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THE RING of INTEGERS in a RADICAL EXTENSION 1. Introduction the Integers of Q( √ 2) Is Z[ N √ 2] for N = 2,3, 4, and 5. In

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THE RING of INTEGERS in a RADICAL EXTENSION 1. Introduction the Integers of Q( √ 2) Is Z[ N √ 2] for N = 2,3, 4, and 5. In THE RING OF INTEGERS IN A RADICAL EXTENSION KEITH CONRAD 1. Introduction p p n n Thep integers of Q( 2) is Z[ 2]p for n = 2; 3, 4, and 5. In fact this is true for n ≤ 1000. Is Z[ n 2] the ring of integers of Q( n 2) for all n ≥ 2? For comparison, the cyclotomic field Q(ζn) has ring of integers Z[ζn] for all n, so maybe this is something similar. n Let'sp broaden our scope. If n ≥ 2 and T − a is irreducible in Z[T ], or equivalentlyp [Q( n a): Q] = n, we seek necessary and sufficient conditions for the integers of Q( n a) to p p p n be Zp[ a]. Some constraint is needed, since the integers of Q( 5) are more than Z[ 5], as (1+ 5)=2 is an algebraic integer. The conditions we find will turn out to have a connection to old work on Fermat's Last Theorem. 2. Reduction to prime-power n p n n Theoremp 2.1. For n ≥ 2 and T − a irreducible in Z[T ], if Z[ a] is the ring of integers of Q( n a) then a must be squarefree. Proof. We give a proof by example. If α = p3 p2q for different primes p and q, then (α2=p)3 = 2 2 pq , sopα =p is an algebraicp integer not in Z[α]. Use this idea to find an algebraic integer in Q( n a) not in Z[ n a] if a is not squarefree. n n n Since we are assuming T − a is irreducible, a is not 1. If a = −1p then T − a = T + 1 r n is irreducible if and only if n = 2 pis a power of 2, in which case −1 = ζ2r+1 and this n generates the ring of integers of Q( −1) = Q(ζ2r+1 ) since this is a cyclotomic field. So we assume from now on that a is squarefree and is not ±1. Then a has a prime factor and T n −a is automatically irreducible in Z[T ] for all n since it is Eisenstein at each prime factor of a. The next theorem reduces us to the case when n is a prime power. n Lemma 2.2. Let a 6= ±1 be squarefree in Z and T − a be irreduciblep over Q. For each d n positivep integer pd dividing n, T − a is irreduciblep over Q. If Q( a) has ring of integers Z[ n a], then Q( d a) has ring of integers Z[ d a] Proof. Exercise. Theorem 2.3. Let a be a squarefree integer other than ±1. When (m; n) = 1, the following conditions are equivalent: p p mn mn (a) The integers of Q( p a) are Z[ p a]. p p (b) The integers of Q( m a) and Q( n a) are Z[ m a] and Z[ n a]. Proof. That (a) ) (b) follows from Lemma 2.2. We will show (b) ) (a) using properties of discriminants. It is obvious if m or n is 1, so we may assume m > 1 and n > 1. The mth, nth, and mnth roots of a are, as abstract roots, only well-defined up to mul- tiplication by roots of unity. The choice of root doesn't affect the number field up to isomorphism, but it's convenient to give the three roots the obvious multiplicative relation: 1 2 KEITH CONRAD p p p p p take mn a to be a fixed root of T mn − a and then define m a = mn an and n a = mn am. If a > 0 we can usep the positive realp roots. m n Let K = Q( a) and L = Q( a), so [K p: Q] = m and [L : Q] = n. Since (m; n) = 1, mn field theory impliesp [KL : Q] = mnp. Since a is an algebraic integer of degree mn inside 1 mn mn of KL, KL = Q( a) and Z[ pa] has finite indexp in OKL. We will show the index is 1. m n From the hypotheses OK = Z[ a] and OL = Z[ a], disc(K) = disc(T m − a) = ±mmam−1 and disc(L) = disc(T n − a) = ±nnan−1: Let d be the greatest common divisor of these discriminants.p p Since (pm; n) = 1, each prime m n mn factor of d is a factor of a. The ring OK OL = Z[ a; a] lies in Z[ a]. Then 1 1 p O ⊂ O O ⊂ Z[ mn a]; KL d K L d where the first containmentp is a general