1. ## Galois Theory

1. The polynomail $\displaystyle f(x) = x^5-5x+1$ is irreducible in $\displaystyle \mathbb Q [x]$. Find the number if real roots of $\displaystyle f(x).$

2. Let M be the splitting field for $\displaystyle f(x)$ over $\displaystyle \mathbb Q.$ Show that $\displaystyle M/\mathbb Q$ is a Galois extension
and determine its Galois group $\displaystyle Gal(M/\mathbb Q).$ Is $\displaystyle f(x)$ solvable by root?

In the following it can be used without a proof that the Discrimentant of $\displaystyle f(x)$
is $\displaystyle -3\cdot5^6\cdot 17.$

3. Show that $\displaystyle \mathbb Q(\sqrt{-51})$ is contained in $\displaystyle M$ and determine that $\displaystyle f(x)$
is irreducible in $\displaystyle \mathbb Q(\sqrt{-51})[x].$

4. Determine the Galois group $\displaystyle Gal(M/ \mathbb Q(\sqrt{-51})).$

2. Originally Posted by peteryellow
1. The polynomail $\displaystyle f(x) = x^5-5x+1$ is irreducible in $\displaystyle \mathbb Q [x]$. Find the number if real roots of $\displaystyle f(x).$

2. Let M be the splitting field for $\displaystyle f(x)$ over $\displaystyle \mathbb Q.$ Show that $\displaystyle M/\mathbb Q$ is a Galois extension
and determine its Galois group $\displaystyle Gal(M/\mathbb Q).$ Is $\displaystyle f(x)$ solvable by root?

In the following it can be used without a proof that the Discrimentant of $\displaystyle f(x)$
is $\displaystyle -3\cdot5^6\cdot 17.$

3. Show that $\displaystyle \mathbb Q(\sqrt{-51})$ is contained in $\displaystyle M$ and determine that $\displaystyle f(x)$
is irreducible in $\displaystyle \mathbb Q(\sqrt{-51})[x].$

4. Determine the Galois group $\displaystyle Gal(M/ \mathbb Q(\sqrt{-51})).$
1. use rational root test

3. rational root test to find real root? and then whT bout others?

4. Originally Posted by peteryellow
1. The polynomail $\displaystyle f(x) = x^5-5x+1$ is irreducible in $\displaystyle \mathbb Q [x]$. Find the number if real roots of $\displaystyle f(x).$
Using analytic methods (or just graphing it) we see that $\displaystyle f$ has three real roots, the other two roots are complex conjugates.

2. Let M be the splitting field for $\displaystyle f(x)$ over $\displaystyle \mathbb Q.$ Show that $\displaystyle M/\mathbb Q$ is a Galois extension
and determine its Galois group $\displaystyle Gal(M/\mathbb Q).$ Is $\displaystyle f(x)$ solvable by root?
Obviously, $\displaystyle M/\mathbb{Q}$ is a Galois extension since it is a splitting field over a perfect field. We want to compute $\displaystyle \text{Gal}(M/\mathbb{Q})$. First of all we can identify the Galois group as a subgroup of $\displaystyle S_5$ because each automorphism of $\displaystyle M$ is also a permutation of the roots of $\displaystyle f$. To show that $\displaystyle \text{Gal}(M/\mathbb{Q})\simeq S_5$ all we need to do is show that $\displaystyle \text{Gal}(M/\mathbb{Q})$ contains a permutation of order 5 and a transpotion.* Let $\displaystyle \alpha$ be a root of $\displaystyle f$ in $\displaystyle M$ then $\displaystyle \mathbb{Q}\subseteq \mathbb{Q}(\alpha)\subseteq M$. Notice that $\displaystyle [\mathbb{Q}(\alpha):\mathbb{Q}] = 5$, therefore $\displaystyle 5$ divides $\displaystyle |\text{Gal}(M/\mathbb{Q})$ by the fundamental theorem of Galois theory. Thus, by Cauchy's theorem there is an element in $\displaystyle \text{Gal}(M/\mathbb{Q})$ of order $\displaystyle 5$ i.e. a $\displaystyle 5$-cycle if viewed as an element in $\displaystyle S_5$. And surly there is a transposition in $\displaystyle \text{Gal}(M/\mathbb{Q})$, just take $\displaystyle \sigma: M\to M$ defined by $\displaystyle \sigma (x) = \overline{x}$ i.e. complex conjugation. Then $\displaystyle \sigma$ is a transposition if viewed as an element of $\displaystyle S_5$. Therefore, $\displaystyle \text{Gal}(M/\mathbb{Q}) \simeq S_5$. This symmetric group is not solvable therefore the polynomial is not solvable by radicals.

*)A subgroup of $\displaystyle S_5$ generated by a 5-cycle and a 2-cycle must be $\displaystyle S_5$ itself.

3. Show that $\displaystyle \mathbb Q(\sqrt{-51})$ is contained in $\displaystyle M$ and determine that $\displaystyle f(x)$
is irreducible in $\displaystyle \mathbb Q(\sqrt{-51})[x].$

4. Determine the Galois group $\displaystyle Gal(M/ \mathbb Q(\sqrt{-51})).$
Let $\displaystyle \Delta = \sqrt{-3\cdot 5^6\cdot 17}$ then $\displaystyle \mathbb{Q}(\Delta)$ corresponds to subgroup $\displaystyle S_n\cap A_n = A_n$. Notice that $\displaystyle \mathbb{Q}(\Delta) = \mathbb{Q}(\sqrt{-51})$, therefore $\displaystyle \text{Gal}(M/\mathbb{Q}(\sqrt{-51})) \simeq A_n$ under this Galois correspondence.