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Let $K/F$ be a field extension and let $f(x)\in F[x]$. I know $f(x)$ have a splitting field, i.e. a field $E$ that $f(x)$ splits in ($E/F$ and $f(x)$ doesn't split in any proper subfield of $E$).

I heard the term "the splitting field of $f(x)$ over $K$" - but what does this mean ?

I realize that it is also true that $f(x)\in K[x]$, but I still don't understand the looks like we want a field $L$ s.t $L/K$ is the minimal field extension s.t $f(x)$ splits in $L$ but this like the composition of $E,L$ , but they arn't both subfields of a field I know...

I'm confused, can someone please explain the term ?

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up vote 3 down vote accepted

Let $K$ be a field and let $f(x)$ be a nonconstant polynomial in $K[x]$. A splitting field of $f$ over $K$ is a field extension $L$ of $K$ such that:

  1. $f(x)$ splits into linear factors in $L$; and
  2. $L = K(\{u\in L\mid f(u) = 0\})$; that is, $L$ is generated over $K$ by the roots of $f(x)$.

One can prove by induction on $\deg(f)$ that splitting fields always exist, and moreover that two splitting fields for $f$ over $K$ are isomorphic over $K$.

One can extend this to arbitrary sets of polynomials: let $K$ be a field, and let $S\subseteq K[x]$ be a set of nonconstant polynomials. A splitting field of $S$ over $K$ is a field extension $L$ of $K$ such that:

  1. For every $f(x)\in S$, $f(x)$ splits in $L$; and
  2. If $U=\{ a\in L \mid \text{there exists }f\in S\text{ such that }f(a)=0\}$, then $L=K(U)$. That is, $L$ is generated over $K$ by the roots of the polynomials in $S$.

It is easy to see that if $S$ is finite, $S=\{f_1,\ldots,f_n\}$, then a splitting field for $S$ over $K$ is the same thing as a splitting field for $g(x)$ over $K$, where $g(x) = f_1(x)\cdots f_n(x)$. Thus, splitting fields are generally only interesting for either single polynomials, or infinite sets of polynomials. The fact that every set of nonconstant polynomials in $K[x]$ has a splitting field over $K$, and moreover that any two such splitting fields are isomorphic over $K$, can be established by applying Zorn's Lemma.

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Given a field $K$ and a polynomial $f\in K[x]$, a splitting field for $f$ over $K$ is a field $L$ such that $f$ splits into linear factors when considered as an element of $L[x]$, i.e. $$f=(x-a_1)\cdots(x-a_n)\;\;\text{ for some }a_1,\ldots,a_n\in L,$$ and $L=K(a_1,\ldots,a_n)$ - note that this second condition is equivalent to saying that $f$ doesn't split in any proper subfield $F\subsetneq L$ containing $K$.

When people say the splitting field of $f$ over $K$, there are two possible meanings:

  1. It is a theorem that any two splitting fields of a polynomial $f$ are isomorphic. In particular, if $E_1$ and $E_2$ are two splitting fields of $f$ over $K$, then there is an isomorphism $\phi: E_1\to E_2$ such that $\phi(a)=a$ for all $a\in K$. By a common abuse of terminology, we might therefore talk about "the" splitting field, when really there are many, because we can always identify them.

  2. If there is an algebraically closed field $M$ of which all fields under discussion (including $K$) are implicitly assumed to be subfields (a common choice, if we are dealing with characteristic 0 fields, is $M=\mathbb{C}$), then there is a unique splitting field of $f$ which is containined in $M$, and so we might use the article "the" since there is only one.

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$E$ is the minimal field extension of $F$ in which $f(x)$ splits.

$L$ is the minimal field extension of $K$ in which $f(x)$ splits.

$L$ is a field extension of $F$ in which $f(x)$ splits, and therefore $E$ is a subfield of $L$ -- i.e. $E/F$ is isomorphic as an extension of $F$ to a unique intermediate field $E'$ of $L/F$.

$L = KE'$, because $L$ is the only intermediate field of $L/F$ that contains both $K$ and $E'$.

Edit: it's worth pointing out that one isn't often thinking of fields but of field extensions. Thinking of $E/F$ is different than thinking of $E$ on its own. Asking for the splitting field of $f(x)$ over $K$ means we don't want $L$, or even $L/F$, but the extension $L/K$.

