# Proving that kernels of evaluation maps are generated by the $x_i - a_i$

I want to prove that the kernel of the evaluation map $s_a : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto a_i$ where $a = (a_1,\dots, a_n) \in \mathbb{C}^n$ is the ideal generated by $\{x_1 - a_1, \dots, x_n - a_n\}$.

The proof in the book first shows it easily for the case $a = 0$, and says that by substitutions of variables $x_i' = x_i - a_i$, you can show it for the rest of the cases. I know that $f \in \ker s_a$ iff $f(a) = 0$. And $\phi : \mathbb{C}[x_1,\dots, x_n] \rightarrow \mathbb{C}[x_1 - a_1, \dots, x_n-a_n], x_i \mapsto x_i - a_i$ is a surjective ring homomorphism. Please give a hint and not the full answer. Thanks.

• Hint: It certainly contains the $x_i-a_i$. What kind of ideal is $(x_1-a_1,\ldots,x_n-a_n)$? – Alex Youcis Sep 21 '13 at 6:56
• You mean it's a maximal ideal? But that is proved first knowing that it's a kernel of a ring-hom to a field. I haven't proved that the kernel is indeed it. – StudySmarterNotHarder Sep 21 '13 at 6:59
• Okay, so you're saying that alternatively I can prove this by noticing that $\ker s_a$ definitely contains the ideal generated the $x_i - a_i$ and since the latter is also a maximal ideal, they coincide. Hmmm... – StudySmarterNotHarder Sep 21 '13 at 7:02
• Correct. It's obvious that $(x_1-a_1,\ldots,x_n-a_n)$ is maximal. – Alex Youcis Sep 21 '13 at 7:02
• I don't understand how that is obvious. – StudySmarterNotHarder Sep 21 '13 at 7:05

## 2 Answers

First of all $\mathbb{C}[x_1 - a_1, \dots, x_n-a_n]=\mathbb{C}[x_1,\dots, x_n]$ and I think in the book you are reading it must be $$\phi : \mathbb{C}[x_1,\dots, x_n] \rightarrow \mathbb{C}[x_1, \dots, x_n], x_i \mapsto x_i - a_i.$$ This is easily seen an isomorphism. You also have two maps: $s_a : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto a_i$ whose kernel you want to determine and $s_0 : \mathbb{C}[x_1,\dots,x_n] \rightarrow \mathbb{C}, x_i \mapsto 0$ whose kernel you already know. Now check that $s_a\circ\phi=s_0$ and let's find out the kernel of $s_a$: $f\in\ker s_a$ iff $s_a(f)=0$ iff $s_0\circ\phi^{-1}(f)=0$ iff $\phi^{-1}(f)\in\ker s_0$ iff $f\in \phi(\ker s_0)$, so $$\ker s_a=\phi(\ker s_0).$$ Since we know that $\ker s_0=(X_1,\dots,X_n)$ it follows $$\ker s_a=\phi((X_1,\dots,X_n))=(\phi(X_1),\dots,\phi(X_n))=(X_1-a_1,\dots,X_n-a_n).$$

The rings $\mathbb{C}[x_i]$ and $\mathbb{C}[x_i - a_i] = \{f(x-a) : f\in\mathbb{C}[x_i]\}$ are equal for any $a\in \mathbb{C}^n$. Proof: for any $f \in$ the first ring, there exists $f(x+a)$ also in the first ring, and so $f(x + a - a) = f(x)$ exists in the second ring. The opposite inclusion is more clear. To answer the question, for any $a$, $f \in \ker s_a$ iff $f(a) = 0$. Choose $g (x-a) = f(x)$. Then $f(a) = 0$ iff $g(0) = 0$, which happens iff the const part of $g(x)$ is zero, so $f(x)$ equals a combination of multiples of the $(x_i - a_i)$, or in other words is in the ideal $(x_1 - a_1, \dots, x_n-a_n)$. Thus that ideal is the kernel which is maximal being the kernel of a ring homomorphism onto a field. QED