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(Weak form of) Hilbert's nullstellensatz: In $\mathbb{C}[x_1,\cdots,x_n]$, every maximal ideal is of the form $(x_1-a_1,\cdots,x_n-a_n)$ for some $(a_1,\cdots,a_n)\in\mathbb{C}^n$.

Proof: (1) The set $\{\frac{1}{x-a}:a\in\mathbb{C}\}$ is linearly independent in $\mathbb{C}(x)$. (This is true if $\mathbb{C}$ is replaced by any field; am I right?)

(2) Let $M$ be a maximal ideal in $\mathbb{C}[x_1,\cdots,x_n]$.

(3) Let $\pi_1: \mathbb{C}[x_1]\rightarrow \mathbb{C}[x_1,\cdots,x_n]/M$, $f(x_1)=x_1+M$ be natural ring homomorphism.

(4) Suppose if possible $\ker\pi_1=0$. Then $\mathbb{C}[x_1]$ is isomorphic to subring of (quotient) field, so it should be field, i.e. $\mathbb{C}[x_1]=\mathbb{C}(x_1)$. Now

  • $\mathbb{C}(x_1)$ is of uncountable dimension over $\mathbb{C}$ (by (1)).

  • But $\mathbb{C}[x_1,\cdots,x_n]/M$ is of countable dimension over $\mathbb{C}$, so $\mathbb{C}(x_1)$ has countable dimension over $\mathbb{C}$.

(5) Thus in (4) $\ker\pi_1$ is non-zero.

(6) Then one shows that $\ker\pi_1=M\cap \mathbb{C}[x_1]$ contains some polynomial $x_1-a_1$; similarly, we obtain $M$ contains $x_i-a_i$ for some $a_i\in\mathbb{C}$ ($i=1,2,\cdots,n$). This completes proof.


Question: The weak form is till valied if $\mathbb{C}$ is replaced by any algebraically closed field. Take $\overline{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$. Then $\overline{\mathbb{Q}}$ is countable, so $\{\frac{1}{x-a}:a\in\overline{\mathbb{Q}}\}$ has countable dimension; thus in (4) we do not arrive directly at contradiction, and so above proof doesn't work.

My question is can we modify the above proof for $\overline{\mathbb{Q}}$ instead of $\mathbb{C}$?

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Just because (the image of) $x_1$ is invertible in $\mathbb C[x_1,\ldots,x_n]/M$, this doesn't imply that this inverse can be written as a polynomial in $x_1$.

Rather, the argument is that since $\mathbb C[x_1]$ embeds via $\pi_1$ into the field $\mathbb C[x_1,\ldots,x_n]/M$, this embedding extends to an embedding of $\mathbb C(x_1) into $\mathbb C[x_1,\ldots,x_n]/M$.

Now the dimension argument applies.


If we replace $\mathbb C$ by a countable field such as $\overline{\mathbb Q}$, then this dimension argument doesn't apply, and one needs a more subtle argument.

(A similar situation occurs in other contexts where one can make a countability argument over $\mathbb C$, e.g. with Quillen's Lemma, which follows from the same sort of dimension argument with $\mathbb C$ coefficients, but needs a more subtle argument when the coefficient field is countable.)

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