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3 2 < = 0 > 8 7 t @ l u v s s w bje?Proof Let u 2R be nonzero Since u 2F is algebraic over K the extension K(u)=K is nite with basis B = f1;u;u2;;ukgfor some k Since R is a ring containing< = > 6 $ ?

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Start studying Logic Final (Quizzes 4 & 5) Learn vocabulary, terms, and more with flashcards, games, and other study toolsSection 74 Part III (1) 1 (S • K) ⊃ R 2 K 4 T 2, 3, MP/ S ⊃ R 3 (K • S) ⊃ R 1, Com 4 K ⊃ (S ⊃ R) 3, Exp 5 S ⊃ R 2, 4, MPConsequently, if 2 and 3 are not squares in Fp, it follows that 2 ≡ y2k1( mod p) and 3 ≡ y2l1( mod p), for some k,l ∈ Z Hence 6 ≡ y2(kl1)( mod p) is a

L m hi p n l qk j ri ^ _ ` a b ` _ b rs k wl n l p qh j k j l n l l x i o vqw k oq rs _ c d e l is rp qj ix \ y qr j oi ql zj k w o j y y qr n n j ri rp ml p f m oi g p j l s j i n h\ \ rp k " ( # 0 !B) a b = p a2L We conclude that K L x62 #10 Let F be an extension field of K Let a 2F be algebraic over K, and let t 2F be transcendental over K2 0 2 1 2 0 2 2 S t u d e n t S u p p l y L i s t K i n d e r g a r t e n We e n co u r a g e yo u t o r e u se su p p l i e s w h e n e ve r p o ssi b l e • B a ckp a ck (label with your child's name) • 1 b o x o f b e g i n n e r p e n ci l s ( t h e se a r e t h i cke r p e n ci l s)

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