Distinct zeros of irreducible polynomial

[jostpuur]

This claim is supposed to be true. Assume that $p\in\mathbb{F}[X]$ is an irreducible polynomial over a field $\mathbb{F}\subset\mathbb{C}$. Also assume that

$$p(X)=(X-z_1)\cdots (X-z_N)$$

holds with some $z_1,\ldots, z_N\in\mathbb{C}$. Now all $z_1,\ldots, z_N$ are distinct.

Why is this claim true?

For example, if $z_1=z_2$, then $(X-z_1)^2$ divides $p$, but I see no reason to assume that $(X-z_1)^2\in\mathbb{F}[X]$, so the claim remains a mystery to me.

 

[wisvuze]

This is because every field F of characteristic zero is a perfect field. This means that every irreducible polynomial over F is separable ( splits in the splitting field with distinct linear factors ).

A polynomial p(x) has multiple roots in the splitting field, iff the polynomials p(x) and Dp(x) [its formal derivative ] have a root in common. In other words, if c is the root of p(x), the minimal polynomial of c over F will divide both p(x) and Dp(x). So, if p(x) has multiple roots in the splitting field, p(x) and Dp(x) are not co-prime.

Assume p(x) in F[x] is irreducible, and has degree n. Then, Dp(x) is degree n - 1; however, p(x) is irreducible, so its only factors in a factorization are p( x ) and constants. So, since Dp(x) is degree n -1 < n, its non-constant factors must be degree d: 1 <= d <= n - 1. So, p(x ) and Dp(x) do not have any common non-unit factors, and hence the polynomials are co-prime. So, p(x) is separable

 

[jostpuur]

Thank you for the good answer, but I did not understand the purpose of every part of it. Is there something wrong with proof like this:

Assume that $p\in\mathbb{F}[X]$ has a zero $x$ of higher degree than one. Now $Dp\in\mathbb{F}[X]$ has the same zero. Let $m$ be the minimial polynomial of $x$ in $\mathbb{F}$. Now $m|p$ and $m|Dp$, so in particular $m|p$ with such $m$ that its degree is less than the degree of $p$, which is a contradiction.

 

[wisvuze]

There isn't anything wrong with saying it like that. That's the idea: if p is irreducible over F, then p is a minimal polynomial ( once you scale it to be monic ), of some x which is a root in the splitting field. If x is also a root of Dp, then the minimal polynomial of x has to divide Dp. As Dp has a smaller degree, this is impossible.
This is exactly what was going on in the last paragraph. Actually it's a proof (although it's pretty trivial ) of the obvious fact that since Dp has smaller degree, the minimal polynomial cannot divide Dp. the irreducible factors of Dp must have degree <= n - 1. But the minimal polynomial must be degree n, and if the min polynomial divided Dp, it would appear in the decomposition of Dp into irreducibles

 

[blue_raver22]

The claim is true because of the fundamental theorem of algebra, which states that every polynomial of degree n over the complex numbers has exactly n complex roots, counting multiplicities. In this case, since p(X) is an irreducible polynomial of degree n, it must have exactly n distinct complex roots. Therefore, the equation p(X)=(X-z_1)\cdots (X-z_N) must hold with distinct values for z_1,\ldots, z_N.

In the example given, if z_1=z_2, then the polynomial (X-z_1)^2 would indeed divide p(X), but this would contradict the assumption that p(X) is irreducible. Therefore, it must be the case that z_1 and z_2 are distinct, and this holds for all values of z_i, ensuring that all z_1,\ldots, z_N are distinct.

In summary, the claim is true because of the fundamental theorem of algebra, and the assumption that p(X) is an irreducible polynomial over a field \mathbb{F}\subset\mathbb{C} ensures that all z_1,\ldots, z_N are distinct roots.