I'm having trouble coming up with a sufficiently patriarchal argument. Does "these parts" refer to the pre-suffrage era?

Oops, I guess I was supposed to submit a single piece. I'd say "Oeuvres de Serre" but it doesn't contain his books.

Yeah, see Lemma 1.2 and its proof in section 4.1 in Hartshorne. Your divisor has to be effective to get h^0 positive, and if it has degree 0, then it is the zero divisor. I believe this extends to all proper varieties.

I'd like to emphasize that I'm not necessarily looking for holistic, coherent answers. Fragmentary brain dumps are also welcome (and may receive up-votes).

I used to spend a lot of time looking at the K-theory archive (www.math.uiuc.edu/K-theory/) but I have yet to find enlightenment.

A group of permutations of a nonempty set is transitive if the set has one orbit, i.e., if for any pair of elements of the set, there is an element of the group taking one to the other.

I don't really think ordinary tensor products are boring. I should change it to a more judgment-neutral word.

As far as I can tell, 31427 was chosen because it is the smallest number whose square is larger than 987654321. This covers all cases where a,b, and c are nonnegative. I would not be surprised if there were additional solutions where a and b have opposite signs.

Minor corrections: the morphism is of filtered vector spaces, and the degrees of the associated graded map should be shifted up.

The bilinear form should still be positive definite.