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I am studying algebraic curves but I have no background of commutative algebra.An important theorem in this topic is the weak Nullstellensatz which states that:

Any maximal ideal of $K[X_1,...,X_n]$ where $K$ is algebraically closed is of the form $\langle X_1-a_1,...,X_n-a_n\rangle$.

This follows from Zariski's lemma.The statement is as follows:

If the field $L$ is a finitely generated $K$-algebra where $K\subset L$ is a subfield,then $L/K$ is a finite extension,hence algebraic.

Where the definition of a finitely generated $K$-algebra is as follows:

Any quotient of the form $K[X_1,X_2,...,X_n]/I$ for some $n\in \mathbb N$ (and some ideal $I$ of $K[X_1,...,X_n]$) is called a finitely generated $K$-algebra.

Now I am looking for a proof of Zariski's lemma which does not use much of commutative algebra.But first I want some intuition about what we are trying to do and why this theorem should be true and what it really means because just by looking at the statement,it is not possible for me to understand its significance.I would also like some analogy with some known fact from field theory.Can someone provide me with an answer?

Kishalay Sarkar
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    I don't think there is much intuition here, this is indeed a surprising result. Its significance is clear though, as many times we can say a field extension is finite when we know very little about that extension. For example, if $m$ is a maximal ideal of $F[x_1,...,x_n]$ then the quotient $F[x_1,...,x_n]/m$ is a finite (and hence algebraic) extension of $F$, which is very nontrivial. Not sure there is a proof which avoids using at least some commutative algebra though. After all, this theorem is more about ring theory than field theory. – Mark Jan 26 '23 at 13:52

2 Answers2

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Zariski's lemma and hence the weak Nullstellensatz has a "cheap proof" in the special case that $K$ is uncountable, which is one of the most relevant cases in practice anyway because it covers $K = \mathbb{C}$.

Here is a sketch, details left as exercises: let $L/K$ be a field extension where $K$ is uncountable. Then $L$ is either algebraic or transcendental. If $L$ is transcendental it has some transcendental element $t$. Now I claim that the elements $\frac{1}{t - a}, a \in K$ of $L$ are all linearly independent over $K$; this follows from the uniqueness of partial fraction decomposition in $K(t)$ (exercise #1). Since $K$ is uncountable, there are uncountably many of these, and so $L$ has uncountable dimension over $K$.

But if $L$ is finitely generated over $K$ as a $K$-algebra then it must have at most countable dimension over $K$ (exercise #2). So in this case $L$ cannot contain an element transcendental over $K$, and so must be algebraic. Since it is finitely generated it is generated by a finite number of elements algebraic over $K$ which implies that it is finite-dimensional over $K$ (exercise #3).

So, the basic idea is that transcendental extensions are "too big" to be finitely generated as $K$-algebras. There is actually a way to deduce the general case from the special case that $K$ is uncountable, which you can see at this MO question.

Qiaochu Yuan
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See Theorem 2.11 in https://kconrad.math.uconn.edu/blurbs/ringtheory/maxideal-polyring.pdf. Some applications are in Section 3. An analogous result is proved in a similar way as Theorem 5.1: a field that is finitely generated as a $\mathbf Z$-algebra (such as $\mathbf Z[x,y]/M$ where $M$ is a maximal ideal) is a finite field.

KCd
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