Irreducibility after substitution I would like to show that when $f(x,y)$ is irreducible over $\mathbb{C}[x,y]$ then $f(x^2,y)$ is irreducible over $\mathbb{C}[x,y]$.  I know that this is not true in general, for example, $f(x,y) = y^2 - x$.  However, I know that the original polynomial contains only even powers of $x$.  Does this extra condition allow me to conclude $f(x^2,y)$ is irreducible over $\mathbb{C}[x,y]$?  
 A: The answer is yes.  (In what follows, I write $f(x^2,y)$ for your $f(x,y)$.)
1)  By symmetry, if $h(x,y)$ is an irreducible factor of $g(x^2,y)$ then so is $h(-x,y)$.  
2)  In particular, if $h(x,y)$ is an irreducible factor of $f(x^4,y)$ then so is $h(-x,y)$.
3)  Write $f(x^4,y)$ as a product of irreducible factors.  Let $h(x,y)$ be such a factor.  
3a)  Suppose that $h(x,y)=\alpha h(-x,y)$ for some constant $\alpha$.  Comparing coefficients, we see that either $\alpha=1$ and $h$ is an even function of $x$ or $\alpha=-1$ and $h$ is an odd function of $x$.
3aa) Suppose $\alpha=1$ and $h$ is an even function of $x$.  Then $h(\sqrt{x},y)$ is a polynomial factor of $f(x^2,y)$, hence --- by the irreducibility assumption ---$h(\sqrt{x},y)=f(x^2,y)$ up to a constant.  Thus $h(x,y)=f(x^4,y)$ up to a constant, and the latter is irreducible.
3ab)  Suppose $\alpha=-1$ and $h$ is an odd function of $x$.  Then $h$ is divisible by $x$, hence by irreducibility we can assume $h(x,y)=x$.  But if $x$ divides $f(x^4,y)$ then the latter has no constant term in $x$, so $x$ divides $f(x^2,y)$, a contradiction.  [Note:  My original post overlooked this case.  Many thanks to Emil Jerabek for pointing this out and keeping me honest.]
3b)  Suppose instead that $h(x,y)$ is not a constant factor of $h(-x,y)$.  Write $H(x^2,y)=h(x,y)h(-x,y)$.  Then $f(x^4,y)/H(x^2,y)$ is a polynomial in $x^2$ and $y$, so (1) applies and we proceed by induction to write $f(x^4,y)$ as a product of factors $H_i(x^2,y)$.  
4)  It follows that $f(x^2,y)$ is the product of the $H_i(x,y)$.  By assumption $f(x^2,y)$ is irreducible, so there is only one $H_i$.  
5)  Thus $f(x^4,y)=h(x,y)h(-x,y)$ where $h$ is irreducible.  
6)  If $h$ contains terms of odd degree in $x$, let $d$ be the smallest of those odd degrees.  Then $f(x^4,y)$ contains a term of degree $2d$ in $x$, a contradiction.  So  all terms of $h$ are of even degree in $x$.
7)  Thus $f(x^2,y)=h(\sqrt{x},y)h(-\sqrt{x},y)$ is a polynomial factorization, contradicting the assumption that $f(x^2,y)$ is irreducible.
A: Maybe the paper http://math.univ-lille1.fr/~pde/ptmy.pdf by Pierre Debes is relevant. In particular, see Lemma 0.1. 
