a c b b a + c b + c. c b a + c

TRACE AND NORM KEITH CONRAD 1. Introduction Let L/K be a finite extension of fields, with n = [L : K]. We will associate to this extension two important functions L K, called the trace and the norm. For each α L, let m α : L L be multiplication by α: m α (x) = αx for x L. Each m α is a K-linear map from L to L, so choosing a K-basis of L lets us write m α as an n n matrix, which will be denoted [m α ], or [m α ] L/K if we need to emphasize the field extension that is involved. (We need to put an ordering on the basis to get a matrix, but we will often just refer to picking a basis and list it in a definite way instead of saying pick an ordered basis.) Example 1.1. If c K, then with respect to any K-basis of L, [m c ] is the scalar diagonal matrix c I n. Example 1.2. Let K = R, L = C, and use basis {1, i}. For α = a + bi, [m α ] equals ( ) a b. b a If instead we use the basis {i, 1}, in the opposite order, then [m α ] equals ( ) a b, b a which serves as a reminder that [m α ] depends on the choice of K-basis of L. Example 1.3. Let K = Q and L = Q( r) for r a nonsquare rational number. Use the basis {1, r}. For α = a + b r, [m α ] equals ( ) a rb. b a Example 1.4. Let K = Q and L = Q(γ) for γ a root of X 3 X 1. {1, γ, γ 2 }. For α = a + bγ + cγ 2, [m α ] equals a c b b a + c b + c. c b a + c Use the basis Example 1.5. Let K = Q and L = Q(γ) for γ a root of X 4 X 1 (which is irreducible over Q since it s irreducible mod 2). Use the basis {1, γ, γ 2, γ 3 }. For α = a+bγ +cγ 2 +dγ 3, [m α ] equals a d c b b a + d c + d b + c c b a + d c + d d c b a + d 1.

2 KEITH CONRAD Definition 1.6. The trace and norm of α from L to K are the trace and determinant of m α as a K-linear map: Tr L/K (α) = Tr(m α ) K, N L/K (α) = det(m α ) K. The trace and determinant of m α can be computed from any matrix representation of m α as a K-linear map on L. By Example 1.1, for any c K the matrix [m c ] is the n n diagonal matrix with c on the main diagonal, where n = [L : K], so Tr L/K (c) = nc, N L/K (c) = c n. In particular, Tr(1) = [L : K]. By Example 1.2, By Example 1.3, Tr C/R (a + bi) = 2a, N C/R (a + bi) = a 2 + b 2. Tr Q( r)/q (a + b r) = 2a, N Q( r)/q (a + b r) = a 2 rb 2. By Example 1.4, Tr Q(γ)/Q (a + bγ + cγ 2 ) = 3a + 2c and N Q(γ)/Q (a + bγ + cγ 2 ) = a 3 + b 3 + c 3 ab 2 + ac 2 bc 2 + 2a 2 c 3abc. By Example 1.5, Tr Q(γ)/Q (a + bγ + cγ 2 + dγ 3 ) = 4a + 3d and N Q(γ)/Q (a + bγ + cγ 2 + dγ 3 ) = a 4 b 4 + c 4 d 4 + 3a 3 d 2a 2 c 2 + 3a 2 d 2 + ab 3 + ac 3 +ad 3 + 2b 2 d 2 bc 3 bd 3 + cd 3 3a 2 bc + 4ab 2 c 4a 2 bd 5abd 2 + ac 2 d + 4acd 2 + 3b 2 cd 4bc 2 d 3abcd. Remark 1.7. In the literature you might see S or Sp used in place of Tr since Spur is the German word for trace. 2. Initial Properties of the Trace and Norm The most basic algebraic properties of the trace and norm will follow from the way m α depends on α. Theorem 2.1. The function α m α is a K-linear ring homomorphism L End K (L) and is injective. Concretely, this says the matrices in the previous examples are embeddings of the field L into the n n matrices over K. For instance, from Example 1.2 the 2 2 real matrices of the special form ( a b b ) add and multiply in the same way as complex numbers. a Proof. For α, β, and x in L, and m α+β (x) = (α + β)(x) = αx + βx = m α (x) + m β (x) = (m α + m β )(x) (m α m β )(x) = m α (βx) = α(βx) = (αβ)x = m αβ (x), so m α+β = m α + m β and m αβ = m α m β. Easily m 1 is the identity map on L, so α m α is a ring homomorphism. For c K, m cα (x) = (cα)x = c(αx) = c(m α (x)) = (cm α )(x), so m cα = cm α. Therefore α m α is K-linear. We can recover α from m α by evaluating at 1: m α (1) = α 1 = α, so α m α is injective.

