Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathfrak{m}_P$.

share|improve this question
    
@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.... –  Federica Maggioni Feb 28 '13 at 11:32
    
@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 –  Federica Maggioni Feb 28 '13 at 11:38
    
Then use a $K$-automorphism of $K[X_1,\dots,X_n]$ and reduce the problem to the case $b_i=0$ for all $i$. –  user26857 Feb 28 '13 at 11:48
    
@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 –  Federica Maggioni Feb 28 '13 at 11:59

3 Answers 3

up vote 2 down vote accepted

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 $f\in(X_1,\dots,X_n)$.

share|improve this answer

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 $f\in \ker v_P,$ $f\in \mathfrak{m}_P.$

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.