DISCRIMINANTS AND RAMIFIED PRIMES. 1. Introduction A prime number p is said to be ramified in a number field K if the prime ideal factorization

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1 DISCRIMINANTS AND RAMIFIED PRIMES KEITH CONRAD 1. Introducton A prme number p s sad to be ramfed n a number feld K f the prme deal factorzaton (1.1) (p) = po K = p e 1 1 peg g has some e greater than 1. If every e equals 1, we say p s unramfed n K. Example 1.1. In Z[], the only prme whch ramfes s 2: (2) = (1 + ) 2. Example 1.2. Let K = Q(α) where α s a root of f(x) = T 3 9T 6. Then 6 = α 3 9α = α(α 3)(α + 3). For m Z, α + m has mnmal polynomal f(t m) n Q[T ], so N K/Q (α + m) = f( m) = m 3 9m + 6 and the prncpal deal (α m) has norm N(α m) = m 3 9m + 6. Therefore N(α) = 6, N(α 3) = 6, and N(α + 3) = 6. It follows that (α) = p 2 p 3, (α 3) = p 2 p 3, and (α + 3) = p 2 p 3 (so, n partcular, α + 3 and α 3 are unt multples of each other). Thus (2)(3) = (6) = (α)(α 3)(α + 3) = p 2 p 2 2 p 3 3, so (2) = p 2 2 p 2 and (3) = p3 3. Ths shows 2 and 3 are ramfed n K. Note that one of the exponents n the factorzaton of (2) exceeds 1, whle the other equals 1. One way to thnk about ramfed prmes s n terms of the rng structure of O K /(p). By (1.1) and the Chnese Remander Theorem, (1.2) O K /(p) = O K /p e 1 1 O K/p eg g. If some e s greater than 1, then the quotent rng O K /p e has a nonzero nlpotent element (use the reducton modulo p e of any element of p p e ), so the product rng (1.2) has a nonzero nlpotent element. If each e equals 1, then O K /(p) s a product of (fnte) felds, and a product of felds has no nonzero nlpotent elements. Thus, p ramfes n K f and only f O K /(p) has a nonzero nlpotent element. Our goal n ths handout s to prove the followng result, whch characterzes the prme numbers ramfyng n a number feld n terms of the dscrmnant. Theorem 1.3. For a number feld K, the prmes whch ramfy are those dvdng the nteger dsc Z (O K ). Snce dsc Z (O K ) 0, only fntely many prmes ramfy n K. 1