TRACE AND NORM 3 Corollary 2.2. The trace Tr L/K : L K is K-linear and the norm N L/K : L K is multiplicative. Moreover, N L/K (L ) K. Proof. We have equations of linear maps m α+β = m α + m β and m cα = cm α. Taking the trace of both sides, Tr L/K (α + β) = Tr L/K (α) + Tr L/K (β) and Tr L/K (cα) = ctr L/K (α). So the trace is K-linear. Taking the determinant of both sides of the equation m αβ = m α m β, we get N L/K (αβ) = N L/K (α)n L/K (β). Finally, since N L/K (1) = 1, for nonzero α in L we take norms of both sides of α (1/α) = 1 to get N L/K (α)n L/K (1/α) = 1, so N L/K (α) 0. The linearity of the trace means its calculation is reduced to finding its values on a basis. Example 2.3. For the extension Q(γ)/Q where γ 3 γ 1 = 0, use the basis {1, γ, γ 2 }. We have Tr(a + bγ + cγ 2 ) = atr(1) + btr(γ) + ctr(γ 2 ) for a, b, c Q, where Tr = Tr Q(γ)/Q. Thus knowing the traces of 1, γ, and γ 2 will tell us the trace down to Q of a general element of Q(γ). We have Tr(1) = [Q(γ) : Q] = 3, but to find the traces of γ and γ 2 it appears we need to compute matrices for multiplication by γ and γ 2. We will see later (Corollary 3.14) that there is a connection between traces and the coefficients of minimal polynomials, which will let us compute the traces of γ and γ 2 (in Example 3.16) without computing matrices for multiplication by γ or γ 2. In all of our previous calculations of the trace and norm, the formulas are polynomials in the coefficients of the basis we used. For instance, N C/R (a + bi) = 2a and N C/R (a + bi) = a 2 + b 2. Let s see this is a general feature. Theorem 2.4. The trace and norm of α are both polynomial functions in coordinates of α: if we pick a K-basis e 1,..., e n of L then there are polynomials P and Q in K[x 1,..., x n ] such that Tr L/K (c 1 e 1 + + c n e n ) = P (c 1,..., c n ), N L/K (c 1 e 1 + + c n e n ) = Q(c 1,..., c n ) for all c i K. More specifically, P and Q are homogeneous polynomials of degrees 1 and n, respectively. Proof. Since the trace is K-linear, Tr L/K (c 1 e 1 + + c n e n ) = c 1 Tr L/K (e 1 ) + + c n Tr L/K (e n ) = P (c 1,..., c n ), where P (x 1,..., x n ) = n i=1 Tr L/K(e i )x i. For the norm, N L/K (c 1 e 1 + + c n e n ) = det(m c1 e 1 + +c ne n ) = det(c 1 m e1 + + c n m en ). For indeterminates x 1,..., x n, the determinant Q(x 1,..., x n ) := det(x 1 [m e1 ] + + x n [m en ]) is a homogeneous polynomial in K[x 1,..., x n ] of degree n from the expansion formula for determinants as a sum of products, since each entry of the matrix x 1 [m e1 ] + + x n [m en ] is a K-linear combination of x 1,..., x n, hence a homogeneous polynomial of degree 1. A product of n homogeneous polynomials of degree 1 is a homogeneous polynomial of degree n and a sum of homogeneous polynomials of degree n is a homogeneous polynomial of degree n or is 0, and Q(1,..., 1) = 1, so Q(x 1,..., x n ) is not the zero polynomial and thus is homogeneous of degree n. Substituting c i for x i shows N L/K ( n i=1 c ie i ) = Q(c 1,..., c n ) for all c i K.

