By Takashi Ono (auth.)

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**Extra resources for An Introduction to Algebraic Number Theory**

**Sample text**

19). 22), we have aba = baa E 7L; similarly we have abf3 E 7L. Therefore, ab(a ± (3) E 7L. 41). Next, since 7L is a ring, we have aabf3 = abaf3 E 7L. 3. 37 To the Gauss Reciprocity Law have a[3 E (IlZ c 0, again. Finally, we must verify that a-I EO if a E 0 and a =t= 0. To do this, let ao«' + al a n - I + ... + an-I a + an = 0, ai E (Il, n 2': 1, ao=t= 0, be a relation for a E O. Multiplying by a- n on both sides, we get aO+alCl' -1 + ... +an-l (-l)n-l Cl' +an(-l)n Cl' =0 Since at least one of aI, ...

Q n Z = Z. PROOF. Since a- E Z is a root of f(X) = X - a- we have a- E Z and so Z c Q c Z. Now take any r E Q n Z. Write r = c/d, d 2::: 1, (c, d) = 1. Since r E Z, we find f(X) = xn + a1Xn-l + ... , (c/dt + a1(c/dt- 1+ ... + an = O. Multiplying d n on both sides, we have c n + a1cn-1d + ... + and n = 0, which implies that d I c n. , r E lL. Therefore we have Q n lL = lL. D. 19. For any algebraic number a- E (J), there is a natural number a E N such that aa- is an algebraic integer: aa- E lL. PROOF.

Since both sides of (p) are ± 1 and p is odd, we must have the equality. D. 26 (The second supplementary law). = e and 'Y/ = ,+ ,-1. Since ,4 = -1, we get ,2 + PROOF. ,-2 = 0 Put, and 'Y/ = (' + ,-1)2 = ,2 + 2 + ,-2 = 2. 19). 21, we have Ap(2)1] == 1]P C-1)P == CP + C-P (p). Now, we have p == ±l (8):} ~p P == ±3 (8):} ~p = (C + + ~-p = ~ + ~-l = 1], + ~-p = ~-3 + ~-3 = _~-l_ ~ = -1], and hence we have ~p + ~-p = { p==±l p == ±3 1], -1], (8), (8). In other words, we get (-1)"'1] == Ap(2) 1] (p), with {jJ = (p2_ 1)/8, and then (-1)"'2 == AA2)2 (p) because 1]2 = 2.