2 2 KEITH CONRAD 2. The power-bass case We wll frst consder Theorem 1.3 when O K has a power bass over Z. The treatment of the general case n Secton 3 wll not rely on the power-bass case at all, but the power-bass case s techncally smpler. We wll just sketch the basc deas behnd ths specal case. Proof. Assume O K = Z[α] for some α O K. Let f(t ) be the mnmal polynomal of α over Q, so f(t ) s monc n Z[T ]. Wth po K factorng nto prmes as n (1.1), by defnton p ramfes n O K f and only f some e > 1. The factorzaton of the mod p reducton f(t ) n (Z/pZ)[T ] s f(t ) = π e 1 1 πeg g for some dstnct monc rreducbles π (Z/pZ)[T ]. The π (T ) s are separable (all rreducbles over a fnte feld are separable), so some e > 1 f and only f f(t ) has a repeated root n a splttng feld over Z/pZ. Ths s equvalent to f(t ) havng dscrmnant 0: p ramfes n O K f and only f dsc(f) = 0 n Z/pZ. Snce the dscrmnant of a monc polynomal s a unversal polynomal n ts coeffcents (consder the quadratc case, where T 2 + bt + c has dscrmnant b 2 4c), dscrmnants of monc polynomals behave well under reducton: dsc(f(t ) mod p) = dsc(f(t )) mod p. Therefore dsc(f(t )) = 0 n Z/pZ f and only f dsc(f) 0 mod p. Thus p ramfes n O K f and only f p dsc(f). Snce O K = Z[α] = Z[T ]/(f(t )), dsc(f) = dsc Z (Z[T ]/(f)) = dsc Z (O K ). 3. The general case To prove Theorem 1.3 for every O K, we wll examne dscrmnants of rng extensons to show computng the dscrmnant commutes wth reducton mod p: dsc Z (O K ) mod p = dsc Z/pZ (O K /(p)). Then we wll use (1.2) to wrte dsc Z/pZ (O K /(p)) as a product of dscrmnants of rngs of type O K /p e and compute the dscrmnants of these partcular rngs. Defnton 3.1. Let A be a commutatve rng and B be a rng extenson of A whch s a fnte free A-module: B = Ae 1 Ae n. Then we set dsc A (e 1,..., e n ) = det(tr B/A (e e j )) A. Remark 3.2. The dscrmnant of a bass s an algebrac concept of volume. To explan ths vewpont, we should thnk about Tr B/A (xy) as an analogue of the dot product v w n R n. For a bass v 1,..., v n n R n, the ordnary Eucldean volume of the parallelotope { } a v : 0 a 1 havng edges v s det(v v j ). The dscrmnant of an A-bass of B uses the A-valued parng x, y = Tr B/A (xy) on B n place of the R-valued dot product on R n and we just drop the absolute value and the square root when we make the algebrac analogue. How are the dscrmnants of two A-module bases for B related? Pck a second bass e 1,..., e n of B as an A-module. Then e = a j e j, j=1

3 DISCRIMINANTS AND RAMIFIED PRIMES 3 where a j A and the change of bass matrx (a j ) has determnant n A. Then ( ) Tr B/A (e e j) = Tr B/A a k e k a jl e l k=1 so Therefore We set = k=1 l=1 l=1 a k Tr B/A (e k e l )a jl, (Tr B/A (e e j)) = (a j )(Tr B/A (e e j ))(a j ). dsc A (e 1,..., e n) = (det(a j )) 2 dsc A (e 1,..., e n ). dsc A (B) = dsc A (e 1,..., e n ) A for any A-module bass {e 1,..., e n } of B. It s well-defned up to a unt square. In partcular, the condton dsc A (B) = 0 s ndependent of the choce of bass. Gven a number feld K, ramfcaton of the prme p n K has been lnked to the structure of the rng O K /(p) n Secton 1. Let s look at the dscrmnant of ths rng over Z/pZ. Lettng K have degree n over Q, the rng O K s a free rank-n Z-module, say O K = Zω. Reducng both sdes modulo p, O K /(p) = (Z/pZ)ω, so O K /(p) s a vector space over Z/pZ of dmenson n. The dscrmnant of O K s dsc Z (O K ). The next lemma says reducton modulo p commutes (n a sutable sense) wth the formaton of dscrmnants. Lemma 3.3. Choosng bases approprately for O K and O K /(p), dsc Z (O K ) mod p = dsc Z/pZ (O K /(p)). Proof. Pck a Z-bass ω 1,..., ω n of O K. The reductons ω 1,..., ω n n O K /(p) are a Z/pZbass, so the multplcaton matrx [m x ] for any x O K, wth respect to the bass {ω }, reduces modulo p to the multplcaton matrx [m x ] for x on O K /(p) wth respect to the bass {ω }. Therefore Tr (OK /(p))/(z/pz)(x) = Tr(m x ) = Tr(m x ) mod p = Tr OK /Z(x) mod p. Thus, the mod p reducton of the matrx (Tr OK /Z(ω ω j )) s (Tr (OK /(p))/(z/pz)(ω ω j )). Now take determnants. Lemma 3.4. Let A be a commutatve rng and B 1 and B 2 be commutatve rng extensons of A whch are each fnte free A-modules. Then, choosng A-module bases approprately, dsc A (B 1 B 2 ) = dsc A (B 1 ) dsc A (B 2 ).

