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François G. Dorais
François G. Dorais
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I know a way to do this, but it involves some very heavy machinery...

The first component are effective bounds on the degrees of the polynomials in the conclusion of the Weak Nullstellensatz. Such bounds are not that easy to get and there has been a lot of literature on the Effective Nullstellensatz. Perhaps the earliest effective bounds were found by Grete Hermann Die Frage der endlich vielen Schritte in der Theorie der Polynomideale (Mathematische Annalen 95, 1926), but there has been a lot of work on improving these bounds and also obtaining lower bounds over the years. [E.g., D. W. Brownawell, Bounds for the degrees in the Nullstellensatz, Ann. of Math. (2) 126 (1987), 577-591] It's interesting to read these papers, but I will only use the fact that effective bounds do exist.

Using these bounds it is possible to find a sequence of first-order sentences $\phi_{n,k,r}$, which together are equivalent to the Weak Nullstellensatz; the sentence $\phi_{n,k,r}$ is a first order rendition of the following statement.

If $p_1(\bar{x}),\dots,p_k(\bar{x})$ ($\bar{x} = x_1,\ldots,x_r$) are polynomials of degree at most $n$ without common zeros, then there are polynomials $q_1(\bar{x}),\dots,q_k(\bar{x})$ of degree at most $b(n,k,r)$ such that $p_1(\bar{x})q_1(\bar{x})+\cdots+p_k(\bar{x})q_k(\bar{x}) = 1$.

The bounds $n$ and $b(n,k,r)$ are necessary so that the $p_i(\bar{x})$ and $q_i(\bar{x})$ have a bounded number of coefficients. Otherwise, we could not use a fixed number of variables for these coefficients.

That said, the other piece of heavy machinery is the fact that the theory of algebraically closed fields of a given characteristic is complete, i.e. every first-order sentence is decided by the axioms. Therefore, if the above sentences $\phi_{n,k,r}$ are true in any algebraically closed field of a given characteristic, then they must be true in all algebraically closed fields of the same characteristic. In particular, the Weak Nullstellensatz for $\mathbb{C}$ implies the Weak Nullstellensatz for all algebraically closed fields of characteristic zero.

From here, you can use the Rabinowitsch trick to get the Strong Nullstellensatz...

PS: You do not need the Nullstellensatz to prove that the theory of algebraically closed fields of a given characteristic is complete. You implicitly need the Nullstellensatz to prove the effective upper bounds, but you only need them for the one field and you can think of them as wild guesses that turn out to be right.

I know a way to do this, but it involves some very heavy machinery...

The first component are effective bounds on the degrees of the polynomials in the conclusion of the Weak Nullstellensatz. Such bounds are not that easy to get and there has been a lot of literature on the Effective Nullstellensatz. Perhaps the earliest effective bounds were found by Grete Hermann Die Frage der endlich vielen Schritte in der Theorie der Polynomideale (Mathematische Annalen 95, 1926), but there has been a lot of work on improving these bounds and also obtaining lower bounds over the years. It's interesting to read these papers, but I will only use the fact that effective bounds do exist.

Using these bounds it is possible to find a sequence of first-order sentences $\phi_{n,k,r}$, which together are equivalent to the Weak Nullstellensatz; the sentence $\phi_{n,k,r}$ is a first order rendition of the following statement.

If $p_1(\bar{x}),\dots,p_k(\bar{x})$ ($\bar{x} = x_1,\ldots,x_r$) are polynomials of degree at most $n$ without common zeros, then there are polynomials $q_1(\bar{x}),\dots,q_k(\bar{x})$ of degree at most $b(n,k,r)$ such that $p_1(\bar{x})q_1(\bar{x})+\cdots+p_k(\bar{x})q_k(\bar{x}) = 1$.

The bounds $n$ and $b(n,k,r)$ are necessary so that the $p_i(\bar{x})$ and $q_i(\bar{x})$ have a bounded number of coefficients. Otherwise, we could not use a fixed number of variables for these coefficients.

That said, the other piece of heavy machinery is the fact that the theory of algebraically closed fields of a given characteristic is complete, i.e. every first-order sentence is decided by the axioms. Therefore, if the above sentences $\phi_{n,k,r}$ are true in any algebraically closed field of a given characteristic, then they must be true in all algebraically closed fields of the same characteristic. In particular, the Weak Nullstellensatz for $\mathbb{C}$ implies the Weak Nullstellensatz for all algebraically closed fields of characteristic zero.

From here, you can use the Rabinowitsch trick to get the Strong Nullstellensatz...

PS: You do not need the Nullstellensatz to prove that the theory of algebraically closed fields of a given characteristic is complete. You implicitly need the Nullstellensatz to prove the effective upper bounds, but you only need them for the one field and you can think of them as wild guesses that turn out to be right.

I know a way to do this, but it involves some very heavy machinery...

