Polynomials/Field/Euclidean division/Introduction/Section

From Wikiversity
Jump to navigation Jump to search

Let be a field and let be the polynomial ring over . Let be polynomials with . Then there exist unique polynomials such that

We prove the statement about the existence by induction over the degree of . If the degree of is larger than the degree of , then and is a solution.

Suppose that . By the remark just made also holds, so is a constant polynomial, and therefore (since and is a field) and is a solution.

So suppose now that and that the statement for smaller degrees is already proven. We write and with . Then setting we have the relation

The degree of this polynomial is smaller than and we can apply the induction hypothesis to it. That means there exist and such that

From this we get altogether

so that and is a solution.

To prove uniqueness, let , both fulfilling the stated conditions. Then . Since the degree of the difference is smaller than , this implies and so .

The computation of the polynomials and is also called long division. The polynomial is a factor of if and only if the division of by yields the remainder . The proof of this theorem is constructive, meaning it can be used to do the computation effectively. For this, one has to be able to do the computing operations in the field . We give an example.

We want to apply the Euclidean division (over )

So we want to divide a polynomial of degree by a polynomial of degree , hence the quotient and also the remainder have (at most) degree . For the first step, we ask with which term we have to multiply to achieve that the product and have the same leading term. This is . The product is

The difference between and this product is

We continue the division by with this polynomial, which we call . In order to get coincidence with the leading coefficient we have to multiply with . This yields

The difference between this and is therefore

This is the remainder and altogether we get

Let be a field and let be the polynomial ring over . Let be a polynomial and . Then is a zero of if and only if is a multiple of the linear polynomial .

If is a multiple of , then we can write

with another polynomial . Inserting yields

In general, there exists, due to fact, a representation

where either or the degree of is , so in both cases is a constant. Inserting yields

So if holds, then the remainder must be , and this means .

Let be a field and let be the polynomial ring over . Let be a polynomial () of degree . Then has at most zeroes.

We prove the statement by induction over . For the statement holds. So suppose that and that the statement is already proven for smaller degrees. Let be a zero of (if does not have a zero at all, we are done anyway). Hence, by fact and the degree of is , so we can apply to the induction hypothesis. The polynomial has at most zeroes. For we have . This can be zero, due to fact, only if one factor is , so the zeroes of are or a zero of . Hence, there are at most zeroes of .