- In Particular All Powers 1 2 Are Units Even Negative Powers Of 1 2 Are Units Because They Are Positive Po 1 (114.91 KiB) Viewed 43 times
] - In particular, all powers (1 + √2)" are units. Even negative powers of 1 + √2 are units because they are positive powers of the inverse -1+√2. Since these powers give distinct values, Z[√2] has infinitely many units. (By the way, there are infinitely many non-units too. An element of Z[√2] taken at random will likely be a non-unit.) (a) If z = a+b√2 € Z[√2], the conjugate of z is the number z = a-b√2 also in Z[√2]. The norm N(z) is an integer defined by the rule: N(2) = zz = (a+b√2)(a - b√2) = a² - 26². For example, N(1+√2) = -1, N(3 + 2√2) = 1, N(3+√2) = 7. As verified in the study of Pell's equation, N is a multiplicative func- tion. That is N(zw) = N(z)N(w), for all z, w in Z[√2]. If u € Z[√2], prove that u is a unit if and only if N(u) = ±1. (b) Part (a) informs us that in the search for units in Z[√2], we need to consider those elements v that satisfy N(v) = 1, and also those w that satisfy N(w) = -1. Now let u be the special unit 3 + 2√2. This unit played a key role in solving Pell's equation. Suppose that v E Z[√2] and that N(v) = 1. Show that u must be an integer power of u or the negative of an integer power of u. That is v=tu" where n is an integer. Hint. The requirement N(v) = 1 is nothing but Pell's equation, which was solved in an earlier exercise. It comes down to proper application of that solution. If v = a +b√2, explain why a is never 0 and then discuss separately the five cases which follow: 3 1 S 4 2 a 0,b=0; a > 0, b>0; a < 0,6 <0; a > 0, b<0; a<0, b>0. (c) Let z be the special unit 1 + √2. Observe that N(z) = -1 and also that 2² = 3+2√/2, which was the special unit u of part (b). 17