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You can use Jacobson, Dummit and Foote, and Kaplansky's Fields and Rings. Pages 50-54 tells how to find the Galois group of a cubic and quartic. You can use the fact that \( \mathbb F_{p^n} \) is Galois over \( \mathbb F_p \) with a cyclic Galois group generated by the Frobenius automorphism: \(\alpha \mapsto \alpha^p \). You are not required to use TeX but if you don't write very neatly.

1. Let \(f(x) = x^4 + ax^2 + b\) be an irreducible polynomial over a field \( F \) whose characteristic is not 2. Show the Galois group of \( f \) has order 4 or 8. Show the Galois group is the Klein 4-group \( V \) if and only if \( \sqrt b \in F \). Hint: Find the resolvent cubic and show it is always reducible.

2. Let \( \alpha = \sqrt{2+\sqrt 2} \). Show that the minimum polynomial of \( \alpha \) over \( \mathbb Q \) is \( f(x) = x^4 - 4x^2 + 2 \). Show that \( \mathbb Q(\alpha)/\mathbb Q \) is Galois with Galois group \( C_4 \) (the cyclic group). Hint: show that \( \beta \in \mathbb Q(\alpha) \) where \( \beta = \sqrt{2-\sqrt 2} \). (Hint for the hint: evaluate \( \alpha \beta \).)

3. Let \( x^3 + px + q \in F[x] \) be an irreducible cubic, where \( F \) is a finite field. Show \( -4p^3 - 27 q^2 \) is a square in \( F \).

4. Let \( f(x) \in F[x] \) be an irreducible, separable polynomial of degree 4 and let \( G \) be the Galois group of \( f \). Let \( \alpha \) be a root of \( f \). Show there is no field properly between \( F \) and \( F(\alpha) \) if and only if \( G = S_4 \) or \( G = A_4 \).

5. (Bonus problem) As in problem 1, let \(f(x) = x^4 + ax^2 + b\) be an irreducible polynomial over a field \( F \) whose characteristic is not 2. The roots of \( f \) are the square roots of the quadratic \( u^2 + au + b \). Let \( \alpha \), \( -\alpha \), \( \beta \) and \(-\beta \) be these roots. Show that \( \beta \in F(\alpha) \) (equivalently \( F(\alpha) = F(\alpha,\beta) \)) if and only if either \( b \) or \( b(a^2-4b) \) is a square in \( F \). Use this to show that the Galois group of \( f \) is \( C_4 \) if and only if \( b(a^2-4b) \) is a square in \( F \) and \( \sqrt b \notin F \).

Solutions to 1 and 5) The resolvent cubic is \( y^3 - ay^2 -4by + 4ab = (y-a) (y^2 - 4b) \). Using Theorem 43 of Kaplansky's Fields and Rings this gives that the Galois group of \( f(x) \) is either \( V, C_4 \) or \( D_8 \) and that it is \( V \) if and only if \( \sqrt b \in F \). (Actually you need to show that \( f(x) \) is separable which you can do by just finding the roots using the quadratic formual.)

Of course \( F(\alpha,\beta) \) is the splitting field. Now \( \beta \in F(\alpha) \) iff \( F(\alpha) = F(\alpha,\beta) \) and \( [F(\alpha) : F] = 4 \). So \( \beta \in F(\alpha) \) iff the Galois group, \( G \), is of order 4. An easy calculation shows \( \alpha \beta = \sqrt b \). So if \( b \) is a square in \( F \) the \( \beta = \sqrt b / \alpha \in F(\alpha) \). If \( r = \sqrt{b(a^2-4b)} \) is in \( F \), then (following Mr. Sanford whose proof was nicer), using \( 2\alpha^2 + a = \sqrt{a^2-4b} \) we get \[ \frac{r}{2\alpha^2 + a} = \sqrt b = \alpha \beta \] showing that \( \beta \in F(\alpha) \) is this case as well. This proves one direction of the first statement.

Now suppose \( \beta \in F(\alpha) \) and \( \sqrt b \notin F \). So \( G \) has order 4 and so must be \( C_4 \), since it's not \( V \). Since \( C_4 \) has only one subgroup of order 2, \( F(\alpha) \) has only one intermediate field of degree 2. If \( a^2 - 4b \) is also not a square in \( F \) then the intermediate fields \( F(\sqrt b) = F(\alpha\beta) \) and \( F(\sqrt{a^2-4b}) = F(\alpha^2) \) must be equal. So there must be \( x, y \in F \) such that \[ \sqrt{a^2-4b} = x + y\sqrt b \] Squaring both sides shows \( x = 0 \) and multipling by \( \sqrt b \) gives \( \sqrt{b(a^2 - 4b)} = yb \), showing \( \sqrt{b(a^2 - 4b)} \) is in \( F \). All the statements of the problem follow.