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[SOLVED] splitting field does not seem to exist.. :O

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
(Herstein Pg 222) DEFINITION: If $f(x) \in F[x]$, a finite extension $E$ of $F$ is said to be a splitting field over $F$ for $f(x)$ if over $E$(that is, in $E[x]$), but not over any proper sub-field of $E$, $f(x)$ can be factored as a product of linear factors.

Now here's my question. Take $p(x)=x^2-4 \in F[x]$, $F$ is the field of Complex Numbers. What is the splitting over $F$ for $p(x)$??

I would be tempted to say that $F$ itself is the splitting field over $F$ for $p(x)$. But then $\mathbb{R}$, the field of real numbers, would be a proper sub-field of $F$ in which $p(x)$ can be factored as a product of linear factors, viz, p(x)=(x-2)(x+2).

What have I missed?
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
Re: splitting field soes not seem to exist.. :O

there is a slight omission in the definition. recall that E is a splitting field over F.

that means that the proper subfields referenced in the definition, are to be understood as "sub-fields over F", that is, extensions K, F ≤ K < E.

since x2-4 splits in C, the splitting field of x2-4 over C, is C itself.
of course we can find smaller fields in which this polynomial splits. but since C already contains these smaller fields, we really aren't "extending anything".

for emphasis, the definition of a splitting field for a polynomial p(x) in F[x], depends on F.

for example, the splitting field of x2+1 over Q is Q(i), the gaussian rationals. the splitting field of x2+1 over R, is C, the complex numbers. these are *not* the same fields.

in other words, we are talking about a unique (up to isomorphism) extension of F, not a unique (up to isomorphism) field.