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
(This may not be in F, but this is why we
dealt with .) Once we have found
, we can solve forOne of
them will satisfy (* ).
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
known to satisfy
When n = 0,
Now, for ease of notation, let
Then, for some b ∈ F, the induction hypothesis gives
So, does the trick. We
now have our power series expansion
of y in terms of x. Use this method to compute yn 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
Claim. The P of minimal degree m found in Step 4 is an irreducible factor
Assuming the claim, then P | Q, so if Q ≠ P, we replace Q
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) = x2 − y2.
Then is a point on the curve Q = 0.
Expanding y in a power series: We have
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
mod pn, 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.
Start solving your Algebra Problems
in next 5 minutes!
Download (and optional CD)
Click to Buy Now:
2Checkout.com is an authorized reseller
of goods provided by Sofmath
Attention: We are
currently running a special promotional offer
for Algebra-Answer.com visitors -- if you order
Algebra Helper by midnight of
you will pay only $39.99
instead of our regular price of $74.99 -- this is $35 in
savings ! In order to take advantage of this
offer, you need to order by clicking on one of
the buttons on the left, not through our regular
If you order now you will also receive 30 minute live session from tutor.com for a 1$!
You Will Learn Algebra Better - Guaranteed!
Just take a look how incredibly simple Algebra Helper is:
: Enter your homework problem in an easy WYSIWYG (What you see is what you get) algebra editor:
Step 2 :
Let Algebra Helper solve it:
Step 3 : Ask for an explanation for the steps you don't understand:
Algebra Helper can solve problems in all the following areas:
simplification of algebraic expressions (operations
with polynomials (simplifying, degree, synthetic division...), exponential expressions, fractions and roots
(radicals), absolute values)