0

Let $K$ be an infinite field, then I’d like to show that the morphism $\phi : K[x_1, \ldots, x_n] \rightarrow K$ such that $ P \mapsto ev(P)$ (mapping polynomials to polynomial functions) is injective. It is suggested in this answer to do this by induction, but I’m stuck in $n=1$:

If $n=1$, $P\in \ker(\phi)$ iff $P(k)=0$ ($k\in K$) iff $x_1-k$ divides $P(x_1)$, but I don’t see how this implies that $P=0$. Any help would be appreciated.

Rodrigo de Azevedo
  • 23,223
Yamido
  • 43
  • 4
    The morphism should be $\phi : K [x_1, \dots , x_n] \to K^{(K^n)}$, not $\phi : K [x_1, \dots , x_n] \to K$ (the latter is not injective indeed). For $n=1$ you have that a nonzero polynomial has only finitely many zeroes, hence the kernel has only the zero polynomial. – Crostul Sep 23 '22 at 17:48
  • @Crostul Just making sure, what do you mean by the notation $K^{(K^n)}$ exactly? – Yamido Sep 23 '22 at 17:55
  • @Crostul “you have that a nonzero polynomial has only finitely many zeroes…” Oh, and then you’d have a contradiction because $k$ could be any element of $K$ and $K$ is an infinite field? – Yamido Sep 23 '22 at 17:57
  • 1
    You got it. Here $K^{(K^n)}$ denotes the set of functions $K^n \to K$, which is a $K$-vector space. – Crostul Sep 23 '22 at 18:00
  • @Crostul All is clear, thank you! – Yamido Sep 23 '22 at 18:00

1 Answers1

1

The base case comes from the following well known fact: if $P, Q$ are polynomials of degree at most $m$, and $P(a)=Q(a)$ for $m+1$ different points $a$, then $P=Q$. You should try to prove this (hint: consider the polynomial $R = P - Q$).

Fred T
  • 674
  • 3
  • 11