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Then it's not really $L=KE$ but $L=KE'$ where $E'$ is a subfield of $L$ that isomorphic to $E$, right ? – Belgi Jun 6 '12 at 23:09
Sure, if you're requiring subfields to be actual subsets, rather than given by monic maps. Of course $E/F$ might embed more than one way into $L/F$, so one should probably make that a choice is involved explicit anyways. At least I think it could. It's probably unique if $K/F$ is Galois or when $K \cap E = F$ (as extension of $F$ -- I think this makes sense because $E/F$ is Galois), but I'd have to think about it. – Hurkyl Jun 6 '12 at 23:14
@Belgi: There, I think I've arranged my answer to contain true statements if you have $F \subseteq E$ as part of the meaning of $E/F$. I still maintain that "up to isomorphism" notation is often useful, though -- since $E'$ is unique, IMO it often isn't worth distinguishing between $E/F$ and $E'/F$. – Hurkyl Jun 6 '12 at 23:32

Perhaps by looking at an example you can understand what we mean by the splitting field. Suppose you look at $x^3 - 2$ over $\Bbb{Q}$. Then this polynomial is irreducible by Eisenstein's Criterion with $p = 2$. Now I claim that in the field $\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2})$ where $\omega = e^{2\pi i/3}$ our polynomial $x^3 - 2$ factors into linears. But this is clear from the fact that $$x^3 - 2 = (x - \sqrt[3]{2})(x - \omega\sqrt[3]{2})(x - \omega^2 \sqrt[3]{2}).$$

Hence we can make an ansatz that in general if a field extension $E/F$ contains all the roots of some polynomial $f(x) \in F[x]$, then we should be able to factorise $f(x)$ over $E$ into linears. Now several people have given you the definition of a splitting field of a polynomial. Let us see why $\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2})$ is a splitting field for $x^3 - 2$. On one hand by minimality of the splitting field $K$, we must have that $K \subset \Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2})$.

On the other,because each term that we adjoined to $\Bbb{Q}$ to get our extension is a root of $x^3 - 2$, they are all in the splitting field and hence $\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}) \subset K$. It follows that

$$\Bbb{Q}(\sqrt[3]{2},\omega\sqrt[3]{2},\omega^2\sqrt[3]{2}) = K.$$

Now if we view everything as sitting inside the big algebraically closed field $\Bbb{C}$, one can speak of the splitting field. Otherwise, if you are over a finite field, it is not so easy to see what big algebraically closed field all the finite fields sit inside. In fact if we wanted to even speak of the algebraic closure of a finite field, we need to set up some directed system and take a direct limit. So in this case, by the splitting field we have to mean the one upto isomorphism.

Let's turn our attention to what happens over finite fields. By Lagrange's theorem, the polynomial $x^{p^n}- x$ splits completely over the finite field $\Bbb{F}_{p^n}$. Since any two splitting fields are isomorphic, we conclude that any two finite fields with the same number of elements are isomorphic. For example, weird as it may seem consider the polynomials

$$x^4 + x^3 + 1, \hspace{2mm} x^4 + x + 1$$

over $\Bbb{F}_2$ that are irreducible. Then we have a very interesting phenomena: In $\Bbb{F}_2[x]/(x^4 + x^3 + 1)$, the polynomial $x^4 + x^3 +1$ splits completely! In fact this always happens: If you have an extension $\Bbb{F}_{p^n}(a)$ obtained by adjoining some root $a$ of $f(x) \in \Bbb{F}_{p^n}[x]$, then suppose that $\deg f = n$. Then

$$ \Bbb{F}_{p^n}(a) \cong \Bbb{F}_{p^{n^2}}.$$

But the guy on the right is a separable extension and by what I said above is also a normal extension so is Galois. Hence $\Bbb{F}_{p^n}(a)$ is a Galois extension of $\Bbb{F}_{p^n}$, so in particular is normal so that $f(x)$ splits completely in here! Similary we see that $x^4 + x + 1$ splits completely in $\Bbb{F}_2[x]/(x^4 + x + 1)$ so that since

$$\Bbb{F}_2[x]/(x^4 + x + 1) \cong \Bbb{F}_2[x]/(x^4 + x^3 + 1),$$

both of these are splitting fields for $x^4 + x + 1$ and $x^4 + x^3 + 1$!

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