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Totally Ramified Primes and Eisenstein Polynomials

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Totally Ramified Primes and Eisenstein Polynomials TOTALLY RAMIFIED PRIMES AND EISENSTEIN POLYNOMIALS KEITH CONRAD 1. Introduction A (monic) polynomial in Z[T ], n n−1 f(T ) = T + cn−1T + ··· + c1T + c0; is Eisenstein at a prime p when each coefficient ci is divisible by p and the constant term c0 is not divisible by p2. Such polynomials are irreducible in Q[T ], and this Eisenstein criterion for irreducibility is the way essentially everyone first meets Eisenstein polynomials. Here we will show Eisenstein polynomials at a prime p give us useful information about p-divisibility of coefficients of a power basis and ramification of p in a number field. Let K be a number field, with degree n over Q. A primep number p is said to be totally n 2 ramified in Kpwhen pOK = p . For example, in Z[i] and Z[ p−5] we have (2) = (1 + i) and (2) = (2; 1 + −5)2, so 2 is totally ramified in Q(i) and Q( −5). 2. Eisenstein polynomials and p-divisibility of coefficients Our first use of Eisenstein polynomials will be to extract information about coefficients for algebraic integers in the power basis generated by the root of an Eisenstein polynomial. Lemma 2.1. Let K=Q be a number field with degree n. Assume K = Q(α), where α 2 OK and its minimal polynomial over Q is Eisenstein at p. For a0; a1; : : : ; an−1 2 Z, if n−1 (2.1) a0 + a1α + ··· + an−1α ≡ 0 mod pOK ; then ai ≡ 0 mod pZ for all i. Proof. Assume for some j 2 f0; 1; : : : ; n − 1g that ai ≡ 0 mod pZ for i < j (this is an empty condition for j = 0). We will prove aj ≡ 0 mod pZ. Since ai ≡ 0 mod pZ for i < j,(2.1) implies j j+1 n−1 (2.2) ajα + aj+1α + ··· + an−1α ≡ 0 mod pOK : n−1−j n−1 Multiply through this congruence by α , making all but the first term ajα a multiple n n of α . Since α is the root of an Eisenstein polynomial at p, we have α ≡ 0 mod pOK , so n−1 (2.3) ajα ≡ 0 mod pOK : n−1 Write this congruence as an equation, say ajα = pγ with γ 2 OK . Now take norms of both sides down to Q: n n−1 n aj NK=Q(α) = p NK=Q(γ): The right side is an integral multiple of pn. On the left side the norm of α is, up to sign, the constant term of its minimal polynomial for K=Q since α generates K=Q. The integer NK=Q(α) is divisible by p exactly once (Eisenstein condition!), so divisibility of n n−1 n n aj NK=Q(α) by p implies p j aj , so p j aj. Thus ai ≡ 0 mod pZ for i < j + 1. Repeat this for j = 0; 1; : : : ; n − 1 to get p j ai for all i. 1 2 KEITH CONRAD Theorem 2.2. Let K=Q be a number field with degree n. Assume K = Q(α), where α is an algebraic integer whose minimal polynomial over Q is Eisenstein at p. If n−1 r0 + r1α + ··· + rn−1α 2 OK with ri 2 Q, then each ri has no p in its denominator. Proof. Assume some ri has a p in its denominator. Let d be the least common denominator of the ri's, so p j d. Write ri = ai=d where ai 2 Z, so some ai is not divisible by p (otherwise d, being divisible by p, would not be the least common denominator). Then a + a α + ··· + a αn−1 r + r α + ··· + r αn−1 2 O =) 0 1 n−1 2 O : 0 1 n−1 K d K Multiply through by the integer d to get n−1 a0 + a1α + ··· + an−1α 2 dOK ⊂ pOK : Lemma 2.1 tells us ai 2 pZ for every i. This is a contradiction. Theorem 2.3. Let K = Q(α) where α 2 OK is the root of an Eisenstein polynomial at p, with degree n. Then p - [OK : Z[α]]. Proof. We argue by contradiction. Suppose p j [OK : Z[α]]. Then OK =Z[α], viewed as a finite abelian group, has an element of order p: there is some γ 2 OK such that γ 62 Z[α] but pγ 2 Z[α]. Using the basis f1; α; : : : ; αn−1g for K=Q, write n−1 γ = r0 + r1α + ··· + rn−1α with ri 2 Q. Since γ 62 Z[α], some ri is not in Z. Since pγ 2 Z[α] we have pri 2 Z. Hence ri has a p in its denominator, which contradicts Theorem 2.2. p p 3 3 Example 2.4. We show the ringp of algebraicp integers of Q( 2) is Z[ 2]. Let O be the full ring of algebraic integers of Q( 3 2), so Z[ 3 2] ⊂ O and p p disc(Z[ 3 2]) = [O : Z[ 3 2]]2 disc(O): p 3 2 3 By an explicit calculation,p discZ(Z[ p2]) = −108 = −2 3 , so 2 and 3 are the only primes 3 3 3 that could divide [O : pZ[ 2]]. Since 2 is the root of T − 2,p which is Eisenstein at 2, 2 3 3 3 does not divide [O : Z[ 2]] by Theorem 2.3. The number 1 + 2 is a root ofp (T − 1) − 2 = 3 2 3 T − 3pT + 3T − 3, whichp is Eisensteinp at 3, so 3 does not divide [O : Z[1 + 2]]. The ring 3 3 3 Z[1 + p2] equals Z[ 2], so [O : Z[ 2]] is not divisible by 3. Therefore this index is 1, so O = Z[ 3 2]. p p Example 2.5. We show the ring O of algebraic integers of Q( 4 2) is Z[ 4 2]. Since p p disc(Z[ 4 2]) = [O : Z[ 4 2]]2 disc(O) p p p 4 11 4 4 and the discriminant of Z[ 2] is −2 ,[O : Z[ 2]] is ap power of 2. Because 2 is a root of T 4 − 2 that is Eisenstein at 2, 2 does not divide [O : Z[ 4 2]] by Theorem 2.3. Therefore the index is 1. p p 5 5 Examplep 2.6. We show the ring O of algebraic integers ofpQ( 2) is Z[ 2]. The discrimi-p 5 4 5 5 5 nant of Z[ 2] is 2 5 , so the only prime factors ofp [O : Z[ 2]] could be 2 and 5. Since 2 is a root of T 5 − 2, which is Eisenstein at 2, and 5 2 − 2 is a root of (T + 2)5 − 2 = T 5 + 10T 4 + 40T 3 + 80T 2 + 80T + 30; p p which is Eisenstein at 5, neither 2 nor 5 divides the index since Z[ 5 2 − 2] = Z[ 5 2]. TOTALLY RAMIFIED PRIMES AND EISENSTEIN POLYNOMIALS 3 p p Remark 2.7. It is natural to ask if the ring of integers of Q( n 2) is Z[ n 2] for all n. It is true for all n ≤ 1000, it is not always true! See https://kconrad.math.uconn.edu/ blurbs/gradnumthy/integersradical.pdf, p p Example 2.8. We show the ring O of algebraic integers of Q( 3 5) is Z[ 3 5]. Since p p disc(Z[ 3 5]) = [O : Z[ 3 5]]2 disc(O) p p 3 3 2 3 3 and the discriminant of Z[ 5] is 3 · 5 ,[pO : Z[ 5]] is a factor of 15. Since T p− 5 is Eisenstein at 5, 5 does not divide [O : Z[ 3 5]] by Theorem 2.3. The number 1 + 3 5 is a 3 3 2 root of (T p− 1) − 5 = T p− 3T − p3T − 6, whichp is Eisenstein at 3, so 3 does not divide [O : Z[1 + 3 5]], and Z[1 + 3 5] = Z[ 3 5], so [O : Z[ 3 5]] is not divisible by 3. Therefore this index is 1. Example 2.9. As a final use of Theorem 2.3, we compute the ring of integers of 3 cubic fields. For i = 1; 2; 3, define three number fields Ki = Q(αi) where αi is the root of the cubic polynomial fi(T ): 3 3 3 (2.4) f1(T ) = T − 18T − 6; f2(T ) = T − 36T − 78; f3(T ) = T − 54T − 150: These polynomials are each Eisenstein at both 2 and 3, so they are irreducible over Q. Each polynomial has the same discriminant: 22356 = 22 · 35 · 23. (Recall disc(T 3 + aT + 3 2 b) = −4a − 27b .) Let's show Z[αi] is the ring of integers of Ki in each case. Since 2 2 22356 = disc(Z[αi]) = [OKi : Z[αi]] disc(OKi ), [OKi : Z[αi]] divides 2 · 3 . Since all the polynomials are Eisenstein at 2 and 3, neither 2 nor 3 divides the index of Z[αi] in OKi by Theorem 2.3. That proves the index is 1 in all three cases. Therefore disc(OKi ) = disc(Z[αi]) = disc(Z[T ]=(fi(T ))) = disc(fi(T )) = 22356 for i = 1; 2; 3. The fields K1, K2, and K3 are all cubic extensions of Q with the same discriminant and the ring of integers of Ki has a power basis. The cubic polynomials fi each have 3 real roots. So far the Ki's seem to be quite similar. Are they isomorphic fields? No. To prove this, we show some prime numbers factor differently in the fields.
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