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Let $K$ be the splitting field of an irreducible polynomial over $F$. Let $f$ be a polynomial which has a root in $K$. I am trying to prove that $f$ splits over $K$, using only basic field theory, without the Galois group. Is this possible?

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Presumably $f$ is irreducible over $F$. – André Nicolas Jan 2 '12 at 20:32
The full proof takes a bit of work. Unless someone is kind enough to type it all out here, I'd suggest you look for it in a Galois Theory textbook such as Stewart's book. – Gerry Myerson Jan 2 '12 at 21:28
up vote 3 down vote accepted

Yes. The immediate thing I can think of is proving the standard equivalence of the definitions of a normal extension. Try to prove the equivalence of the following statements for fields $L/K$:

  1. Every $K$-embedding of $L$ into $K^a$ induces an automorphism of $L$.

  2. $L$ is the splitting field of a family of polynomials in $K[X]$.

  3. Every irreducible polynomial of $K[X]$, which has a root in $L$ splits in $L$.

Now prove that $(1)\Leftrightarrow (2)$ and $(1)\Leftrightarrow (3)$. These are all elementary and imply what you want.

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The statements are equivalent, and the proof can be done without Galois Theory, but if I remember right the part that OP wants is complicated enough to take a full page to prove in, say, Stewart's Galois Theory textbook, and is not completely straightforward. – Gerry Myerson Jan 2 '12 at 21:27
The OP is looking for the implication $(2)\Rightarrow (3)$ which is easiest to prove by doing $(2)\Rightarrow (1)\Rightarrow (3)$. This doesn't take even half a page. – pki Jan 3 '12 at 8:09

$\mathbb Q(\sqrt 2)$ is the splitting field of $x^2 - 2$ which is irreducible over $\mathbb Q$, and $x^4 -4 = (x^2-2)(x^2 +2)$ has clearly a root in $\mathbb Q(\sqrt 2)$ but doesn't split over $\mathbb Q(\sqrt 2)$. I think you may have something there but you might need more constraints on $f$. (One necessary condition would be that $f$ is irreducible, because if not, you can consider $f$ that is just the product of irreducibles in $K[x]$.)

Hope that helps,

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I will just use the the primitive element theorem (which can be proved elementarily). Let $\alpha\in K$ such that $K = F[\alpha]$, let $P\in F[X]$ such that $P(\alpha)=0$.

Let $q \in F[X]$ such that $q(\alpha)$ is a root of $f$.

Let $$T = \prod_{\alpha \text{ root of } P} (X-q(\alpha)) \in F[X] $$ or in more conventional notation, denoting $\alpha_1 = \alpha, \alpha_2, \dots, \alpha_n$ the roots of $P$, $$T = \prod_{i=1}^n (X-q(\alpha_i)).$$ This is the Tschirnhaus transformation of $P$ by $q$. You can see easily that this is in $F[X]$ because it is invariant by permutation of the roots of $P$. It can be computed easily eg using a resultant.

Compute the gcd of $T$ and $f$. It is clearly non trivial because thy share a common linear factor. As $f$ is irreducible, it is equal to $f$. It is now clear that $f$ splits in $K$.

PS If $f$ is not irreducible, the above procedure gives you a non trivial factor of $f$. In fact this proves constructively that either $f$ splits in $K$, or $f$ has a non trivial factor.

PPS In this proof you can already see the "reason" why this is true: because the roots of $P$ are "undistinguishable", as soon as $q(\alpha)$ is a root of $f$ for $\alpha$ a root of $P$, this is true for all roots of $P$. In some more abstract proofs this intuition can be difficult to retrieve...

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