property of all pairs ofp number fields K and L. mn mn Thus dOKL ⊂ Z[ a]. Therefore the quotient group OKL=Z[ a] is killed by d. mn Since a is squarefree,p for each prime p dividingpa the polynomialpT − a is Eisenstein mn 2 mn mn at p, so p - [OK : Z[ a]]. Therefore [OK : Z[ a]] = jOKL=Z[ a]j is relatively prime to a, and thus also to d. A group that is killed by an integer relatively prime to its size is trivial (the order of each elementp of the group divides two relatively prime integers, so the mn order is 1), so OKL = Z[ a]. Corollary 2.4.pLet a bep a squarefree integer other than ±1. For n ≥ 2p, if the ringp of integers of Q( pr a) is Z[ pr a] for all pr jj n then the ring of integers of Q( n a) is Z[ n a]. Proof. Induct on the number of prime factors of n. The base case is a hypothesis and Theorem 2.3 provides the inductive step. p p We have reduced ourselves to figuring when Q( n a) has ring of integers Z[ n a] for square- free a 6= ±1 and n a prime power. 3. Prime-power n p Let n = pr for prime p and r ≥ 1. To study the integers of Q( n a) for squarefree a 6= ±1, we treat separately the cases that p j a and p - a. When p j a we will have a definitive solution, while the case p - a will be solved only conditionally. Theorem 3.1. Whenp a is squarefreep other than ±1 and p is a prime dividing a, the ring of integers of Q( pr a) is Z[ pr a] for all r ≥ 1. p p r r r pr pr p r p p −1 Proof. Let K = Q( a). The index [OKp: Z[ a]] divides disc(T − a) = ±(p ) a . pr Since p j a, the prime factors of [OK : Z[ a]] all divide a. r p p pr For everyp prime q dividing a, T − a is Eisenstein at q, so q - [OK : Z[ a]]. Therefore pr [OK : Z[ a]] has no prime factors, so it is 1. Theorem 3.2. If a is squarefree other than ±1 and p is a prime not dividing a such that (3.1) ap−1 6≡ 1 mod p2; p p then the ring of integers of Q( pr a) is Z[ pr a] for all r ≥ 1. p p p 1Here we use our convention that m a and n a are powers of mn a. 2 For every root α of a polynomial that's Eisenstein at a prime p, the index [OQ(α) : Z[α]] is not divisible by p. THE RING OF INTEGERS IN A RADICAL EXTENSION 3 Fermat's little theorem tells us ap−1 ≡ 1 mod p for all prime p not dividing a. The noncongruence in (3.1) has modulus p2 and may or may not hold, depending on p. p pr Proof. As in the previous proof, a prime factor of [OK : Z[ a]] is p or is a factor of a, and p r pr p no prime factor of a divides [OK : Z[ a]] since T − a is Eisenstein at every prime factor of a. This time the prime p does not divide a, so we need a new argument to show p is not p r pr p a factor of [OK : Z[ a]], thus making the index 1. We can't use the polynomial T − a pr directly, since it is not Eisenstein at p. But perhaps itp has a translatep (T + c) − a that is Eisenstein at p. That would be good enough, since Z[ pr a] = Z[ pr a − c]. What c could be used? In order for (T + c)pr − a to be Eisenstein at p, there is no problem with the inner coefficients, which are all multiples of p. Is the constant term cpr − a a multiple of p exactly once? This is equivalent to r r (3.2) cp ≡ a mod p and cp 6≡ a mod p2: We will show the choice c = a fits (3.2) when (3.1) holds. For all integers x, xp ≡ x mod p, so by repeatedly taking the pth power we have apr ≡ a mod p, which is the first condition in (3.2) with c = a. For all integers x and y, x ≡ y mod pj =) xp ≡ yp mod pj+1; so starting from ap ≡ a mod p we get apr+1 ≡ apr mod pr+1 by raising to the pth power r times and increasing the modulus each time. Therefore apr ≡ ap mod p2, so checking that apr 6≡ a mod p2 is the same as checking ap 6≡ a mod p2. This last noncongruence is equivalent to ap−1 6≡ 1 mod p2 since (p; a) = 1, so (3.1) implies (T + a)pr − a is Eisenstein at p.
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