Polynomial ring/Field/Zeroes/Introduction/Section

A zero of a polynomial ${}P$ is an element ${}a\in K$ such that ${}P(a)=0$. A polynomial does not necessarily have zeroes, and this depends also on the base field. The polynomial ${}X^{2}+1$ has no real zero, but it has the complex zeroes ${}{\mathrm {i} }$ and ${}-{\mathrm {i} }$. As an element in ${}\mathbb {R} [X]$, the polynomial ${}X^{2}+1$ can not be written as a product of simpler polynomials. However, in ${}\mathbb {C} [X]$, it has the factor decomposition

${}X^{2}+1=(X-{\mathrm {i} })(X+{\mathrm {i} })\,.$

Remark

Let ${}K$ be a field, let ${}K[X]$ be the polynomial ring over ${}K$ and ${}a\in K$. Then the evaluation mapping

$K[X]\longrightarrow K,P\longmapsto P(a),$

is ${}K$-linear. Moreover, we have

${}(PQ)(a)=P(a)Q(a)\,,$

see exercise.

Lemma

Let ${}K$ be a field and let ${}K[X]$ be the polynomial ring over ${}K$. Let ${}P\in K[X]$ be a polynomial and ${}a\in K$. Then ${}a$ is a zero

of ${}P$ if and only if ${}P$ is a multiple of the linear polynomial ${}X-a$.

Proof

If ${}P$ is a multiple of ${}X-a$, then we can write

${}P=(X-a)Q\,$

with another polynomial ${}Q$. Inserting ${}a$ yields

${}P(a)=(a-a)Q(a)=0\,.$

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

${}P=(X-a)Q+R\,,$

where either ${}R=0$ or the degree of ${}R$ is ${}0$, so in both cases ${}R$ is a constant. Inserting ${}a$ yields

${}P(a)=R\,.$

So if ${}P(a)=0$ holds, then the remainder must be ${}R=0$, and this means ${}P=(X-a)Q$.

$\Box$

Corollary

Let ${}K$ be a field and let ${}K[X]$ be the polynomial ring over ${}K$. Let ${}P\in K[X]$ be a polynomial (${}\neq 0$) of degree

${}d$. Then ${}P$ has at most ${}d$ zeroes.

Proof

We prove the statement by induction over ${}d$. For ${}d=0,1$ the statement holds. So suppose that ${}d\geq 2$ and that the statement is already proven for smaller degrees. Let ${}a$ be a zero of ${}P$ (if ${}P$ does not have a zero at all, we are done anyway). Hence, ${}P=Q(X-a)$ by fact and the degree of ${}Q$ is ${}d-1$, so we can apply to ${}Q$ the induction hypothesis. The polynomial ${}Q$ has at most ${}d-1$ zeroes. For ${}b\in K$ we have ${}P(b)=Q(b)(b-a)$. This can be zero, due to fact  (5), only if one factor is ${}0$, so the zeroes of ${}P$ are ${}a$ or a zero of ${}Q$. Hence, there are at most ${}d$ zeroes of ${}P$.

$\Box$