4 KEITH CONRAD Example 2.5. For the extension Q(γ)/Q where γ 3 γ 1 = 0, and basis {1, γ, γ 2 }, we have by Example 1.4 x 1 [m 1 ] + x 2 [m γ ] + x 3 [m γ 2] = x 1 x 3 x 2 x 2 x 1 + x 3 x 2 + x 3. x 3 x 2 x 1 + x 3 Each entry of this matrix is a homogeneous polynomial of degree 1 in the x i s. The determinant is, up to sign, a sum of products of one term from each row and column, such as x 1 (x 1 + x 3 ) 2 using the main diagonal, and these terms are all homogeneous of degree 3. Our next result is called the transitivity of the trace and norm in towers of field extensions. Theorem 2.6. Let L/F/K be a tower of finite extensions. For α L, Tr L/K (α) = Tr F/K (Tr L/F (α)), N L/K (α) = N F/K (N L/F (α)). Remark 2.7. Don t write the right sides as Tr L/F (Tr F/K (α)) or N L/F (N F/K (α)), which make no sense: the first map on α should go from L down to F and the second map should go from F down to K, not the other way around. Proof. Let m = [L : F ] and d = [F : K], as in the field diagram below. L F K To prove transitivity of the trace, let {e 1,..., e m } be an F -basis of L and {f 1,..., f d } be a K-basis of F. Then a K-basis of L is For α L, let {e 1 f 1,..., e 1 f d,..., e m f 1,..., e m f d }. αe j = m d m c ij e i, c ij f s = i=1 for c ij F and b ijrs K. Thus α(e j f s ) = i F/K, and L/K, we have d b ijrs f r, r=1 r b ijrse i f r. Using the above bases for L/F, [m α ] L/F = (c ij ), [m cij ] F/K = (b ijrs ), [m α ] L/K = ([m cij ] F/K ), where the last matrix is a block matrix. Thus Tr F/K (Tr L/F (α)) = Tr F/K ( i = i Tr F/K (c ii ) = i r = Tr L/K (α). b iirr c ii )

TRACE AND NORM 5 The proof of transitivity of the trace was essentially a straightforward calculation. By comparison, the proof of transitivity of the norm is much more delicate! A proof is in Section 4. We will not use the transitivity formulas, except in Remark 3.12, but they are important. 3. The Trace and Norm in terms of Roots The numbers Tr L/K (α) and N L/K (α) can be expressed in terms of the coefficients or the roots of the minimal polynomial of α over K. To explain this we need another polynomial in K[X] related to α besides its minimal polynomial over K. Definition 3.1. For α L, its characteristic polynomial relative to the extension L/K is the characteristic polynomial of m α : L L as a K-linear map: χ α,l/k (X) = det(x I n [m α ]) K[X]. Example 3.2. For c K, m c : L L has matrix representation ci n, so χ c,l/k (X) = det(xi n ci n ) = (X c) n = X n ncx n 1 + + ( 1) n c n. Example 3.3. For the extension C/R, the characteristic polynomial of the matrix in Example 1.2 is χ a+bi,c/r (X) = X 2 2aX + a 2 + b 2. so For any n n square matrix A, its characteristic polynomial has the form det(xi n A) = X n Tr(A)X n 1 + + ( 1) n det A, (3.1) χ α,l/k (X) = X n Tr L/K (α)x n 1 + + ( 1) n N L/K (α). This tells us the trace and norm of α are, up to sign, coefficients of the characteristic polynomial of α, which can be seen in Examples 3.2 and 3.3. Unlike the minimal polynomial of α over K, whose degree [K(α) : K] varies with K, the degree of χ α,l/k (X) is always n, which is independent of the choice of α in L. Theorem 3.4. Every α in L is a root of its characteristic polynomial χ α,l/k (X). Proof. This is a consequence of the Cayley-Hamilton theorem, which says any linear operator is killed by its characteristic polynomial. For the linear operation m α on L, this means χ α,l/k (m α ) = O. Since α m α is a K-linear ring homomorphism L End K (L), for any polynomial f(x) K[X] we have f(m α ) = m f(α). Therefore O = χ α,l/k (m α ) = m χα,l/k (α). The only β in L such that m β = O is 0, so χ α,l/k (α) = 0. Example 3.5. The complex number a + bi is a root of the real polynomial χ a+bi,c/r (X) = X 2 2aX + a 2 + b 2, which can be seen by direct substitution of a + bi into this polynomial. Although α is a root of χ α,l/k (X), which is in K[X], this does not mean χ α,l/k (X) is the minimal polynomial of α in K[X]. For example, we observed already that the degree of χ α,l/k (X) is always n = [L : K], whereas the degree of the minimal polynomial of α in K[X] varies with α. We will see next that the minimal and characteristic polynomials of α are closely related to each other: the latter is a power of the former. Theorem 3.6. For α L, let π α,k (X) be the minimal polynomial of α in K[X]. If L = K(α) then χ α,l/k (α) = π α,k (X). More generally, χ α,l/k (X) = π α,k (X) n/d where d = deg π α,k (X) = [K(α) : K].