4 4 KEITH CONRAD Proof. Pck A-module bases for B 1 and B 2 : m B 1 = Ae, B 2 = Af j. As an A-module bass for B 1 B 2 we wll use the m + n elements e 1,..., e m, f 1,..., f n. Snce e f j = 0 n B 1 B 2, the matrx whose determnant s dsc A (B 1 B 2 ) s a block dagonal matrx ( ) (Tr(B1 B 2 )/A(e e k )) O. O (Tr (B1 B 2 )/A(f j f l )) For any x B 1, multplcaton by x on B 1 B 2 klls the B 2 component and acts on the B 1 -component n the way x multples on B 1, so a matrx for multplcaton by x on B 1 B 2 s a matrx whose upper left block s a matrx for multplcaton by x on B 1 and other blocks are 0. Thus Tr (B1 B 2 )/A(x) = Tr B1 /A(x) for x B 1. Smlarly, Tr (B1 B 2 )/A(x) = Tr B2 /A(x) for x B 2. Thus ( ) ( (Tr(B1 B 2 )/A(e e k )) O (TrB1 = /A(e e k )) O O (Tr (B1 B 2 )/A(f j f l )) O (Tr B2 /A(f j f l )) and takng determnants gves Now we prove Theorem 1.3. j=1 dsc A (B 1 B 2 ) = dsc A (B 1 ) dsc A (B 2 ). Proof. We have p dsc Z (O K ) f and only f dsc Z (O K ) 0 mod p. By Lemma 3.3 dsc Z (O K ) mod p = dsc Z/pZ (O K /(p)), so p dsc Z (O K ) f and only f dsc Z/pZ (O K /(p)) = 0 n Z/pZ. In (1.2), each factor O K /p e s a Z/pZ-vector space snce p p e. Usng (1.2) and Lemma 3.4, g dsc Z/pZ (O K /(p)) = dsc Z/pZ (O K /p e ). Therefore we need to show for any prme number p and prme-power deal p e such that p e (p) that dsc Z/pZ (O K /p e ) s 0 n Z/pZ f and only f e > 1. (Recall that the vanshng of a dscrmnant s ndependent of the choce of bass.) Suppose e > 1. Then any x p p e s a nonzero nlpotent element n O K /p e. By lnear algebra over felds, such an x can be used as part of a Z/pZ-bass of O K /p e, say {x 1,..., x n } wth x = x 1. Wrtng the trace map Tr (OK /p e )/(Z/pZ) as Tr for short, the frst column of the matrx (Tr(x x j )) contans the numbers Tr(x x). These traces are all 0: x x s nlpotent, so the lnear transformaton m x x on O K /p e s nlpotent and thus ts egenvalues all equal zero. Snce one column of the trace-parng matrx (Tr(x x j )) s all 0, dsc Z/pZ (O K /p e ) = 0. Now suppose e = 1. Then O K /p e = O K /p s a fnte feld of characterstc p. We want to prove dsc Z/pZ (O K /p) 0. If ths dscrmnant s 0, then (because O K /p s a feld) the trace functon Tr: O K /p Z/pZ s dentcally zero. However, from the theory of fnte felds, ths trace functon can be wrtten as a polynomal functon: Tr(t) = t + t p + t p2 + + t pr 1, ),

5 DISCRIMINANTS AND RAMIFIED PRIMES 5 where #(O K /p) = p r. Snce the degree of ths polynomal s less than the sze of O K /p, ths functon s not dentcally zero on O K /p. Therefore the dscrmnant of a fnte extenson of Z/pZ does not equal zero.

THE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens

THE CHINESE REMAINDER THEOREM. We should thank the Chinese for their wonderful remainder theorem. Glenn Stevens THE CHINESE REMAINDER THEOREM KEITH CONRAD We should thank the Chnese for ther wonderful remander theorem. Glenn Stevens 1. Introducton The Chnese remander theorem says we can unquely solve any par of