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Let $k$ be a field, then I want to prove the following statement: for every $P=(b_1,\ldots,b_n)\in K^n$, the ideal $\mathfrak{m}_P=(x_1-b_1,\ldots,x_n-b_n)$ is maximal in the polynomial ring $k[x_1,\ldots,x_n]$.

To prove this, I consider the evaluation map $$v_P:k[x_1,\ldots,x_n]\longrightarrow k$$ sending a polynomial $f(x_1,\ldots,x_n)$ to $f(b_1,\ldots,b_n)$. Then $v_P$ is a surjective morphism of rings. So we have that the quotient of $k[x_1,\ldots,x_n]$ by the kernel of $v_P$ is isomorphic to $k$, which is a field, thus is a field itself and $\ker v_P$ is maximal. So we are left to prove that $\mathfrak{m}_P=\ker v_P$. One of the inclusions is obvious, by definition of $\mathfrak{m}_P$. On the other side, I don't know how to prove that $\ker v_P$ is contained in $\mathfrak{m}_P$.

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  • $\begingroup$ @YACP I really don't know....it is surjective, hence $\ker v_P$ is maximal, and if i knew $ker v_P\subseteq \mathfrak{m}_P$ i could conclude for the equality, but the inclusion i know is the opposite one.... $\endgroup$ Commented Feb 28, 2013 at 11:32
  • $\begingroup$ @YACP No, but i only know that $\ker v_P$ is maximal....maximality of $\mathfrak{m}_P$ is actually what i'm going to prove $\endgroup$ Commented Feb 28, 2013 at 11:38
  • $\begingroup$ Then use a $K$-automorphism of $K[X_1,\dots,X_n]$ and reduce the problem to the case $b_i=0$ for all $i$. $\endgroup$
    – user26857
    Commented Feb 28, 2013 at 11:48
  • $\begingroup$ @YACP In the case $P=(0,\ldots,0)$, the map $v_P$ sends a polynomial to its constant term, thus the kernel is the set of polynomial with zero constant term, which coincides with the set of polynomial divisible by $X_i$ for all $i$, so i think it's okay, thank you $\endgroup$ Commented Feb 28, 2013 at 11:59

3 Answers 3

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Let $\varphi:K[X_1,\dots,X_n]\to K[X_1,\dots,X_n]$ defined by $\varphi(X_i)=X_i+b_i$ for all $i$. Then $\varphi$ is a $K$-automorphism of $K[X_1,\dots,X_n]$ and $\mathfrak m_P=(X_1-b_1,\dots,X_n-b_n)$ is maximal iff $\varphi(\mathfrak m_P)$ is maximal. But $\varphi(\mathfrak m_P)=(X_1,\dots,X_n)$. It remains to prove that $(X_1,\dots,X_n)$ is maximal. Now use the evaluation homomorphism $K[X_1,\dots,X_n]\to K$ sending $X_i$ to $0$ for all $i$. Then its kernel coincide to $(X_1,\dots,X_n)$. Why? Since $f(0,\dots,0)=0$ iff $f\in(X_1,\dots,X_n)$.

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As $x_i + \mathfrak{m}_P = b_i +\mathfrak{m}_P, \ f(x_1,\cdots, x_n) + \mathfrak{m}_P = f(b_1, \cdots, b_n) + \mathfrak{m}_P.$ If $f\in \ker v_P,$ $f\in \mathfrak{m}_P.$

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An idea: avoiding Hilbert's Nullstellensatz and stuff, you can try a direct approach:

$$g\in\ker v_P\Longrightarrow g(b_1,...,b_n)=0$$

Let us look at things as $\,t(x_n):=g(b_1,\ldots,b_{n-1})(x_n)\in k(b_1,\ldots,b_{n-1})[x_n]\,$ . Thus, the polynomial $\,t(x_n)\,$ has $\,b_n\,$ as one of its roots, so using the division lemma for the (rational functions) field$\,k(b_1,...,b_{n-1})\,$ , we get

$$g(b_1,...,b_{n-1})(x_n)=t(x_n)=(x_n-b_n)r(x_n)\;\;,\;\;r(x_n)\in k(b_1,...,b_n)[x_n]$$

Try now, perhaps, some inductive argument here.

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