# Thread: Galois Theory

1. ## Galois Theory

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

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

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

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

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

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

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

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

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

4. Determine the Galois group $Gal(M/ \mathbb Q(\sqrt{-51})).$
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 $f(x) = x^5-5x+1$ is irreducible in $\mathbb Q [x]$. Find the number if real roots of $f(x).$
Using analytic methods (or just graphing it) we see that $f$ has three real roots, the other two roots are complex conjugates.

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

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

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

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