By Grauert G. (ed.)

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**Example text**

An algebraic extension of Q of degree 2 is traditionally called a √ quadratic ﬁeld. Such a ﬁeld is given as F = Q( α ) with an element α of Q× that is not a square in Q. We call F a real or imaginary quadratic ﬁeld according as α > 0 or α < 0. Replacing α by its suitable integer √ multiple, we may assume that F = Q( m ) with a square-free positive or negative integer m = 1. Let J = JF as before. 10a) J = Z[μ], μ = (1 + m )/2 if m − 1 ∈ 4Z. 10b) J = Z[ m ] if m − 1 ∈ / 4Z. √ To prove these, take ξ = α+β m ∈ J with α, β ∈ Q.

Let K be the splitting ﬁeld of xm − 1 over F contained in F with m as above. Observe that the characteristic of F is either 0 or the prime number that divides q. Since xm − 1 has no multiple root, we have K = F (α) m−1 with a primitive m-th root of unity α. Then xm − 1 = i=0 (x − αi ). Let g be the minimal polynomial of α over F. 7, we see that g is irreducible. Now xm − 1 = i=0 (x − αi ) in the residue ﬁeld, and since xm − 1 is a separable polynomial over k, we see that α is of order m. 7 shows that n = deg(g) = deg(g) = [K : F ].

Also every element of L times a suitable nonzero element of R is integral over R. Thus we obtain our proposition. 5. Let L be as above, and B a subring of L containing R; let α ∈ L. If α is integral over B and every element of B is integral over R, then α is integral over R. Consequently the integral closure of R in L is integrally closed. Proof. 1) with ci ∈ B. 4 shows that R[c1 , . . , cn ] = j=1 Rdj n m with some dj . Then R[c1 , . . 3, α is integral over R. Thus we obtain our lemma. 6. Suppose R is integrally closed; let f (x) ∈ R[x] and f = gh with monic g and h in F [x].