Irreducibility of polynomials in $\mathbb{F}_p[x][y]$ Suppose we have a polynomial $P(x,y)$ in $\mathbb{F}_p[x][y]$ that is irreducible as a polynomial in $y$. I guess that $P(x^p,y)$ is also irreducible as a polynomial in $y$. How can I prove it?
Thank you in advance!
Edit, after Kenta's comment: What if I add the condition that
there exists an $n$ not divisible by $p$
such that the coefficient of $y^n$ in $P$
is nonzero?
 A: Assume the field characteristic $p>0$.
Suppose ${\bar x}^p=x$ and $P(x,\bar y)=0$ for some element $\bar x$ and $\bar y$ in some fixed algebraic closure of $\mathbb F_p(x)$, where  $P(x,y)\in\mathbb{F}_p[x][y]$ is irreducible as a polynomial in $y$ with coefficient in $\mathbb F_p(x)$. 

Is $P(x^p, y)$ also irreducible as a polynomial in $y$ with coefficient in $\mathbb F_p(x)$? In other words, is $P(x, y)$ is also irreducible as a polynomial in $y$ with coefficient in $\mathbb F_p(\bar x)$?
As Atticus Stonestrom remarked, it is not necessarily true. For example, $p=2$, $P(x,y)=y^2-x = y^2-{\bar x}^2=(y-\bar x)^2$.

Suppose there exists an $n$ not divisible by $p$ such that the coefficient of $y^n$ in $P$ is nonzero. Then $P(x, y)$ is also irreducible as a polynomial in $y$ with coefficient in $\mathbb F_p(\bar x)$.
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Proof:  Since the coefficient of $y^n$ in $P(x,y)$ is nonzero for some $n$ not divisible by $p$, $P'_y(x,y)\not=0$. Hence $\mathbb{F}_p(x)[\bar y]\e\mathbb{F}_p(x)$ is a separable extension of $\deg_y(P)$.
Since degrees of separability are multiplicative in field extensions, the degree of separability of $\mathbb{F}_p(\bar x)[\bar y]\e\mathbb{F}_p(x)$ is at least that of  $\mathbb{F}_p(x)[\bar y]\e\mathbb{F}_p(x)$, $\deg_y(P)$.
Extension $\mathbb{F}_p(\bar x)[\bar y]\e\mathbb{F}_p(x)$ is

*

*extension $\mathbb{F}_p(\bar x)\e\mathbb{F}_p(x)$, which is a pure inseparable extension,

*followed by extension $\mathbb{F}_p(\bar x)[\bar y]\e\mathbb{F}_p(\bar x)$.

Since degrees of separability are multiplicative in field extensions, the degree of separability of $\mathbb{F}_p(\bar x)[\bar y]\e\mathbb{F}_p(\bar x)$ is the same as the degree of separability of $\mathbb{F}_p(\bar x)[\bar y]\e\mathbb{F}_p(x)$.
The degree of separability of $\mathbb{F}_p(\bar x)[\bar y]\e\mathbb{F}_p(\bar x)$ is at most the degree of $\bar y$ over $\mathbb F_p(\bar x)$, which is at most $\deg_y P$ since $P(x, \bar y)= 0$.
Hence the degree of $\bar y$ over $\mathbb F_p(\bar x)$ is $\deg_y P$. So $P(x, y)$ is irreducible over $\mathbb F_p(\bar x)$. $\quad\checkmark$.
