User jürgen böhm - MathOverflow

Answer by Jürgen Böhm for which homogeneous polynomials split into linear factors?

2012-10-10T22:38:54Z

The questions 1. to 3. can be answered by using the theory of Gröbner bases. Let

(1) $f(x_1,\ldots,x_d) = \sum a_{i_1\ldots i_d}\, x^{i_1} \cdots x^{i_d}$

with indeterminate coefficients $a_{i_1\ldots i_d}$.

Now assume there is a factorization

(2) $f(x_1,\ldots,x_d) = \prod_{i=1}^n (b_{i1} x_1 + \cdots + b_{id} x_d)$

again with indeterminate coefficients $b_{ij}$

Multipliying (2) out and equating coefficients of like monomials $x_1^{i_1} \cdots x_d^{i_d}$ in (1) and (2) gives a set of polynomials

(3) $G_\nu(\ldots,a_{i_1\ldots i_d},\ldots, b_{ij}, \ldots) = 0$

Eliminating from the $G_\nu$ all the $b_{ij}$ by Gröbner basis methods gives a set of equations $F_\mu(\cdots,a_{i_1\ldots i_d},\cdots) = 0$. These describe $S$ as an algebraic variety.

Milne, Etale Cohomology, mistake in proposition 2.5?

2012-08-09T01:06:53Z

In Milne, Etale Cohomology, Proposition 2.5 (§2) is stated as follows:

(All rings noetherian.)

Let $B$ be a flat $A$--algebra, and consider $b \in B$. If the image of $b$ in $B/\mathfrak{m} B$ is not a zero-divisor for any maximal ideal $\mathfrak{m}$ of $A$, then $B/(b)$ is a flat $A$--algebra.

Now consider $A = k[[x_1,x_2]]$, $\mathfrak{m}=(x_1,x_2)$ and $\mathfrak{p} = (x_1)$ as $A$-ideals and $B = A_\mathfrak{p}$. Then $B/\mathfrak{m} B = 0$. Let $b = x_1$. As $B/\mathfrak{m} B$ vanishes, it has no zero-divisors, so $b$ is not a zero-divisor there.

But $B/b B \neq 0$ is obviously not $A$-flat, as it does not preserve the injectiveness of $A \overset{\cdot b}{\hookrightarrow} A$.

Of course, one could object that implicitly all $B \otimes_A k(\mathfrak{m})$ should be taken nonzero, but in this case $B$ would be faithfully flat over $A$ so as to strongly narrow the scope of the proposition.

A comparison with Kurke, Pfister, Roczen, Henselsche Ringe und algebraische Geometrie led me to suppose, that the proposition should be worded slightly different, namely:

"$b$ is not a zero-divisor in $B \otimes_A k(A \cap \mathfrak{n})$ for all $\mathfrak{n} \subseteq B$, maximal"

(see, 1.4.5. Korollar, there)

Is the remark above correct?

Answer by Jürgen Böhm for Projective spaces as affine varieties

2012-07-10T14:25:46Z

The example ${\mathbb R}{\mathbb P}^1 = V(x^2+y^2 -1)$ seems compelling as are the other arguments, that real projective space can have nonconstant regular functions.

But what about Hartshorne, Algebraic Geometry, (II, Theorem 5.19) which says, that for $k$ a field, $A$ a finitely generated $k$-Algebra, $X$ a projective scheme over $A$ and $\mathcal F$ a coherent ${\mathcal O}_X$-module on $X$, the $A$-module $\Gamma(X,{\mathcal F})$ is finitely generated.

There seems to be no restriction to $k$ algebraically closed there, and also such an restriction is not made in (I, Theorem 3.9A) on which the proof of (II, 5.19) depends.

Comment by Jürgen Böhm

2012-03-08T14:35:20Z

Unfortunately you are right with your objection in the first comment: It can not be concluded from what was presumed, that H^i_x(V,F) = 0. So the claim (*) is wrong and, as the whole argument is based on it, also the rest of the proof. I consider the question as open again and apologize for not having read the cited Ex. 3.4 carefully enough. I will work further on it, but at the moment can not present a solution or correction.