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

write

f(x) = lim

a0→0

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

2a0

.

There are now several possibilities:

(1) a0 = 0. There are several well-known subcases. Let D =

a1

2

− 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

inﬁnity”.

See Exercise 1. The situation will clearly only get worse as d

increases.

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.

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

write

f(x) = lim

a0→0

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

2a0

.

There are now several possibilities:

(1) a0 = 0. There are several well-known subcases. Let D =

a1

2

− 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

inﬁnity”.

See Exercise 1. The situation will clearly only get worse as d

increases.

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.