1. Warming up to Enumerative Geometry 3
can certainly be written in the form a0x + a1. There are two cases
to consider. In the first case, a1 = 0, we see that f(x) has no roots.
The root can be viewed as having “gone off to ∞” as before: we can
f(x) = lim
a0x + a1,
and then the root −a1/a0 goes to or −∞ as a0 approaches 0 from
one direction or the other. If a1 = 0 as well, then there are infinitely
many solutions: any number x trivially satisfies the equation 0·x = 0.
The situation already gets much richer if d = 2. The quadratic
formula gives
x =
−a1 ± a1 2 4a0a2
There are now several possibilities:
(1) a0 = 0. There are several well-known subcases. Let D =
4a0a2 be the discriminant of f(x).
(a) D 0. There are two roots.
(b) D 0. There are no real roots.
(c) D = 0. There is one root.
(2) a0 = 0. There are several possibilities:
(a) a1 = 0. In this case, there are no solutions (unless
additionally a2 = 0, in which case there are infinitely
many roots).
(b) a1 = 0. There is now exactly one solution x = −a2/a1.
We can think that the “other root” has “gone off to
See Exercise 1. The situation will clearly only get worse as d
We can simplify the answer by changing the question. Essentially
all of these cases can be unified neatly by a few simple changes in the
enumerative problem:
Use complex coefficients and solutions.
Count solutions with multiplicity.
Include infinity.
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