6 KEITH CONRAD In other words, χ α,l/k (X) is the power of the minimal polynomial of α over K that has degree n. As a simple example, for c K its minimal polynomial in K[X] is X c while its characteristic polynomial for L/K is (X c) n, which we already saw in Example 3.2. Proof. A K-basis of K(α) is {1, α,..., α d 1 }. Let m = [L : K(α)] and β 1,..., β m be a K(α)-basis of L. Then a K-basis of L is {β 1, αβ 1,..., α d 1 β 1,..., β m, αβ m,..., α d 1 β m }. Let α α j = d 1 i=0 c ijα i for 0 j d 1, where c ij K. The matrix for multiplication by α on K(α) with respect to {1, α,..., α d 1 } is (c ij ), so χ α,k(α)/k (X) = det(x I d (c ij )). This polynomial has α as a root by Theorem 3.4 (using K(α) in place of L) and it has degree d. Therefore det(x I d (c ij )) = π α,k (X) because π α,k (X) is the only monic polynomial in K[X] of degree d = [K(α) : K] with α as a root. Since α α j β k = d 1 i=0 c ijα i β k, with respect to the above K-basis of L the matrix for m α is a block diagonal matrix with s repeated d d diagonal blocks (c ij ), so χ α,l/k (X) = det(x I d (c ij )) m = π α,k (X) m = π α,k (X) n/d. If L = K(α) then n/d = 1, so the characteristic and minimal polynomials of α coincide. Remark 3.7. If L is larger then K(α) then χ α,l/k (X) and π α,k (X) definitely are not equal since they don t even have the same degree. Corollary 3.8. Write the minimal polynomial π α,k (X) for α over K as X d + c d 1 X d 1 + + c 1 X + c 0. Then and more generally Tr K(α)/K (α) = c d 1, N K(α)/K (α) = ( 1) d c 0, Tr L/K (α) = n d c d 1, N L/K (α) = ( 1) n c n/d 0. Proof. We will prove the general formula at the end. The special case L = K(α) then arises from setting n = d. By Theorem 3.6, χ α,l/k (X) = π α,k (X) n/d = (X d + c d 1 X d 1 + + c 1 X + c 0 ) n/d = X n + n d c d 1X n 1 + + c n/d 0. From this formula and (3.1), Tr L/K (α) = n d c d 1 and N L/K (α) = ( 1) n c n/d 0. Example 3.9. If γ is a root of X 3 X 1, then Tr Q(γ)/Q (γ) = 0, N Q(γ)/Q (γ) = 1. Example 3.10. If γ is a root of X 4 X 1, then Tr Q(γ)/Q (γ) = 0, N Q(γ)/Q (γ) = 1. Example 3.11. If γ is a root of X 5 + 6X 4 + X 3 + 5 (which is irreducible over Q, since it s irreducible mod 2), then Tr Q(γ)/Q (γ) = 6, N Q(γ)/Q (γ) = 5.

