2

I see this theorem in several places

Let F be a field and $f(x)\in F[x]$ have degree $n\geq2$.Let $E/F$ be a splitting field of $f(x)$. Then $[E:F]$ divides $n!$

The proof involves some induction. It assumes the $\alpha$ is a root of $f(x)$, and so we can form the extension $F(\alpha)\cong F[x]/ \langle f(x) \rangle$. Then it is stated that we can factor $f(x)=(x-\alpha)g(x)$ with $g(x)\in F(\alpha)$.

I can't see why $g(x)\in F(\alpha)$ is true. Can you help me with that?

[Edit: Copied from an answer accidentally posted by the same (?) user. JL]

Let me give an example:

Suppose $F=\mathbb{Q}$ and that we have $f(x)=x^3-3$. then $f(x)$ can be decomposed as $f(x)=(x-\sqrt[3]{3})(x^2+\sqrt[3]{3}+(\sqrt[3]{3})^2)$

And so we have $g(x)=x^2+\sqrt[3]{3}+(\sqrt[3]{3})^2$

If we denote $E=F(\sqrt[3]{3})$ then we have $g(x)\in E[x]$

The claim is that $g(x)\in E[x]$ is always true. I just can't see why this is generally true

vonbrand
  • 28,394
Nadavb
  • 21
  • 1
    Hint: Each polynomial of degree $n$ over a field has at most $n$ distinct roots. – sqtrat Feb 21 '16 at 18:31
  • 2
    The familiar division algorithm for polynomials produces coefficients that are in $F(\alpha)$. – André Nicolas Feb 21 '16 at 18:33
  • What Tobias writes in his answer (+1) is all you need. Because $E=F(\alpha)$ is an extension field of $F$, we can think that both $f(x)\in E[x]$ and $(x-\alpha)\in E[x]$. The polynomial division then produces a quotient $g(x)\in E[x]$. – Jyrki Lahtonen Feb 21 '16 at 22:54
  • Related: https://math.stackexchange.com/questions/62762 – Watson Jun 16 '16 at 12:02

1 Answers1

2

If $\alpha$ is a root of any polynomial $f(x)$ then $x-\alpha$ divides $f(x)$.

The reason for that is that we can use the "euclidean division algorithm" to express $f(x) = Q(x)(x-\alpha) + R(x)$ for some polynomials $Q,R$, with $R$ having degree smaller than the divisor $x-\alpha$.

That means that $R(x)$ is a constant $R(x) = c$. Now using the fact $\alpha$ is a root we find that $0 = Q(\alpha)(\alpha-\alpha) + c$ forces $c = 0$ which proves that $x-\alpha$ divides $f(x)$.

Brennan.Tobias
  • 959