**Rational place curves**

Let φ = (G_0, G_1, G_2) : **P**^2 —> **P**^3 be a morphism of degree d. We claim
that φ always maps into a plane curve C ⊂ **P**^2.

First method: Choose an integer e. Consider the map

K[X_0, X_1, X_2]_e —→ K[Y_0, Y_1]_{de}

which assigns to the homogeneous polynomial F of degree e in X_0, X_1, X_2 the homogeneous degree de polynomial F(G_0, G_1, G_2). We computed in the lecture that this linear map has a kernel as soon as e » d namely as soon as

(e + 2 choose 2) > de + 1

Thus we see in this manner that φ always maps into *some* plane curve.

Second method: general theory tells us that we can in fact find a degree **d** curve.
To find such a curve we use a bit of field theory. Let us dehomogenize the situation.
Namely, on affine pieces the map φ on points looks like this

(1 : t) —→ (G_0(1, t) : G_1(1, t) : G_2(1, t)) = (1, f(t), g(t))

where f(t) = G_1(1, t)/G_0(1, t) and g(t) = G_2(1, t)/G_3(1, t). We are going to assume that f is not a constant rational function (if this happens you have to chance coordinates to get to a situation where it doesn't happen). Now we want to find a degree d polynomial Q in x, y such that Q(f, g) = 0. Let L = K(x) and M = K(t). These are purely transcendental field extensions of K. Consider the map

L —> M, x —> f

This is well defined as f is not a constant; namely, given any rational function h(x) the substitution h(f) is a rational function in t. Moreover, since L is a field this map is injective. Hence we can think of M as a field extension of L. Since M is generated by t over K, we see that M is generated by t over L as well. And we see that t statisfies the equation

x = f(t) ⇔ xG_0(1, t) = G_1(1, t)

over L. This equation has degree d! Hence the field extension M/L has degree at most d. Ok, now consider the element g in M. By general field theory, we see that there exist a_0(x), …, a_d(x) in L not all zero such that

a_0(x) + a_1(x) g + … + a_d(x) g^d = 0

in M. Clearing denominators we may assume that a_i ∈ K[x]. Thus we see that

Q = a_0(x) + a_1(x)y + … + a_d(x)y^d

is an element of the ideal I of polynomials Q in K[x, y] for which Q(f, g) = 0. Hence this ideal contains an element Q with deg_y(Q) ⇐ d. To finish the proof you can argue using the following steps:

**Exercise 23.5:**
Show, using the fact that K[x, y] is a UFD that the ideal I is a principal ideal.

**Exercise 23.6 (optional):**
Let Q in I be a generator. If a monial x^iy^j with i + j > d occurs find a linear change of
coordinates such that y^{i + j} occurs.

**Exercise 23.7 (optional):**
Finish the argument.

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