User:Michellepoliseno /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.