galois -check solution

**dopi** · March 23rd 2009, 06:11 PM

can someone check my solutions please

f is poly t^5 -3 in Q[t], where alpha= 3^1/5 in R, epsilon = e^(2pi/5)

(a) need to find zeros of f in terms of a and epsilon

my solution to (a)

alpha, alpha*epsilon, alpha*epsilon^2, alpha*epsilon^3, alpha*epsilon^4

(b) L = (alpha, epsilon). want to show that L is a splitting field of f over Q.

my solution to (b)
zeros of f are given by (t -alpha)*(t -alpha*epsilon)*(t -alpha*epsilon^2)*(t -alpha*epsilon^3)*(t -alpha*epsilon^4)

so L = (alpha, epsilon) is splitting field

(c) want to state min polynomial of alpha over Q ane epsilon over Q(alpha), and from this i want to write down the bases of the extensions Q(alpha):Q and L:Q(alpha) and then find the value of [L:Q]

my solution to (c)

min poly of alpha t^5 - 3
min poly of epsilon t^4 + t^3 + t^3 + t^2 + t + 1

a basis for Q(alpha) over Q is {1, alpha, alpha^2, alpha^3,alpha^4}
a basis for Q(alpha, epsilon) over Q(alpha) is {1, epsilon, epsilon^2, epsilon^3}

therefore [L:Q] = 9

thank you

**siclar** · March 23rd 2009, 07:41 PM

(a) is correct.

For (b), you need to invoke more than the factorization. Namely, $\mathbb{Q}(\alpha)$ is not the splitting field because there are complex roots of the polynomial and this field is real. The only other possible proper subfield is $\mathbb{Q}(\epsilon)$ , but if it is indeed a proper subfield of $L$ , then it does not contain $\alpha$ and so cannot split the polynomial. Thus $L$ is the splitting field.

For (c), everything but your calculation of $[L:\mathbb{Q}]$ is correct. You seem to have added instead of multiplied. $[L:\mathbb{Q}]=[L:\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$ . But $[L:\mathbb{Q}(\alpha)]=4$ since the minimal polynomial for epsilon is still of degree 4 and $[\mathbb{Q}(\alpha):\mathbb{Q}]=5$ similarly. Thus $[L:\mathbb{Q}]=4*5=20$ .

You multiply the degrees because the basis is basically the product of bases, i.e. $1, \alpha, \ldots, \alpha^4, \epsilon, \epsilon\alpha, \ldots, \epsilon\alpha^4, \ldots, \epsilon^3\alpha^4$ .

**ThePerfectHacker** · March 23rd 2009, 08:04 PM

Originally Posted by dopi

can someone check my solutions please

f is poly t^5 -3 in Q[t], where alpha= 3^1/5 in R, epsilon = e^(2pi/5)

(a) need to find zeros of f in terms of a and epsilon

my solution to (a)

alpha, alpha*epsilon, alpha*epsilon^2, alpha*epsilon^3, alpha*epsilon^4

This is good.

(b) L = (alpha, epsilon). want to show that L is a splitting field of f over Q.

my solution to (b)
zeros of f are given by (t -alpha)*(t -alpha*epsilon)*(t -alpha*epsilon^2)*(t -alpha*epsilon^3)*(t -alpha*epsilon^4)

so L = (alpha, epsilon) is splitting field

This is good.

(c) want to state min polynomial of alpha over Q ane epsilon over Q(alpha), and from this i want to write down the bases of the extensions Q(alpha):Q and L:Q(alpha) and then find the value of [L:Q]

my solution to (c)

min poly of alpha t^5 - 3
min poly of epsilon t^4 + t^3 + t^3 + t^2 + t + 1

a basis for Q(alpha) over Q is {1, alpha, alpha^2, alpha^3,alpha^4}
a basis for Q(alpha, epsilon) over Q(alpha) is {1, epsilon, epsilon^2, epsilon^3}

therefore [L:Q] = 9

thank you

There is a problem here.

You need to verify that these are minimal polynomials by showing they are irreducible. It is easy to see that $t^5-3$ is irreducible over $\mathbb{Q}$ by Eisenstein. Now since $[\mathbb{Q}(\alpha):\mathbb{Q}]=5$ and $f(t) =t^4+t^3+t^2+t+1$ is irreducible over $\mathbb{Q}$ with $\gcd(4,5)=1$ it follows that $f(t)$ is irreducible over $\mathbb{Q}(\alpha)$ . Therefore, $t^4+t^3+t^2+t+1$ is minimal polynomial over $\mathbb{Q}(\alpha)$ . We have proven that $\alpha$ has degree $5$ over $\mathbb{Q}$ and $\varepsilon$ has degree $4$ over $\mathbb{Q}(\alpha)$ . Therefore, a basis for $L/\mathbb{Q}(\alpha)$ is $\{1,\varepsilon,\varepsilon^2,\varepsilon^3\}$ and a basis for $\mathbb{Q}(\alpha)/\mathbb{Q}$ is $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ . Thus, $L/\mathbb{Q}$ has a basis consisting of all these possible products between these bases and so there are $4\cdot 5=20$ elements which means $[L:\mathbb{Q}]=20$ .

Just in case you are interested the Galois group is the Frobenius group $F_{20}$ .

**dopi** · March 24th 2009, 09:54 AM

Originally Posted by ThePerfectHacker

This is good.

This is good.

There is a problem here.

You need to verify that these are minimal polynomials by showing they are irreducible. It is easy to see that $t^5-3$ is irreducible over $\mathbb{Q}$ by Eisenstein. Now since $[\mathbb{Q}(\alpha):\mathbb{Q}]=5$ and $f(t) =t^4+t^3+t^2+t+1$ is irreducible over $\mathbb{Q}$ with $\gcd(4,5)=1$ it follows that $f(t)$ is irreducible over $\mathbb{Q}(\alpha)$ . Therefore, $t^4+t^3+t^2+t+1$ is minimal polynomial over $\mathbb{Q}(\alpha)$ . We have proven that $\alpha$ has degree $5$ over $\mathbb{Q}$ and $\varepsilon$ has degree $4$ over $\mathbb{Q}(\alpha)$ . Therefore, a basis for $L/\mathbb{Q}(\alpha)$ is $\{1,\varepsilon,\varepsilon^2,\varepsilon^3\}$ and a basis for $\mathbb{Q}(\alpha)/\mathbb{Q}$ is $\{1,\alpha,\alpha^2,\alpha^3,\alpha^4\}$ . Thus, $L/\mathbb{Q}$ has a basis consisting of all these possible products between these bases and so there are $4\cdot 5=20$ elements which means $[L:\mathbb{Q}]=20$ .

Just in case you are interested the Galois group is the Frobenius group $F_{20}$ .

what would be the order of $\Gamma Q(t^5 - 3)$

i thought it might be 5, because the basis of the roots has exactly 5 elements? is this correct

**ThePerfectHacker** · March 24th 2009, 07:12 PM

Originally Posted by dopi

$\Gamma Q(t^5 - 3)$

How is this defined?

**dopi** · March 24th 2009, 07:34 PM

Originally Posted by ThePerfectHacker

How is this defined?

f = t^5 - 3 in Q[t] (Q is rationals)
$<br /> \Gamma Q(f)<br />$
this is all i was given, apart from all the other parts answered above
thanks

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