Algebraic Geometry Jottings 8

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This will be a “chewing” post.

It would be pointless in these posts just to repeat the definitions and proofs from Kendig and Fulton. My aim is satori, where chewing over the material makes it seem inevitable rather than contrived. (If I get halfway there I’ll be happy.) Pursuing this goal, I’ve adopted (temporarily) a cavalier attitude towards rigor. That will continue in this post, with even less in the way of caveats. I want to see why stuff should be true, before worrying about how to show rigorously that it is. So (for example) when I say something is a basis, I mean it looks like a basis, and in fact is a basis, but not that we’ve proved it’s a basis.

We’ve seen three things going on with Bézout’s theorem: (a) algebraic closure; (b) points at infinity; (c) multiplicity. Now for a new one: (d) local vs. global. The dimension of the coordinate ring Γ(EF) tells us the total number of intersections (if (a)–(c) are respected). The dimension of the modded local ring, as I’m calling it (local ring mod the curves= (\mathcal{O}_P/(E,F)_P), tells us the number of intersections residing at P.

Let’s dwell on how this works for a couple of cases where (a)–(c) present no difficulties—all points real, no points at infinity, all multiplicities are 1. Start by noticing that k[x,y] has infinite dimension over k. The ideal (E,F) plays the role of a hitman, killing off basis elements. Let’s call (E,F) “The Eraser”. Write π for the canonical homomorphism from k[x,y] to Γ(EF). With all multiplicities being 1, the kernel of π (that is, (E,F)) consists of all the polynomials that take the value zero at all intersections. “The Eraser” erases all information about a polynomial except its values at the intersections.

Example: E(x,y) = yx2 and F(x,y) = y–1. As we saw last time, we have yx2 and y ≡ 1, so x2 ≡ 1, so (x+1)(x—1) ≡ 0. (In terms of π, π(y) = π(x2), etc.) The basis of k[x,y] is {1, x, yx2, xy, y2, x3, x2y, …}. Since y ≡ 1, we erase all basis elements with a y in them; “The Eraser” can erase an element not by zeroing it out, but by making it redundant. Likewise for x2, x3, etc. We are left with the basis {1, π(x)} for Γ(EF). For any pk[x,y], π(p) is essentially the pair (p(–1,1), p(1,1)) and so Γ(EF) is kk.

If P is the point (1,1), then the basis of \mathcal{O}_P/(E,F)_P is just {1}. That’s because (E,F)P is a bigger eraser than (E,F): (x+1)(x—1) ≡ 0 mod (E,F), so x—1 ≡ 0 mod (E,F)P. If πP is the canonical homomorphism from \mathcal{O}_P to \mathcal{O}_P/(E,F)_P, then πP erases all info except for the value at P.

Example: the cubics of post 3, say y = x3x. So we have E(x,y) = y–(x3x) and F(x,y) = y, hence yx3x and y ≡ 0, hence x3x ≡ x(x+1)(x–1) ≡ 0. All elements with y‘s disappear from the basis, likewise all xn with n≥3. We’re left with {1, π(x), π(x)2}. Passing to the three local rings at (–1,0), (0,0), and (1,1), we have x ≡ –1, 0, 1 respectively; for example, at (0,0) we can divide x(x+1)(x–1) ≡ 0 by (x+1)(x–1).

So when all multiplicities are 1, piece of cake! How about when we have two intersections, each with multiplicity 2? Like this one:

x2+y2 = 1
2x2+y2 = 1

We have x2+y2 ≡ 1 and 2x2+y2 ≡ 1, so x2 ≡ 0 and so y2 ≡ 1. It looks like {1, π(x), π(y), π(xy)} is a basis for the coordinate ring. Let’s see if this makes sense.

The polynomial x is zero at the two intersections, but nonetheless “The Eraser” does not erase it. Being zero at all the intersections isn’t enough to get you whacked. Of course we’ve already seen this situation with the E(x)=xn example in the last post. Multiplicity takes this form: the erasing homomorphism π preserves the first n coefficients, or equivalently, info about the value and the first n–1 derivatives. Here n=2, because the polynomials E and F agree to second order.

Both 1 and y take nonzero values at the intersections, so both are safe. As for x and xy, these have nonzero derivatives at the intersections. It looks like Γ(EF) has dimension 4. Hurrah!

How about the two modded local rings? We have y2 ≡ 1, so (y+1)(y–1) ≡ 0, so at the top y ≡ 1 mod (E,F)P, and at the bottom y ≡ –1 mod (E,F)P. So xy  ≡ x or xy  ≡ –x, and {1, πP(x)} is a basis at both intersections. Exercise: how come {1, π(x)} isn’t then a basis for the coordinate ring? Hint: when is y redundant?

I’d like to finish this post by tackling our running example. Can we find a basis for the modded local ring at O with


So far I can’t. I can make some progress. Recall the resultant from post 4, computed with SageMath:

resx(E,F) = 16y14(16y2-5)2

It so happens that the resultant belongs to the ideal (E,F). So in the modded local ring, πP(y)14 = 0. Is {1, πP(y), πP(y)2, … ,πP(y)13} a basis? If so, we’d get the right dimension. But I don’t see how to show that x is dependent on this basis (if it is a basis).

More generally, I’d like to see how Kendig’s resultant-based definition of multiplicity agrees with Fulton’s explicit formula. The obvious approach: show that the Kendig definition satisfies Fulton’s seven properties. The demonstration might not even be that hard.

It would be nice to see an explicit construction. In post 3, we looked at the branches of E (at O) intersecting the branches of F, and counted up to multiplicity 14. Is there some way to harvest the right number of basis elements from each branch-branch intersection? That would make the equivalence of the two definitions truly transparent.

To conclude, I’ll note two other unfinished aspects of this post. We didn’t do an example with intersections at infinity. Also, what would it take to bring the arguments up to snuff, rigor-wise? Mainly showing that the elements of each alleged basis are independent. Topics for the future.

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