Let F be a field and consider
. Our task is to factor

Q(x, y) into a product of irreducible polynomials.

**Remark.**

1. F[x, y] is a UFD.

2. Q(x, y) may be ir reducible over F but may factor over some extension

field L/F.

**Factoring a Polynomial in Two Variables **

**Step 0.** Write compute the gcd of the
, and

factor it out. WLOG So, content(Q) = 1 as a polynomial in

y with coefficients in F [x].

**Step 1. **Regard Q as a polynomial in y with coefficients in F(x). Take

to remove multiple factors .

Combining the previous two steps we can assume that Q has no factors

in F[x], and is square -free.

**Step 2.** Find with

and

The purpose of this is to find a point on the plane curve Q = 0 which

is non-singular and such that the tangent line at this point is not vertical.

This will allow us to write y as a power series in x. Now we justify the

existence of such a point before moving on to finding the power series.

Only finitely many points will satisfy the first equation in (* ) and not

the second. First, compute (by step 1). We
want

such that
(This may not be in F, but this is why we
have

dealt with .) Once we have found
, we can solve for
One of

them will satisfy (* ).

W LOG assume , by extending F if necessary.

**Step 3.** Expand y as a power series:

Now we use a recursive algorithm to compute
(You may have

seen this as Newton’s algorithm or Hensel’s lemma). Suppose
are

known to satisfy

When n = 0,

Now, for ease of notation, let

Then, for some b ∈ F, the induction hypothesis gives

To find
,
let
and compute

So, does the trick. We
now have our power series expansion

of y in terms of x. Use this method to compute y_{n} for

**Step 4.** For m = 1, 2, . . . ,deg Q; try to find P(x, y) ∈ F[x, y] of degree m,

starting with m = 1, with

Stop if you find P, else go to next value of m . This is a system of linear

equations in the coefficients of P. So find the nullspace. If it is zero , go to

the next step.

**Claim.** The P of minimal degree m found in Step 4 is an irreducible factor

of Q.

Assuming the claim, then P | Q, so if Q ≠ P, we replace Q
with Q/P

and repeat steps 2-4.

Why is the claim true? The point about which the power series

expansion of y was given is on the plane curve Q = 0 and also on P = 0. The

conditions on P and Q ensure that the intersection multiplicity of P = 0

and Q = 0 at is at least Bezout’s theorem

then implies that P | Q.

**Example. **Let us illustrate with an easy example. Let Q(x, y) = x^{2} − y^{2}.

Then is a point on the curve Q = 0.
And

Expanding y in a power series: We have

So, giving
.

Then, . Now, find P of degree 1 satisfying

Say, P(x, y) = y − x, so P | Q.

A few weeks ago, we talked about how to factor in Z[x].
So, Q(Y )∈ Z[Y ].

The analogy here is between Z and F[x]. Find a prime p and a y0 such that

This is the same as finding
. Now, use Hensel’s
lemma: find

with
mod p^{n}, for n large in relation to the coefficients of Q. Step

3 is to find the power series expansion of y:

Now find P(Y ) ∈ Z[Y ] of small degree and with small
coefficients such that

(This can no longer be done with linear algebra .) This

congruence defines a lattice in . To find a short vector in a lattice,

there is an algorithm called the LLL-algorithm which we won’t explain.

Then a height calculation will replace Bezout to prove that P | Q.