User:J D Bowen/Math735 HW9

13.1.4 Prove directly that the map $$ a + b \sqrt{2} \to  $$ $$ a - b \sqrt{2} \ $$ is an isomorphism of $$ \Q (\sqrt{2}) \ $$ itself.

Let $$ \phi : \Q (\sqrt{2}) \to \Q (\sqrt{2}) \ $$ be given by $$ \phi (a + b \sqrt{2}) \ $$ = $$ a - b \sqrt{2} \ $$. We want to show that $$ \phi \ $$ is a ring homomorphism. Note that

$$ \phi \ $$ (($$a + b \sqrt{2} \ $$) + ($$ c + d \sqrt{2})) \ $$ =

$$ \phi \ $$ ($$ (a+c) + (b+d) \sqrt{2} \ $$)

= $$a + c - (b+d) \sqrt{2} \ $$

= ($$ a - b \sqrt{2}) \ $$ +  $$ (c - d \sqrt{2}) \ $$

= $$ \phi (a + b \sqrt{2}) \ $$ + $$ \phi (c + d \sqrt{2}) \ $$.

Then, $$ \phi \ $$ ((a + b $$ \sqrt{2} \ $$) (c + d $$ \sqrt{2})) \ $$ =

= $$ \phi (ac + 2bd) + (ad + bc) \sqrt{2} \ $$

= $$ ac + 2bd - (ad + bc) \sqrt{2} \ $$

= $$ (a - b \sqrt{2}) (c - d \sqrt{2}) \ $$

= $$ \phi \ $$ (a + b $$ \sqrt{2} \ $$) $$ \phi \ $$ (c + d $$ \sqrt{2}) \ $$

In $$ \Q (\sqrt{2}) \ $$, 0 = 0 + 0 $$ \sqrt{2} \ $$, and the kernel of $$ \phi \ $$ is 0, which implies that it is injective. Then for any a + b $$ \sqrt{2} \in \Q \sqrt{2},  \exists $$ a - b $$ \sqrt{2} \ $$ such that $$ \phi \ $$ (a - b $$ \sqrt{2} \ $$) = a + b $$ \sqrt{2} \ $$, and thus $$ \phi \ $$ is surjective, and therefore bijective. Thus $$ \phi \ $$ is an isomorphism of $$ \Q (\sqrt{2}) \ $$ with itself.

13.2.3 Determine the minimum polynomial over $$ \Q \ $$ for the element 1 + $$ i \ $$.

Let $$ \alpha \ $$ =  1 + $$ i \ $$. Note that $$ i^2 \ $$ = -1.

Then we have $$ \alpha^2 \ $$ = $$ (1 + i )^2 \ $$ = $$ 1^2 + i^2 + 2i \ $$ = 1 - 1 + 2i = 2($$ \alpha \ $$ - 1).

Thus $$ \alpha \ $$ is a root of $$ x^2 \ $$ -2x + 2 $$ \in \Q \ $$ [X]. According to Eisenstein's criterion, using prime 2, this polynomial is irreducible, and it is monic. So, it is therefore the minimum polynomial of its root, one of which is $$ \alpha \ $$.

13.2.7Prove that $$ \Q (\sqrt{2} + \sqrt{3}) \ $$ = $$ \Q (\sqrt{2}, \sqrt{3}) \ $$.

$$ \sqrt{2} + \sqrt{3} \in \Q (\sqrt{2}, \sqrt{3}) \ $$, so $$ \Q ( \sqrt{2} + \sqrt{3}) \subseteq \Q ( \sqrt{2} , \sqrt{3}) \ $$. Consider $$ \alpha = ( \sqrt{2} + \sqrt{3}) \ $$. Then $$ \alpha^2 = 2 + 3 + 2 \sqrt{2} \sqrt{3} \ $$. But $$ ( \sqrt{2} \sqrt{3} ) = \frac{1}{2} ( \alpha^2 - 5) \in \Q ( \alpha ) \ $$. It follows $$ \alpha \sqrt{2} \sqrt{3} - 2 \alpha \in \Q ( \alpha ) \ $$, but $$ \alpha \sqrt{2} \sqrt{3} - 2 \alpha = (\sqrt{2} + \sqrt{3})\sqrt{2} \sqrt{3} \ $$ = $$ 2\sqrt{3} + 3\sqrt{2} - 2\sqrt{2} - 2\sqrt{3} \ $$ = $$ \sqrt{2} \ $$. Thus, $$ \sqrt{2} \in \Q ( \alpha ) \ $$, which implies that $$ \sqrt{3} = \alpha - \sqrt{2} \in \Q ( \alpha ) \ $$. Thus $$ \Q (\sqrt{2}, \sqrt{3}) \subset \Q ( \sqrt{2} + \sqrt{3} ) \ $$. Therefore $$ \Q (\sqrt{2} + \sqrt{3}) \ $$ = $$ \Q (\sqrt{2}, \sqrt{3}) \ $$. Thus, since [$$ \Q (\sqrt{2}, \sqrt{3}): \Q \ $$] = 4 , [$$ \Q (\sqrt{2} + \sqrt{3}): \Q \ $$] = 4.

13.2.12 Suppose the degree of the extension K/F is a prime $$ p \ $$. Show that any subfield E of K containing F is either K or F.

Note that, $$ p \  $$ = [K: F] = [K: E][E: F], by Theorem 14.

So, [E: F] is a positive integer that divides $$ p \ $$, and therefore must be equal to either 1 or $$ p \ $$. Then if [E: F] = 1, then E = F. And if [E: F] = $$ p \ $$, then [K: E] = 1 and since $$ p \ $$ = [K: F] = [K: E][E: F], K = F. Thus K = E or F = E.