Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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,

share|cite|improve this answer

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...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.