{2}=(x+yi)(x-yi) $$ thus is not absolutely irreducible.

My questions are:

  1. If $K$ is field there are many irreducibility criteria of a polynomial $f\in K[x]$, are there any criteria for polynomials $f\in K[x,y]$ ? (I intend to apply them over the algebraic closure of $K$.)

  2. How can I see that $f(x,y)=y^{2}+x^{2}$and $p(x,y)=x^{2}+y^{2}-1$ are absolutely irreducible as claimed ?

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The polynomials you are considering all have the form $F(x,y):=y^n-\phi(x)$, $\phi\in K[x]$. For such polynomials one can say:

  1. $F$ is irreducible in $K[x,y]$ iff it is irreducible in $k(x)[y]$ (as a polynomial in $y$).

This follows from the Lemma of Gauss.

  1. Assume that $K$ contains the $n$-th roots of unity and let $\phi(x):=\prod\limits_{i=1}^r p_i(x)^{e_i}$ be the prime factorization of $\phi$ in $K[x]$. Then the polynomial $F$ is irreducible in $k(x)[y]$ iff for every prime divisor $q$ of $n$ there exists $e_k$ not divisible by $q$.

Proof: $\Rightarrow$: let $\alpha$ be a root of $F$ as a polynomial in $y$. Then the extension $K(x,\alpha)|k(x)$ by assumption has degree $n$. Assume that $e_i=qf_i$ for every $i$ for some prime divisor of $n$. Then $\alpha^{n/q}=\prod\limits_{i=1}^r p_i(x)^{f_i}$ - contradiction.

$\Leftarrow$: let $\alpha$ be a root of $F$ as a polynomial in $y$. Then by assumption all roots of the minimal polynomial of $\alpha$ over $K(x)$ lie in $K(x)$ and the minimal polynomial has the form $y^d-a$, $a\in k[x]$, where $d$ divides $n$ - see Ch. VIII, Para. 6, Thm. 6.2. in Lang. Since all roots of $F$ have the form $\alpha\zeta$, where $\zeta\in K$ is an $n$-th root of unity, and since $a$ divides $\phi$, the latter being the product of all roots of $F$, one gets that $a$ and $\phi$ must have the same prime divisors in $K[x]$. Therefore $d=n$ under the assumptions made about the exponents of the prime divisors.

One can now apply point 2 to a polynomial $F\in K[x,y]$ but considering it as a polynomial over the algebraic closure: for example all polynomials $y^n-\phi(x)$, where $\phi$ is separable, are absolutely irreducible. This includes the cases $y^2+x^2+1$ and $y^2+x^2-1$ over any field $K$ of characteristic different from $2$.

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  • $\begingroup$ Thank you for posting an answer, I think there is a counter example for your claim: Take $K=\mathbb{R}$ and $\phi=-x^{2}$ which is separable as every irreducible factor is separable. $n=2$ and the $n$-th roots of unity are $\pm1\in\mathbb{R}$. $F(x,y)=y^{2}-\phi(x)=y^{2}+x^{2}=(y+ix)(y-ix)$ is a factorization over $\mathbb{C}$ so that $F(x,y)$ is not irreducible as claimed. Did I miss something ? $\endgroup$ – Belgi Nov 2 '15 at 19:14
  • $\begingroup$ To be separable in my post means to have no multiple roots in the algebraic closure. $\endgroup$ – Hagen Knaf Nov 4 '15 at 20:55

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