By Ash R.B.

It is a textual content for a simple path in algebraic quantity concept.

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7), we found that any nonzero ideal is a factor of a principal ideal. We can sharpen this result as follows. 2 Proposition Let I be a nonzero ideal of the Dedekind domain R. Then there is a nonzero ideal I such that II is a principal ideal (a). Moreover, if J is an arbitrary nonzero ideal of R, then I can be chosen to be relatively prime to J. Proof. Let P1 , . . 1). Then a ∈ I, so (a) ⊆ I. 5)], I divides (a), so (a) = II for some nonzero ideal I . If I is divisible by Pi , then I = Pi I0 for some nonzero ideal I0 , and (a) = IPi I0 .

7 Theorem If I and J are nonzero ideals of B, then N (IJ) = N (I)N (J). Proof. By unique factorization, we may assume without loss of generality that J is a prime ideal P . By the third isomorphism theorem, |B/IP | = |B/I| |I/IP |, so we must show that |I/IP | is the norm of P , that is, |B/P |. 6). 8 Corollary Let I be a nonzero ideal of B. If N (I) is prime, then I is a prime ideal. Proof. Suppose I is the product of two ideals I1 and I2 . 7), N (I) = N (I1 )N (I2 ), so by hypothesis, N (I1 ) = 1 or N (I2 ) = 1.

Show that the trace form Ti (x, y) = TFi /Fp (πi (x)πi (y)) is nondegenerate, and conclude that i Ti is also nondegenerate. We have d = det T (ωi ωj ), in other words, the determinant of the matrix of the bilinear form T (x, y) on B, with respect to the basis {ω1 , . . , ωn }. Reducing the matrix entries mod p, we get the matrix of the reduced bilinear form T0 on the Fp -vector space B/(p). 6. Show that T0 coincides with i Ti , hence T0 is nondegenerate. Therefore d = 0 mod p, so p does not divide d.