Question

Problem 5. Let E1 = Q($\sqrt$2,$\sqrt$7 ), E2= ($\sqrt$2,$\sqrt[3]{2}$), $\alpha$ ₁ = 2$\sqrt$2 + 7$\sqrt$7, and $\alpha$ ₂ = 2$\sqrt$2 + 3($\sqrt[3]{2}$)

(i) Determine [E_i : Q] for i = 1, 2.
(ii) Determine a basis of E_i over Q for i = 1, 2.
(iii) Determine the minimal polynomial of $\alpha$ _i over Q for i = 1, 2.
(iv) Determine if each of the extensions E₁ / Q and E₂ / Q is Galois.

Answer 1

(i). Minimal polynomial of $\sqrt{2}$ over Q is
2
and so
(Q(V2): Q) = 2
. Similarly minimal polynomial of $\sqrt{7}$ over
Q(V2)
is
2 _7
and so
(Q(V2, 17): Q(V2)) = 2
. Hence,
(E1 : 0) = (Q(V2, 17): Q(V2))(QV2) : 0) = 2 x 2 = 4
.

Minimal polynomial of $\sqrt[3]{2}$ over $\mathbb{Q}$ is
とで
and so 
(Q(V2, 72), Q(x2)) = 3
. Hence,
(E2 : 0) = (Q(V2, 12): Q(12))(Q(12) : ) = 3 x 2 = 6
.

(ii). A basis of
Q(V2)
over $\mathbb{Q}$ is
{1, 2}
and a basis of
Q[/2, 17)
over
Q(V2)
is
LA'T}
. Hence, a basis of $E_1$ over $\mathbb{Q}$ is
{1, V2, V7, V14)
.

A basis of
Q[/2, 12)
over
Q(V2)
is
{1, 72, 74
. Hence a basis of over $\mathbb{Q}$ is
{1, V2, 12, 1272, 74, 72/4
.

(iii). Let
a = 2V2
and
f(x) = x2 – 343
. Now,
f(x – 2/2) = (x – 272)2 - 7 = x2 – 4/2x+8 – 343 = x2 – 4/22 – 335
. And
f(x+2V2) = (2+2V2)2 – 7 = 22 +4V2c+8343 = + 4V2c - 335
. So,
f (x – 2/2)f(x + 2/2) = (x2 – 4V2.c – 335)(x2 + 4/2x – 335) = (x2 – 335)2 – (472c)2 = x4 – 670.22 +112225 – 32.c
. Since $7\sqrt{7}$ is a root of ,
2V2+7V7
is a root of
f(x - 212)
and hence the minimal polynomial of
2V2+7V7
is the above 4 degree polynomial.

Take
g(2) = 73 – 54
. Then,
g(x - 2V2) = (1 - 2V2)3 – 54 = 73 - 6V2.1? + 241 – 16V2 - 54
and
9(2+2/2) = 2.3 +67222 +24x + 16V2 – 54
. Similarly as above, since $3\sqrt[3]{2}$ is a root of minimal polynomial of
2024332
is
g(x2 - 2V2)9(2+2V2)
.

(iv). Note that, $E_1$ is the splitting field of the polynomial
(22 - 2) (1 - 7)
over $\mathbb{Q}$ . Since a splitting field is always a normal extension, we get that $E_1/\mathbb{Q}$ is a finite normal extension. Also since characteristics of $\mathbb{Q}$ is 0, any field extension of $\mathbb{Q}$ is separable. Hence, $E_1/\mathbb{Q}$ is a finite, normal, separable extension and so it is a Galois extension.

Now note that,
とで
is an irreducible polynomial over $\mathbb{Q}$ and $E_2$ contains a root ( namely, $\sqrt[3]{2}$ ) of this polynomial. Now if this was a normal extension, then $E_2$ have to contain all the roots the above irreducible polynomial. But $\sqrt[3]{2}\omega$ is a root of the polynomial ( $\omega$ is a complex cube root of 1) and we can clearly see that $E_2$ does not contain $\sqrt[3]{2}\omega$ . Hence,
E2/Q
is not a normal extension and so it's not a Galois extension.