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From Serge Lang's Algebra:

## Theorem 9.1.1

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Let $\phi:k\to L$ be an embedding of k into an algebraically closed field $L$. Then there exists an extension of $\phi$ to a homomorphism $\bar{\phi}: k[x] \to L$.

Note: The $x_i$ are not indeterminates.

## Corollary 9.1.2 (Zariski)

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. If $k[x]$ is a field, then $k[x]$ is algebraic over $k$.

## Corollary 9.1.3

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Fix a finite family of elements $(y_i)_{i=1}^m$ of $k[x]$. If $k[x]$ is an integral domain, there exists a homomorphism $\psi:k[x]\to k^a$, where $k^a$ is the algebraic closure of $k$ such that $\psi(y_i)\neq 0$ for all $1\leq i\leq m$.

## Theorem 9.1.4 (Weak Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Then either $\mathfrak{q}$ is the unit ideal, or $\mathfrak{q}$ has a zero in $k^a$.

## Theorem 9.1.5 (Hilbert's Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Let $f\in k[X]$ be a polynomial vanishing on every zero of $\mathfrak{q}$ in $k^a$. Then there exists $m>0$ such that $f^m\in \mathfrak{q}$.

The proof of 9.1.5 follows from the Rabinowitsch trick and 9.1.4, which in turn follows directly from 9.1.2, which is a straighforward application of 9.1.1.

There is an advantage to this proof because it allows us not only to extend the definition of a variety to non-algebraically closed fields, but also to define an algebraic space, which is defined as a functor that takes field extensions of the basefield $k$ to the zero set of the ideal in that extension, with some nice properties. (This is not standard terminology, but I believe that every functor arising this way is in fact an algebraic space.)

In Eisenbud's commutative algebra book, the Nullstellensatz is generalized further from fields to Jacobson rings, which are rings for which any prime ideal is an intersection of some family of maximal ideals (This is a theorem of Bourbaki).

None of these proofs uses Noether normalization.

Corollary 9.1.2 is a lemma of Zariski that he introduced to prove the Nullstellensatz. I believe Lang's proof of Theorem 9.1.1 is similar to Zariski's proof of 9.1.2.

This is Zariski's paper where he introduced the method used by Lang. If you read the introduction, it's really interesting, because all of the previous proofs had been somewhat nontrivial. This was somewhat groundbreaking for proofs of the Nullstellensatz.

Additionally this lemma of Zariski is a special case of Zariski's main theorem for commutative rings. The Nullstellensatz follows with basically no effort. However, Zariski's main theorem is highly nontrivial.

5 added 179 characters in body

From Serge Lang's Algebra:

## Theorem 9.1.1

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Let $\phi:k\to L$ be an embedding of k into an algebraically closed field $L$. Then there exists an extension of $\phi$ to a homomorphism $\bar{\phi}: k[x] \to L$.

Note: The $x_i$ are not indeterminates.

## Corollary 9.1.2 (Zariski)

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. If $k[x]$ is a field, then $k[x]$ is algebraic over $k$.

## Corollary 9.1.3

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Fix a finite family of elements $(y_i)_{i=1}^m$ of $k[x]$. If $k[x]$ is an integral domain, there exists a homomorphism $\psi:k[x]\to k^a$, where $k^a$ is the algebraic closure of $k$ such that $\psi(y_i)\neq 0$ for all $1\leq i\leq m$.

## Theorem 9.1.4 (Weak Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Then either $\mathfrak{q}$ is the unit ideal, or $\mathfrak{q}$ has a zero in $k^a$.

## Theorem 9.1.5 (Hilbert's Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Let $f\in k[X]$ be a polynomial vanishing on every zero of $\mathfrak{q}$ in $k^a$. Then there exists $m>0$ such that $f^m\in \mathfrak{q}$.

The proof of 9.1.5 follows from the Rabinowitsch trick and 9.1.4, which in turn follows directly from 9.1.2, which is a straighforward application of 9.1.1.

There is an advantage to this proof because it allows us not only to extend the definition of a variety to non-algebraically closed fields, but also to define an algebraic space, which is defined as a functor that takes field extensions of the basefield $k$ to the zero set of the ideal in that extension, with some nice properties. (This is not standard terminology, but I believe that every functor arising this way is in fact an algebraic space.)

In Eisenbud's commutative algebra book, the Nullstellensatz is generalized further from fields to Jacobson rings, which are rings for which any prime ideal is an intersection of some family of maximal ideals (This is a theorem of Bourbaki).

None of these proofs uses Noether normalization.

Corollary 9.1.2 is a lemma of Zariski that he introduced to prove the Nullstellensatz. I believe Lang's proof of Theorem 9.1.1 is similar to Zariski's proof of 9.1.2.

Additionally this lemma of Zariski is a special case of Zariski's main theorem for commutative rings. The Nullstellensatz follows with basically no effort. However, Zariski's main theorem is highly nontrivial.

4 edited body

From Serge Lang's Algebra:

## Theorem 9.1.1

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Let $\phi:k\to L$ be an embedding of k into an algebraically closed field $L$. Then there exists an extension of $\phi$ to a homomorphism $\bar{\phi}: k[x] \to L$.

Note: The $x_i$ are not indeterminates.

## Corollary 9.1.2 (Zariski)

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. If $k[x]$ is a field, then $k[x]$ is algebraic over $k$.

## Corollary 9.1.3

Let $k$ be a field, and let $k[x]:=k[x_1,...,x_n]$ be a finitely generated $k$-algebra. Fix a finite family of elements $(y_i)_{i=1}^m$ of $k[x]$. If $k[x]$ is an integral domain, there exists a homomorphism $\psi:k[x]\to k^a$, where $k^a$ is the algebraic closure of $k$ such that $\psi(y_i)\neq 0$ for all $1\leq i\leq m$.

## Theorem 9.1.4 (Weak Nullstellensatz)

Let $k$ be a field, and let $k[X]:=k[X_1,...,X_n]$ be the polynomial ring in $n$ indeterminates over $k$.

Let $\mathfrak{q}$ be an ideal of $k[X]$. Then either $\mathfrak{a}$ \mathfrak{q}$is the unit ideal, or$\mathfrak{q}$has a zero in$k^a$. ## Theorem 9.1.5 (Hilbert's Nullstellensatz) Let$k$be a field, and let$k[X]:=k[X_1,...,X_n]$be the polynomial ring in$n$indeterminates over$k$. Let$\mathfrak{q}$be an ideal of$k[X]$. Let$f\in k[X]$be a polynomial vanishing on every zero of$\mathfrak{q}$in$k^a$. Then there exists$m>0$such that$f^m\in \mathfrak{q}$. The proof of 9.1.5 follows from the Rabinowitsch trick and 9.1.4, which in turn follows directly from 9.1.2, which is a straighforward application of 9.1.1. There is an advantage to this proof because it allows us not only to extend the definition of a variety to non-algebraically closed fields, but also to define an algebraic space, which is defined as a functor that takes field extensions of the basefield$k\$ to the zero set of the ideal in that extension, with some nice properties. (This is not standard terminology, but I believe that every functor arising this way is in fact an algebraic space.)

In Eisenbud's commutative algebra book, the Nullstellensatz is generalized further from fields to Jacobson rings, which are rings for which any prime ideal is an intersection of some family of maximal ideals (This is a theorem of Bourbaki).

None of these proofs uses Noether normalization.

Additionally, Lang's 9.1.2 is a special case of Zariski's main theorem for commutative rings. The Nullstellensatz follows with basically no effort. However, Zariski's main theorem is highly nontrivial.

3 added 111 characters in body