The first component are effective bounds on the degrees of the polynomials in the conclusion of the Weak Nullstellensatz. Such bounds are not that easy to get and there has been a lot of literature on the Effective Nullstellensatz. Perhaps the earliest effective bounds were found by Grete Hermann Die Frage der endlich vielen Schritte in der Theorie der Polynomideale (Mathematische Annalen 95, 1926), but there has been a lot of work on improving these bounds and also obtaining lower bounds over the years. [E.g., D. W. Brownawell, Bounds for the degrees in the Nullstellensatz, Ann. of Math. (2) 126 (1987), 577-591] It's interesting to read these papers, but I will only use the fact that effective bounds do exist.

Using these bounds it is possible to find a sequence of first-order sentences $\phi_{n,k,r}$, which together are equivalent to the Weak Nullstellensatz; the sentence $\phi_{n,k,r}$ is a first order rendition of the following statement.

If $p_1(\bar{x}),\dots,p_k(\bar{x})$ ($\bar{x} = x_1,\ldots,x_r$) are polynomials of degree at most $n$ without common zeros, then there are polynomials $q_1(\bar{x}),\dots,q_k(\bar{x})$ of degree at most $b(n,k,r)$ such that $p_1(\bar{x})q_1(\bar{x})+\cdots+p_k(\bar{x})q_k(\bar{x}) = 1$.

The bounds $n$ and $b(n,k,r)$ are necessary so that the $p_i(\bar{x})$ and $q_i(\bar{x})$ have a bounded number of coefficients. Otherwise, we could not use a fixed number of variables for these coefficients.

That said, the other piece of heavy machinery is the fact that the theory of algebraically closed fields of a given characteristic is complete, i.e. every first-order sentence is decided by the axioms. Therefore, if the above sentences $\phi_{n,k,r}$ are true in any algebraically closed field of a given characteristic, then they must be true in all algebraically closed fields of the same characteristic. In particular, the Weak Nullstellensatz for $\mathbb{C}$ implies the Weak Nullstellensatz for all algebraically closed fields of characteristic zero.

From here, you can use the Rabinowitsch trick to get the Strong Nullstellensatz...

PS: You do not need the Nullstellensatz to prove that the theory of algebraically closed fields of a given characteristic is complete. You implicitly need the Nullstellensatz to prove the effective upper bounds, but you only need them for the one field and you can think of them as wild guesses that turn out to be right.

Source Link
François G. Dorais
François G. Dorais
  • 44.4k
  • 6
  • 150
  • 233

I know a way to do this, but it involves some very heavy machinery...

The first component are effective bounds on the degrees of the polynomials in the conclusion of the Weak Nullstellensatz. Such bounds are not that easy to get and there has been a lot of literature on the Effective Nullstellensatz. Perhaps the earliest effective bounds were found by Grete Hermann Die Frage der endlich vielen Schritte in der Theorie der Polynomideale (Mathematische Annalen 95, 1926), but there has been a lot of work on improving these bounds and also obtaining lower bounds over the years. It's interesting to read these papers, but I will only use the fact that effective bounds do exist.

Using these bounds it is possible to find a sequence of first-order sentences $\phi_{n,k,r}$, which together are equivalent to the Weak Nullstellensatz; the sentence $\phi_{n,k,r}$ is a first order rendition of the following statement.

If $p_1(\bar{x}),\dots,p_k(\bar{x})$ ($\bar{x} = x_1,\ldots,x_r$) are polynomials of degree at most $n$ without common zeros, then there are polynomials $q_1(\bar{x}),\dots,q_k(\bar{x})$ of degree at most $b(n,k,r)$ such that $p_1(\bar{x})q_1(\bar{x})+\cdots+p_k(\bar{x})q_k(\bar{x}) = 1$.

The bounds $n$ and $b(n,k,r)$ are necessary so that the $p_i(\bar{x})$ and $q_i(\bar{x})$ have a bounded number of coefficients. Otherwise, we could not use a fixed number of variables for these coefficients.

That said, the other piece of heavy machinery is the fact that the theory of algebraically closed fields of a given characteristic is complete, i.e. every first-order sentence is decided by the axioms. Therefore, if the above sentences $\phi_{n,k,r}$ are true in any algebraically closed field of a given characteristic, then they must be true in all algebraically closed fields of the same characteristic. In particular, the Weak Nullstellensatz for $\mathbb{C}$ implies the Weak Nullstellensatz for all algebraically closed fields of characteristic zero.

From here, you can use the Rabinowitsch trick to get the Strong Nullstellensatz...

PS: You do not need the Nullstellensatz to prove that the theory of algebraically closed fields of a given characteristic is complete. You implicitly need the Nullstellensatz to prove the effective upper bounds, but you only need them for the one field and you can think of them as wild guesses that turn out to be right.