Checking irreducibility of a polynomial in $\mathbb{K}[x,y]$ and PAC fields 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:


*

*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 $K$.)

*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 ?
 A: 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:


*

*$F$ 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.


*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 $2$. 
