Non-zero polynomial of two variables over field

Question

Let $k$ be a field, $p$-prime number and $P(x,y), Q(x,y)\in k[x,y]$ and consider polynomial $$yQ(x,y^p)-P(x,y^p).$$ I claim that this polynomial is a non-zero polynomial over $k$.

Proof: Consider the following two cases:

1) Suppose that $Q(x,y^p)$ depends on $y$ then $yQ(x,y^p)$ has term $y^{pk+1}$ with $k\geq 1$. But note that $P(x,y^p)$ has not such term. Hence $yQ(x,y^p)-P(x,y^p) \not\equiv 0$.

2) If $Q(x,y^p)$ does not on $y$ then $Q(x,y^p)=F(x)$. Then $yQ(x,y^p)-P(x,y^p)=yF(x)-P(x,y^p)$. But the last polynomial is alzo non-zero because here we also can consider two subcases:

2.1 If $P(x,y^p)$ depends on $y$ then $P(x,y^p)$ has term $y^{pk}$, $k\geq 1$ which $yF(x)$ has not.

2.2. If $P(x,y^p)$ does depend on $y$ then $yF(x)$ has term $y$ which $P(x,y^p)$ has not.

Is this reasoning correct? Would be grateful for any comments.

Answer 1

With the added constraint that $Q(x, y)$ is not identically zero, the reasoning is fine but could be streamlined. It must have at least one non-zero term, which without loss of generality is $a x^b y^c$. Then $yQ(x, y^p)$ has a term $a x^b y^{cp+1}$. But the terms of $P(x, y^p)$ (if there are any - $P$ can be identically zero) are of the form $d x^e y^{pf}$, and we can only have integer solutions to $cp+1 = pf$ if $p = 1$.

Answer 2

You might want to be clearer in what you mean by "a term $y^n$"; is $xy^n$ also "a term $y^n$"? As you're working over the polynomial ring $k[x,y]$ I would say it isn't, but your usage seems to imply it is. Viewing $Q$ and $P$ as polynomials in $L[y]$, where $L:=k(x)$, removes this ambiguity.

Also, your reasoning fails at the very last step; if $F=0$ then $yF(x)$ has no term $y$ in either sense, and correspondingly $P=0$ actually yields a solution; indeed $Q=P=0$ is a solution.

A more concise phrasing of the same argument could be:

If $P,Q\in k[x,y]$ are such that the polynomial is zero, then either $P=Q=0$, or both are non-zero, in which case $$\deg_y yQ(x,y^p)\equiv1\pmod{p} \qquad\text{ and }\qquad \deg_y P(x,y^p)\equiv0\pmod{p},$$ so $yQ(x,y^p)\neq P(x,y^p)$, a contradiction. So if $P$ or $Q$ is non-zero, then so is the polynomial.