TRACE AND NORM 7 Remark 3.12. In the notation of Corollary 3.8, n/d equals [L : K(α)], so we can rewrite the formulas for Tr L/K (α) and N L/K (α) at the end of Corollary 3.8 in terms of Tr K(α)/K (α) and N K(α)/K (α), as follows: (3.2) Tr L/K (α) = [L : K(α)]Tr K(α)/K (α), N L/K (α) = N K(α)/K (α) [L:K(α)]. These formulas for Tr L/K (α) and N L/K (α) are also direct consequences of the transitivity of the trace and norm from Theorem 2.6: for the tower of fields L/K(α)/K, transitivity tells us that Tr L/K (α) = Tr K(α)/K (Tr L/K(α) (α)) = Tr K(α)/K ([L : K(α)]α) = [L : K(α)]Tr K(α)/K (α), and similarly N L/K (α) = N K(α)/K (α) [L:K(α)]. We did not initially derive (3.2) from transitivity because the proof of transitivity of the norm is hard and in fact has been postponed until Section 4. But any mathematician who uses (3.2) almost certainly thinks about these formulas as special cases of transitivity. Example 3.13. If γ is a root of X 3 + X 2 + 7X + 2, which is irreducible over Q, then Tr Q(γ)/Q (γ) = 1, N Q(γ)/Q (γ) = 2. If L is an extension of Q containing γ such that [L : Q] = 12, then we make a field diagram L 4 Q(γ) Q 3 and see that Tr L/Q (γ) = 4( 1) = 4, N L/Q (γ) = ( 2) 4 = 16. Corollary 3.14. Let the minimal polynomial π α,k (X) for α over K split completely as (X α 1 ) (X α d ) over a large enough field extension of K. Then and more generally Tr K(α)/K (α) = α 1 + + α d, N K(α)/K (α) = α 1 α d, Tr L/K (α) = n d (α 1 + + α d ), N L/K (α) = (α 1 α d ) n/d. Proof. The factorization of π α,k (X) = X d + c d 1 X d 1 + + c 1 X + c 0 into linear factors implies c d 1 = d i=1 α i and c 0 = ( 1) d d i=1 α i. Substituting these into the formulas from Corollary 3.8 expresses the trace and norm of α in terms of the roots of π α,k (X). This is not saying the trace and norm of α for L/K are the sum and product of the roots of the minimal polynomial of α over K (which doesn t know about L). Those roots have to be repeated n/d times, where d = [K(α) : K], making a total of n terms in the sum and product. This is already evident in the special case of elements in K: for c K, Tr L/K (c) = nc and N L/K (c) = c n.

8 KEITH CONRAD Theorem 3.15. Suppose in a large enough field extension the characteristic polynomial of α relative to L/K splits completely as Then for any g(x) K[X], so χ α,l/k (X) = (X r 1 ) (X r n ). χ g(α),l/k (X) = (X g(r 1 )) (X g(r n )), Tr L/K (g(α)) = n g(r i ), N L/K (g(α)) = i=1 n g(r i ). In particular, χ α m,l/k(x) = (X r1 m) (X rm n ), so Tr L/K (α m ) = n i=1 rm i. Proof. By Theorem 3.6, χ α,l/k (X) is a power of the minimal polynomial of α in K[X], so every r i has the same minimal polynomial over K as α. Set f(x) = (X g(r 1 )) (X g(r n )). We want to show this is the characteristic polynomial of g(α). The coefficients of f(x) are symmetric polynomials in r 1,..., r n with coefficients in K, so by the symmetric function theorem f(x) K[X]. Let M(X) be the minimal polynomial of g(α) over K, so M(X) is irreducible in K[X]. Since α and each r i have the same minimal polynomial over K, the fields K(α) and K(r i ) are isomorphic over K. Applying such an isomorphism to the equation M(g(α)) = 0 turns it into M(g(r i )) = 0 (because M(X) and g(x) have coefficients in K), so M(X) is the minimal polynomial for g(r i ) over K since M(X) is monic irreducible in K[X]. We have shown all roots of f(x) have minimal polynomial M(X) in K[X], and f(x) is monic, so f(x) is a power of M(X). By Theorem 