Show that f(x) is irreducible over Q and Q(2^(1/5))

[demonelite123]

Show that x^3 + 6x^2 - 12x + 2 is irreducible in \mathbb{Q} as well as in \mathbb{Q}(\sqrt[5]{2}).

the first part i have no trouble with since it follows straight from Eisenstein's Irreducibility Criterion. For the second part i am pretty confused though. i tried looking at the solution to help me understand it but i wasn't sure what the solution did.

They say that [\mathbb{Q}(\sqrt[5]{2}): \mathbb{Q}] = 5 and i understand that since \sqrt[5]{2} is the root of irreducible x^5 - 2 so the degree of the field extension is 5. They then said that if x^3 + 6x^2 - 12x + 2 was reducible, then it would have a linear factor and then there would be a root in \mathbb{Q}(\sqrt[5]{2}). Then since this root has degree 3 and since 3 does not divide 5, then x^3 + 6x^2 - 12x + 2 is irreducible in \mathbb{Q}(\sqrt[5]{2}).

i do not understand the part about the 3 not dividing the 5. I assume they are using the theorem that says [F:K] = [F:E][E:K] if K is a subfield of E and E is a subfield of F. i suspect that this theorem is being used but i don't know how they are using it exactly. can someone help explain? thanks

 

[micromass]

Indeed, since we can find an \alpha\in \mathbb{Q}(\sqrt[5]{2}) that is a root of x^3+6x^2-12x+2, this means that this polynomial is a minimal polynomial of \alpha. Thus [\mathbb{Q}(\alpha),\mathbb{Q}]=3.

Now we have that

[\mathbb{Q}(\sqrt[5]{2}),\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]{2}),\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha),\mathbb{Q}]

This means that

5=[\mathbb{Q}(\sqrt[5]{2}),\mathbb{Q}(\alpha)]*3

Or 3 divides 5. Which is impossible.

 

[demonelite123]

yes that makes sense. i was initially confused on which field to pick but i understand now why Q(a) was chosen. thank you for your reply.