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Polynomial Factorization
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Univariate Factorization – simplifications

When factoring a univariate polynomial over Z, the following simplifications are effective:

removing the integer content of F(Z)

computing square free decomposition (with use of GCD computations or modular interpolation techniques).

one could try to monicize F(Z), but this increases the size of the coefficients of F and in most cases in not worthwhile:

Slide 10

Examples

Examples

Factorization of polynomials over Z will not be more fine-grained, but will only be coarser than factorization over a .

For example, has complex roots and thus it is irreducible over Z. But it is factorizable over any .

For instance,

Slide 11

Univariate Factorization – over

Univariate Factorization – over

Let be a polynomial with coefficients from

First, we get rid of squares:

Slide 12

Univariate Factorization – over

Univariate Factorization – over

Let be a polynomial with coefficients from

First, we get rid of squares:

Slide 13

Factorization over - theoretical basis

Factorization over - theoretical basis

Slide 14

Is there any use of this theorem?

Is there any use of this theorem?

Let us now understand that the equation

is in fact equal to a system of linear equations over

Due to the fact that we are over ,

(because almost all the binomials are divided by p).

Slide 15

And what?

And what?

Also,

and we get a system of linear equations

Slide 16

And what?

And what?

Also,

and we get a system of linear equations

The dimension of its solution space is k, where k is the number of irreducible factors of f.

Slide 17

The last slide about finite fields

The last slide about finite fields

We now know, how many factors there are.

Let to be a basis. If k=1 then the f is irreducible

In the case k>1, we search for , for all .

As a result, we get a number of divisors of f:

If s<k, we calculate and so on.

Slide 18

The last slide about finite fields

The last slide about finite fields

We now know, how many factors there are.

Let to be a basis. If k=1 then the f is irreducible

In the case k>1, we search for , for all .

As a result, we get a number of divisors of f:

If s<k, we calculate and so on.

At the end, we will get all the k factors: for two different factors

there exists an element from the basis such that

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