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Let $R:=\mathbb{C}[X_1,\dots, X_n]$, $a=(a_1,\dots, a_n)\in \mathbb{C}^n$ and $\phi_a:R\rightarrow \mathbb{C}$, $\phi_a(f)=f(a)$. I want to show that $\ker(\phi_a)=(X_1-a_1,\dots, X_n -a_n)$.

I know that to be true for $n=1$ and I also know that $\ker(\phi_a)$ is a maximal ideal in $R$. Moreover, $(X_1-a_1,\dots, X_n -a_n) \subseteq \ker(\phi_a)$ is obvious. I am note sure if I can do an induction in this case. Thank you.

user26857
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Zolf69
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1 Answers1

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Without loss of generality, you can assume that $a=0 \in \mathbb C^n$.

Then you want to show that if $f(0,0,\cdots,0)=0$, then $f \in \langle X_1,\cdots, X_n \rangle= \mathfrak m$.

Write $f$ as a sum of monomials, and observe that the constant terms must vanish. This implies that $f \in \mathfrak m$.

Fredrik Meyer
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  • Okay, thank you. Why isn't generality lost? – Zolf69 Nov 02 '14 at 13:59
  • @Zolf69 Because you can always find an isomorphism $f:\mathbb C[X_1,\cdots,X_n] \to \mathbb C[X_1,\cdots,X_n]$ sending the origin to any other point. Just sends $X_i \mapsto X_i - a_i$. – Fredrik Meyer Nov 02 '14 at 14:00