when $(f+h)^{2}=f^2+h^2$ in a function field setting of characteristic $p\neq2$

Question

I have arrived at a point, where I need to show $(f+h)^{2}$ is equal to $f^2+h^2$, is there any hint. I assume $f,\in \mathbb F_{q}[t]$, $h$ is any polynomial in $\mathbb F_q[t]$ of the form $\sum_{i=0}^{a}a_it^i$, and $a_i$ are variables assuming values from $\mathbb F_q$

Answer 1

Since $(f + h)^2 = f^2 + 2fh + h^2$ and your characteristic is not $2$, this is equivalent to $f h = 0$. Since you're working in an integral domain, this means $f = 0$ or $h = 0$.

The question becomes a bit more interesting if you wonder when $(f + h)^2$ and $f^2 + h^2$ represent the same function from ${\mathbb F}_q$ to itself, i.e., when $(f(\alpha) + h(\alpha))^2 = f(\alpha)^2 + h(\alpha)^2$ for all $\alpha \in {\mathbb F}_q$. It is still equivalent to $fh$ being $0$, but now as functions, i.e., to $f(\alpha) = 0$ or $h(\alpha) = 0$ for all $\alpha \in {\mathbb F}_q$. An obvious example is $f = t$ and $h = t^{q-1} - 1$; here $f(0) = 0$ and $h(\alpha) = 0$ for all $\alpha \in {\mathbb F}_q \setminus \{0\}$.

This is in a certain sense the typical example. Look at the ideal $$I = \{g \in {\mathbb F}_q[t] \;|\; \forall \alpha \in {\mathbb F}_q\colon g(\alpha) = 0\}$$ of all polynomials over ${\mathbb F}_q$ vanishing on ${\mathbb F}_q$. It is generated by the single polynomial $$t^q - t = \prod_{\alpha\in{\mathbb F}_q} (t - \alpha).$$ So, $fh$ being the $0$-function is equivalent to $$t^q - t \;|\; fh,$$ which means you have to distribute the factors $t - \alpha$, $\alpha \in {\mathbb F}_q$, of $t^q - t$ over $f$ and $h$. This, of course, again boils down to $f(\alpha) = 0$ or $h(\alpha) = 0$ for all $\alpha \in {\mathbb F}_q$.

Answer 2

Hint use the perfect square formula $f^2+2fh+h^2=f^2+h^2$ so $fh=0$ after cancelation since $k[x]$ is an integral domain when $k$ is a field what can you conclude?