3.6, χ g(α),l/k (X) is a power of M(X) with degree [L : K] = n = deg f, so χ g(α),l/k (X) = f(x). Example 3.16. Let γ be a root of X 3 X 1. The general trace Tr Q(γ)/Q (a + bγ + cγ 2 ) for a, b, c Q can be computed to be 3a + 2c from the 3 3 matrix in Example 1.4, which is the matrix for multiplication by a + bγ + cγ 2 in the basis {1, γ, γ 2 }. We will now compute this trace without using any matrices. By linearity of the trace, i=1 Tr(a + bγ + cγ 2 ) = atr(1) + btr(γ) + ctr(γ 2 ), where Tr = Tr Q(γ)/Q. The trace of 1 is [Q(γ) : Q] = 3. Since γ generates Q(γ)/Q, χ γ,q(γ)/q (X) = X 3 X 1, whose X 2 -coefficient is 0, Tr(γ) = 0. What about Tr(γ 2 )? Let the roots of X 3 X 1 be r, s, and t. Theorem 3.15 tells us that Tr(γ 2 ) = r 2 +s 2 +t 2, which can be expressed in terms of the coefficients of X 3 X 1: r 2 + s 2 + t 2 = (r + s + t) 2 2(rs + rt + st). From the relations between roots and coefficients of a (monic) polynomial, r + s + t = 0 and rs + rt + st = 1. Thus r 2 + s 2 + t 2 = 0 2 2( 1) = 2, so γ 2 has trace 2. In comparison with the trace, where we can take advantage of linearity, there is no way to compute the norm of a general element without essentially computing a determinant with variable entries. General norm formulas are often quite unwieldy when [L : K] > 2. Example 3.17. If c K then χ α+c,l/k (X) = χ α,l/k (X c) from the definition of the characteristic polynomial. Since N L/K (α) = ( 1) n χ α,l/k (0), replacing α with α + c tells us that N L/K (α + c) = ( 1) n χ α,l/k ( c). If we use c in place of c, then N L/K (α c) = ( 1) n χ α,l/k (c).

TRACE AND NORM 9 4. Transitivity of the Norm We return to the transitivity of the norm, which was left unproved in Theorem 2.6. For a tower of finite extensions L/F/K, we want to show N L/K (α) = N F/K (N L/F (α)) for all α L. The argument we will give is due to Scholl [7]. Case 1: α F. If the transitivity formula N L/K (α) = N F/K (N L/F (α)) were already known, then when α F the right side is N F/K (α [L:F ] ) = N F/K (α) [L:F ], so our goal is to prove N L/K (α) = N F/K (α) [L:F ] when α F. We will first show that χ α,l/k (X) = χ α,f/k (X) [L:F ] when α F. Both χ α,l/k (X) and χ α,f/k (X) are powers of π α,k (X), where the first has degree [L : K] and the second has degree [F : K]. Since χ α,f/k (X) [L:F ] is the power of π α,k (X) with degree [L : F ][F : K] = [L : K], this power must be χ α,l/k (X). Looking at the constant terms on both sides of χ α,l/k (X) = χ α,f/k (X) [L:F ], we have ( 1) [L:K] N L/K (α) = (( 1) [F :K] N F/K (α)) [L:F ]. The power of 1 on the right is ( 1) [L:F ][F :K] = ( 1) [L:K], and canceling this common power on both sides settles Case 1. Case 2: L = F (α). Let m = [F (α) : F ] and d = [F : K]. The field diagram is as follows. F (α) F K Let h(x) be the minimal polynomial of α over F, so h(x) is monic of degree m, say h(x) = X m + c m 1 X m 1 + + c 1 X + c 0. Then (4.1) N F/K (N F (α)/f (α)) = N F/K (( 1) m c 0 ) = ( 1) dm N F/K (c 0 ). We will now compute N F (α)/k (α) and arrive at the same value as in (4.1). Let {f 1,..., f d } be a K-basis of F, so a K-basis of F (α) is {f 1,..., f d, αf 1,..., αf d, α m 1 f 1,..., α m 1 f d }. The number N F (α)/k (α) is the determinant of any matrix for multiplication by α on F (α) as a K-linear map. Using the above basis, the matrix is O O O C 0 I d O O C 1 (4.2) O I d O C 2,....... O O I d C m 1 m d

