IMC 2008 · Day 1 · P2

We can assume that $P \ne 0$.

Let $f \in V$ be such that $P(f) \ne 0$. Then $P(f^2) \ne 0$, and therefore $P(f^2) = a P(f)$ for some non-zero real $a$. Then $0 = P(f^2 - af) = P(f(f - a))$ implies $P(f - a) = 0$, so we get $P(a) \ne 0$. By rescaling, we can assume that $P(1) = 1$. Now $P(X + b) = 0$ for $b = -P(X)$. Replacing $P$ by $\hat{P}$ given as $$\hat{P}(f(X)) = P(f(X + b))$$ we can assume that $P(X) = 0$.

Now we are going to prove that $P(X^k) = 0$ for all $k \ge 1$. Suppose this is true for all $k < n$. We know that $P(X^n + e) = 0$ for $e = -P(X^n)$. From the induction hypothesis we get $$P \bigl( (X + e)(X + 1)^{n-1} \bigr) = P(X^n + e) = 0,$$ and therefore $P(X + e) = 0$ (since $P(X + 1) = 1 \ne 0$). Hence $e = 0$ and $P(X^n) = 0$, which completes the inductive step. From $P(1) = 1$ and $P(X^k) = 0$ for $k \ge 1$ we immediately get $P(f) = f(0)$ for all $f \in V$.