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I'm having a conceptual issue. I know that a splitting field K of p(x) is the smallest field containing both Q and all the roots of p(x)

What about when you aren't given a polynomial? For example:

1) Is $\mathbb{C}$ is a splitting field over $\mathbb{R}$?

2) Is $\mathbb{R}$ is a splitting field over $\mathbb{Q}$?

I am confused how you would begin to answer this. Each field has a lot of different irreducible polynomials.


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Based on the definition you have, which is standard, it only makes sense to speak of the splitting field of a polynomial. Of course $\mathbb{C}$ is a splitting field over $\mathbb{R},$ namely of $x^2+1.$ – Mike B Apr 2 '12 at 4:02
This isn't a conceptual issue, it's an issue of what the definition of "a splitting field" is without further qualifiers. I would guess that it means "the splitting field of some polynomial" (over the base field). – Qiaochu Yuan Apr 2 '12 at 4:14
Another possible meaning is the splitting field of some set of polynomials. This doesn't change the answers here, though. – Chris Eagle Apr 2 '12 at 13:40
My immediate interpretation of the question was the same as Qiaochu's: if $K$ is an extension of $k$, is $K$ the splitting field of some polynomial in $k[x]$? – Chris Leary Apr 2 '12 at 14:55
up vote 3 down vote accepted

Well $\mathbb{C}$, as you may know, is algebraicaly closed. So, every polynomial with coefficients in $\mathbb{R}$ splits in $\mathbb{C}$. Is $\mathbb{C}$ the splitting field of a particular polynomial over $\mathbb{R}$? Sure, just take the polynomial $x^2 +1$ in $\mathbb{R}[x]$, and $\mathbb{C}$ will be the splitting field of this polynomial.

The same cannot be said of $\mathbb{R}$ over $\mathbb{Q}$, because if $\mathbb{R}$ were the splitting field of a particular polynomial with coefficients in $\mathbb{Q}$, then $\mathbb{R}$ would have to be algebraic over $\mathbb{Q}$ (in fact, it would have to be a finite dimensional vector space over $\mathbb{Q}$), but it is not because elements like $\pi$ are not algebraic over $\mathbb{Q}$.

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By the way, to see why $\mathbb{R}$ cannot be a finite dimensional vector space over $\mathbb{Q}$, use a cardinality argument. – Rankeya Apr 2 '12 at 4:06
Why did you take $x^2+1$? Can you choose any irreducible poly in $\mathbb{C}$? – Larry X Apr 2 '12 at 4:11
@Rankeya: What you've said above is incorrect. Linear polynomials are also irreducible (over any field), and the splitting field of a linear polynomial over $\mathbb{R}$ is just $\mathbb{R}$. So not any irreducible polynomial over $\mathbb{R}$ will do; only the quadratic irreducibles will. – Zev Chonoles Apr 2 '12 at 4:56
Yes, any irreducible polynomial of degree > 1 over R will do. Note the the onnly irreducible polynomials over R are quadratic ones, and C is a degree two field extension of R. So, the splitting field of any irreducible quadratic must be C (note that the splitting field of any irredubile quadratic is contained in C) – Rankeya Apr 2 '12 at 4:59
Thanks @ZevChonoles for pointing out an error in my comment. I could not edit my comment, so I just posted a new one and deleted to old, incorrect comment. – Rankeya Apr 2 '12 at 5:50

This is indeed an interesting question. And, yes, there is an intrinsic (or conceptual) characterization of splitting fields of polynomials: they are called normal extensions.

More precisely, given a finite dimensional field extension $F\subset K$ there exists a polynomial $P(X)\in F[X]$ such that $K$ is the splitting field $K=Split_F(P)$ of $P$ iff the extension $F\subset K$ is normal.
Of course I have to tell you what a normal extension is!

Definition An algebraic extension $F\subset K$ is normal iff every irreducible polynomial $f(X)\in F[X]$ which has a root in $K$ actually is a product of linear factors in $K[X]$.

You then have the generalization of the equivalence above to algebraic but not necessarily finite dimensional extensions $F\subset K$:
$K$ is normal over $F$ iff $K$ is the splitting field $K=Split_F((P_i)_{i\in I})$ of a (maybe infinite) family $(P_i)_{i\in I}$ of polynomials $P_i(X)\in F(X)$.

For example an algebraic closure $F^{alg}$ of a completely arbitrary field $F$ is clearly normal over $F$ and it is just as clearly a splitting field for the family of all polynomials in $F[X]$ !

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