10 KEITH CONRAD where C i is the d d matrix for multiplication by c i on F relative to the K-basis {f 1,..., f d }. This is because α α i f j = α i+1 f j for 0 i < m 1, α α m 1 f j = α m f j = α m 1 (c m 1 f j ) α(c 1 f j ) c 0 f j, and expressing c i f j as a K-linear combination of f 1,..., f d involves the matrix C i. In condensed form, the square block matrix (4.2) is ( ) O C (4.3) 0, I (m 1)d B where O is d (m 1)d and B is (m 1)d d. It is left as an exercise to prove that any square block matrix of the form ( ) O A, B I N where O is an M N zero matrix, A is M M and B is N M, has determinant ( 1) MN det A. (Use induction on N and compute the determinant by expansion along the first column.) Therefore the determinant of (4.2), by viewing it as (4.3), equals ( 1) (m 1)d2 det( C 0 ) = ( 1) (m 1)d ( 1) d det(c 0 ) = ( 1) dm det(c 0 ). Since C 0 is a matrix for multiplication by c 0 on F as a K-vector space, its determinant is N F/K (c 0 ), so (4.3) has determinant ( 1) dm N F/K (c 0 ). Thus N F (α)/k (α) = ( 1) dm N F/K (c 0 ). We previously found that N F/K (N F (α)/f (α)) also equals ( 1) dm N F/K (c 0 ), so N F (α)/k (α) = N F/K (N F (α)/f (α)). Case 3: General situation. When α is any element of L, insert F (α) into the tower of field extensions as in the following diagram. L F (α) F Then N L/K (α) = N F (α)/k (α) [L:F (α)] by Case 1 for L/F (α)/k K = (N F/K (N F (α)/f (α))) [L:F (α)] by Case 2 for F (α)/f/k = N F/K (N F (α)/f (α) [L:F (α)] ) by multiplicativity of the norm = N F/K (N L/F (α)) by Case 1 for L/F (α)/f. This completes the proof that the norm map is transitive. Here are references to some other proofs of the transitivity of the norm. Bourbaki [1, p. 548] and Jacobson [5, Sect. 7.4] prove it as a special case of a transitivity formula for determinants of block matrices with commuting blocks.

TRACE AND NORM 11 Lang [6, Chap. VI, Sect. 5] proves it using field embeddings and inseparable degrees. See B. Conrad s handout [2] for a similar argument with more details included. Flanders [3, Theorem 3, Theorem 5], [4, Theorem 3] derives the transitivity of the norm from an interesting characterization of the norm: for any finite extension L/K with degree n, the norm map N L/K is the unique function f : L K such that (i) f(αβ) = f(α)f(β) for all α and β in L, (ii) f(c) = c n for all c K and (iii) f is a polynomial function over K of degree at most n. (A function f : L K is called a polynomial function over K if for some K-basis {e 1,..., e n } of L there is a polynomial P (x 1,..., x n ) with coefficients in K such that f( n i=1 c ie i ) = P (c 1,..., c n ) for all c i K; when this holds for one K-basis of L it holds for any other K-basis of L, with P usually changing.) The transitivity of the norm is an immediate consequence: for K F L, the composite map N F/K N L/F satisfies the conditions that characterize N L/K as a map from L to K. References [1] N. Bourbaki, Algebra I, Springer-Verlag, New York, 1989. [2] B. Conrad, Norm and Trace, math.stanford.edu/ conrad/210bpage/handouts/normtrace.pdf. [3] H. Flanders, The Norm Function of an Algebraic Field Extension, Pacific J. Math 3 (1953), 103 113. [4] H. Flanders, The Norm Function of an Algebraic Field Extension, II, Pacific J. Math 5 (1955), 519 528. [5] N. Jacobson, Basic Algebra I, 2nd ed., W. H. Freeman & Co., New York, 1985. [6] S. Lang, Algebra, 3rd ed., Addison-Wesley, New York, 1993. [7] A. J. Scholl, Transitivity of Trace and Norm, available online at www.dpmms.cam.ac.uk/study/ii/ Galois/handout-3.pdf.