Root field of x^4+ax^2+b

[Kiwi1]

Prove that if \(p(x)=x^4+ax^2+b\) is irreducible in F[x], then \(F[x]/<p(x)>\) is the root field of p(x) over F.

My Attempt:
1. Let F(c) = \(F[x]/<p(x)>\) where c is a root of p(x). Then F(c) is a degree 4 extension over F because c is the root of a 4th order irreducible polynomial in F[x].

2. \(p(x)=x^4+ax^2+b=(x^2+\frac a2)^2-(\frac{a^2}4-b)\)

Form the extension \(F(d_1)\) where \(d_1,-d_1\) are the roots of \(x^2-[\frac{a^2}4-b]\)

Now the four roots (two each) of \(x^2+\frac a2+d_1\) and \(x^2+\frac a2-d_1\) are the four roots of p(x) in a suitable extension of F.

Let \(d_2\) be a root of \(x^2+\frac a2-d_1\) in \(F(d_1,d_2)\). So \(d_2=\sqrt{d_1-\frac a2}\) and \(\pm d_2\) are two roots of p(x). Also \(d_3=\sqrt{-d_1-\frac a2}\) are two more roots this time in \(F(d_1,d_3)\).

3. \(F(d_1,d_2)\) or \(F(d_1,d_3)\) between them contain the four roots of p(x). Choose the one that contains c. Suppose without loss of generality \(c \in F(d_1,d_2)\).

4. \(c \in F(d_1,d_2)\) so \(F(c) \subseteq F(d_1,d_2)\)
5. \(c=\pm d_2\) so \(d_2 \in F(c)\) so \(d_1=d_2^2 +\frac a2\in F(c)\)
6. \(d_1,d_2 \in F(c)\) so \(F(c) \supseteq F(d_1,d_2)\) and using (4) above \(F(c) = F(d_1,d_2)\)Now all I think I need to do is show that \(\pm d_3 \in F(d_1,d_2)\) and conclude that F(c) is a rootfield of p(x)

Unfortunately this last step has me stumped!

 

[Kiwi1]

I think I solved it.

I have
"Theorem 7: Let K be the root field of some polynomial q(x) over F. For every irreducible polynomial p(x) in F[x], if p(x) has one root in K, then p(x) must have all of its roots in K."

In my problem I have \(q(x) = (x-d_2)(x+d_2)=x^2-d_2^2=x^2-d_1+\frac a2 \in F(d_1)[x]\)

So \(F(d_1,d_2)\) is the root field of q(x) over \(F(d_1)\). What's more \(c \in F(d_1,d_2)\) so by Theorem 7 all roots of p(x) (which is irreducible in \(F(d_1)\)) must be in \(F(d_1,d_2)=F(c)\)