By Fokkinga M.M.

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**Extra info for A gentle introduction to category theory. The calculational approach**

**Sample text**

Let a21 and a22 be in Z with a21 ≥ 0, a22 > 0, and a22 as small as possible with β2 = a21 ω1 + a22 ω2 ∈ I (by considering a21 = 0 and a22 = a, we see that such a21 and a22 exist). Note that by considering β2 −kβ1 for some k ∈ Z, we may also suppose that a21 < a11 . In general, for i ∈ {1, 2, . . , n}, define βi = ai1 ω1 + ai2 ω2 + · · · + aii ωi ∈ I with 0 ≤ aij < ajj for j ∈ {1, 2, . . , i − 1} and aii > 0 as small as possible. Observe that aii ≤ a for all i ∈ {1, 2, . . , n} and, hence, aij ≤ a for all i and j with 1 ≤ j ≤ i ≤ n.

I) (j) (1) (1) β1 βn . . det .. (n) (1) βn βn (2) β1 .. (2) βn ... . (n) β1 .. . (n) βn (i) (j) = det β1 β1 + β2 β2 + · · · + βn(i) βn(j) = det T r(β (i) β (j) ) , where the last equation follows from an application of the last lemma. • Integral bases. The numbers 1, α, α2 , · · · , αn−1 form a basis for Q(α) over Q. It follows that every bases for Q(α) over Q consists of n elements. Let R be the ring of algebraic integers in Q(α). We next seek to find a basis for R over Z.

Hence, there exist integers b and c such that g(x) ≡ bxr (mod p) and h(x) ≡ cxs (mod p). Since r > 0 and s > 0, we get that p divides the constant terms of g(x) and h(x). This contradicts that p2 a0 , completing the proof. • The polynomial Φp (x). Using Theorem 45, we can obtain Theorem 46. For every prime p, Φp (x) = xp−1 + xp−2 + · · · + x + 1. Proof. Let f (x) = xp−1 + xp−2 + · · · + x + 1. Since ζp = 1 and ζp is a root of xp − 1 = (x − 1)f (x), we obtain that ζp is a root of f (x). Thus, it suffices to show that f (x) is irreducible over Q.

### A gentle introduction to category theory. The calculational approach by Fokkinga M.M.

by George

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