By Kleshchev A.S.

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**Extra resources for 1-Cohomologies of a special linear group with coefficients in a module of truncated polynomials**

**Example text**

Assume a2 ≡ −1 (mod p) holds for some integer a. Since (p − 1)/2 is odd, ap−1 ≡ (a2 )(p−1)/2 ≡ (−1)(p−1)/2 ≡ −1 (mod p). Fermat’s Little Theorem implies ap−1 is either 0 or 1 modulo p, giving a contradiction. Hence, the lemma follows. Proof of Theorem 62. Assume there are integers x and y satisfying y 2 + 5 = x3 . Then y 2 ≡ x3 − 1 (mod 4) implies that x ≡ 1 (mod 4) (anything else leads to a contradiction). Observe that y 2 + 4 = x3 − 1 = (x − 1)(x2 + x + 1) 49 and x2 + x + 1 ≡ 3 (mod 4). It follows that there must be a prime p ≡ 3 (mod 4) dividing x2 + x + 1 and, hence, y 2 + 4.

Det b1 . .. 0 0 1 .. ... .. 0 0 .. ... .. b2 .. ... . θ .. ... . 0 0 0 0 .. . = θ, bn .. . 1 the lemma implies that ∆(ω (1) , . . , ω (n) ) = θ2 ∆(ω (1) , . . , ω (n) ). Thus, 0 < |∆(ω (1) , . . , ω (n) )| < |∆(ω (1) , . . , ω (n) )|. On the other hand, since ω (k) = β − uω (k) , each ω (j) is an algebraic integer for 1 ≤ j ≤ n. This contradicts the minimality of |∆(ω (1) , . . , ω (n) )|, completing the proof. Homework: (1) Let ω (1) , . . , ω (n) be an integral basis in Q(α).

A > 0 and b > 0. Since a + b N < x + y Now, a + b N > 1 implies 1 1 N , the minimality √ condition on x1 + y1 N now gives a contradiction. The theorem follows. √ Comment: If N is squarefree and N ≡ 1 (mod 4), then the number x + y N is called 1 1 √ the fundamental unit for the ring of algebraic integers in Q( N ). Theorem 33 implies that the fundamental generates all units in the ring in the sense that the units are given √ unit m by ±(x1 + y1 N ) where m denotes an arbitrary integer. • An example.