Without using Hilbert's Nullstelensatz , can we directly prove that all maximal ideals of $\mathbb C[x_1,x_2,...,x_n]$ is of the form $\langle x-a_1,x-a_2,...,x-a_n \rangle$ ? It is easy to prove it for $n=1$ . If it cannot even be proved for general $n$ variables , I would really like to know a direct proof (if any ) for two variables i.e. for $\mathbb C[x,y]$ . Please help . Thanks in advance
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Are you ready to accept weak nullstallensatz? If yes the Rabinowitch's trick is cute and easy. http://en.wikipedia.org/wiki/Rabinowitsch_trick – baharampuri May 28 '15 at 05:19
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Here is a proof, which works over $\mathbb C$ (due to the complex numbers being not countable):
If $\mathfrak m$ is a maximal ideal in $R=\mathbb C[x_1, \dotsc, x_n]$, then the quotient $R/\mathfrak m$ has countable degree over $\mathbb C$, since $R$ itself is a vector space of countable dimension over $\mathbb C$.
The smallest transcendental extension $\mathbb C(X)$ has uncountable degree over $\mathbb C$, since the elements $\frac{1}{X-c},c \in \mathbb C$ are linear independent.
We deduce that $R/\mathfrak m$ is algebraic over $\mathbb C$, hence $R/\mathfrak m = \mathbb C$. This implies the result by considering the map
$$\varphi: \mathbb C[x_1, \dotsc, x_n] \twoheadrightarrow \mathbb C[x_1, \dotsc, x_n]/\mathfrak m = \mathbb C$$
and noting that $x_i-\varphi(x_i)$ are contained in $\ker(\varphi)=\mathfrak m$, in particular $(x_1-\varphi(x_1), \dotsc, x_n-\varphi(x_n)) \subset \mathfrak m$. Equality follows since both ideals are maximal.
