Show that $\mathfrak m_x\subseteq k[T_1,..., T_n]$ is a maximal ideal for $k$ algebraically closed

Question

I am looking for an easy argument to show that for $x:=(x_1,..., x_n)\in k^n$ and $k$ an algebraically closed field $$\mathfrak m_x:=(T_1-x_1,... , T_n-x_n)\subseteq k[T_1,..., T_n]$$ is a maximal ideal.

Note that I am trying to show this in order to prove Hilbert's Nullstellensatz, so I can't use it in the answer.

Answer 1

It's the kernel of the evaluation map $k[T_1,\dots,T_n]\to k$, defined by $P\mapsto P(x_1,\dots,x_n)$. The image of this map is $k$ which is a field, so the kernel is maximal ideal.

Note that this has nothing to do with $k$ being algebraically closed. What the Nullstellensatz says is that all maximal ideals are of this form if $k$ is algebraically closed.

Answer 2

The map $\varphi\colon k[T_1,\dots,T_n]\to k[T_1,\dots,T_n]$ given by $T_i\mapsto T_i+x_i$ is a ring isomorphism and $\varphi(\mathfrak{m}_x)=\mathfrak{m}_0$, so it's sufficient to see that $\mathfrak{m}_0$ is maximal.

Consider $\psi\colon k[T_1,\dots,T_n]\to k$ given by $\psi(T_i)=0$. It is clear that $\mathfrak{m}_0\subseteq\ker\psi$ and that $\ker\psi$ is maximal.

Let $f(T_1,\dots,T_n)\in\ker\psi$. Then the constant term in $f$ is $0$, so $f\in\mathfrak{m}_0$.