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 $\mathfrak{m}_P$.
Best Answer
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 $f\in(X_1,\dots,X_n)$.