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Various properties of groups, rings, and fields, including associativity, identity and inverse elements, abelian groups, cosets, zero divisors, algebraic extensions, and Gaussian sums. It also covers homomorphisms, subrings and subfields, and their kernels and images. definitions, theorems, and proofs to help understand these concepts.
Typology: High school final essays
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A Project Report Submitted in fulfillment of the requirements for the completion of the course MS 515 : Project during the programme M. Sc. in Mathematics.
Submitted by DIPENDRANATH MAHATO Roll No. MSM Supervised by PROF. DHIREN KUMAR BASNET
Submitted to Department of Mathematical Sciences Tezpur University, Napaam Tezpur- Assam, India
In this project, we are discussing the concepts of Basic Field Theory, Characters and finally Gaussian Sums and its few application. In the first chapter we have discussed the preliminary concepts and results of Abstract Algebra. In the second chapter we have discussed the principles of Finite Fields which extends up to the discussion of Traces and Norms of some field element. In the next chapter we have focused on the concept of Characters and the related properties. Finally in the last chapter we have discussed our main topic ’Gaussian Sums’, where we have tried to explore few of the properties and its application in several important theorems like Davenport- Hasse Theorem, Stickelberger’s Theorem and Quadratic Reciprocity Law. Though there is a direct proof of Quadratic Reciprocity Law, but since we are dealing with Gaussian Sums, we have provided only the proof which comes from Gaussian Sums.
In the first place I would like to express my gratitude to my supervisor Prof.Dhiren Kumar Basnet, Department of Mathematical Sciences, Tezpur University for his supervision and guidance from the very early stage of this project. I am thankful to all my friends, and specially Sourav Koner(M.Sc. in Mathematics) and Rabindranath Chakraborty(M.Sc. in Mathematics) for their encouragement and help. Finally, I thank my family for their support and for being the constant source of inspiration.
Thank you all for your insights, guidance and infinite support!
Dipendranath Mahato
× is said to be a Ring, denoted by (R, +, ×) or simply R (when there is no ambiguity in binary operations), if they satisfy the following axioms :
Remarks :
every non-zero element is a unit, i.e., F{ (^0) F} = F×.
Remarks :
(a) for groups Ker f := {x ∈ G : f (x) = e (^) G˜} is a subgroup of G. (b) for rings Ker g := {x ∈ R : g(x) = 0S } is a subring of R. (c) for fields Ker φ := {x ∈ F : φ(x) = 0K} is a subf ield of F.
(a) for groups Imf := {f (x) : x ∈ G} is a subgroup of G˜. (b) for rings Img := {g(r) : r ∈ R} is a subring of S. (c) for fields Imφ := {φ(x) : x ∈ F} is a subf ield of K.
Fields are the most basic algebraic structures dealt in Modern Algebra. There are extensive theories developed by dealing with F ield Extensions, Splitting F ields, Algebraic Closures, Galois T heory and then F inite F ields.
From the notion of characteristic, if a field F have char(F) = 0 then there is a homomorphism from Z → F defined by n 7 → n.1 which provides an embedding of Z in F. Since n. 1 ∈ F, a field, it is invertible and thus we form a sub-ring S generated by 1. This S is the prime subf ield of F, which is clearly isomorphic to Q. Similarly, if char(F) = n then the corresponding prime subf ield is isomorphic to Fp.
Theorem 2.1. If F ⊆ K ⊆ L, then the extension degrees are multiplicative. Therefore
Combining the above two we have [L : K][K : F] = [L : F].
Proof. Since, [L : K] = m and [K : F] = n, we have two bases {k 1 , k 2 , ..., km} of L over K and {f 1 , f 2 , ..., fn} of K over F, respectively. Now let, l ∈ L be arbitrary. Then
l =
∑^ m
i=
αiki (2.1)
For the second statement, if [K : F] is infinite, then there are infinitely many elements in the basis of K over the field F. Since K ⊆ L, each basis element of K is also in L. All these elements are linearly independent, and as basis is the maximal linearly independent set, therefore we have [L : F] is also infinite. Again, if [L : K] is infinite, then there are infinitely many elements in the basis of L over the field K, which are linearly independent over K. Since linearly independence over K implies linearly independence over the field F, we have [L : F] is infinite. Hence the second statement of the proposition is also proved and so the entire proposition is proved.
Lemma 2.2. Let F be a finite field containing a sub-field K with |K| = q (where q is a prime or a prime power) and [F : K] = m, then |F| = qm.
Proof. By the given conditions, F is a vector-space over the field K with dimKF = m. Then b 1 , b 2 , ..., bm is a basis of F over K and for any x ∈ F, it can be uniquely expressed as
x =
∑^ m
i=
ribi
where ri ∈ K, ∀i ∈ { 1 , 2 , ..., m}. Since, |K| = q, then for each ri we have exactly q choices from K. Then
|F| = q × q × ... × q(m − times) = qm.
Lemma 2.3. If F is a finite field, then |F| = pr, where char(F) = p and r = [F : P] (P is the prime subfield of the field F).
Proof. From the given statement F is finite and so char(F) = p. Now P ≃ Fp, from the Remarks of Field under Chapter 1. Since, |Fp| = p, then by Lemma 2.2 we have |F| = pr.
Lemma 2.4. If F is a finite field with |F| = q, then for every a ∈ F satisfies the equation xq^ = x.
Proof. Since F×^ = F{ (^0) F} forms a group with the underlying multiplication of the field, all of it’s members satisfy the equation
xq−^1 = 1F
where x ∈ F∗. Now multiplying both side by x we get xq^ = x. Again 0F satisfies xq^ = x trivially. So for every x ∈ F, the equation xq^ = x is satisfied.
Lemma 2.5. If F be a finite field of cardinality q and K be a subfield of F, then the polynomial xq^ − x ∈ K[x] factors in F[x] in the form
xq^ − x =
α∈F
(x − α).
Proof. Since the polynomial is of degree q, it has at most q roots in F. Also by Lemma 2.4, every element of F satisfies the polynomial in the given lemma. We know that for any field K and for any f ∈ K[x] where deg(f ) > 0, (x − α)^2 |f ⇐⇒ f (α) = f ′(α) = 0 For our case, f (x) = xq^ −x and f ′(x) = qxq−^1 −1 and for any α ∈ F, f (α) = 0 but f ′(α) = − 1 6 = 0. So all the factors of xq^ − x are distinct and hence the result.
Corollary 2.5.1. F is a splitting field of xq^ − x over the field K.
Corollary 2.5.2. Let F be the field extension of the field K, then an element α ∈ K is in F ⇐⇒ αq^ = α.
Theorem 2.6. (Existence and Uniqueness Theorem). For each prime p and each n ∈ N there exists exactly one field K(up to isomorphism) of cardinality q = pn, namely the minimal splitting field of xq^ − x over the prime field of Kp.
Proof. (Proof of Existence) Let xq^ −x ∈ Kp[x] and K be the minimal splitting field of xq^ − x over Kp. Now this polynomial has distinct factors(by Lemma 2.5). Now consider the set S := {α ∈ K|αq^ = α}. We claim that, S is a subfield
Properties of M¨obius μ function :
d|n
μ(d) = 0.
Theorem 2.8. (M¨obius Inversion Formula) For an arithmetic function f and its summatory function F , we have
f (n) =
d|n
μ(d)F (
n d ), for all n ∈ N
Proof. Since F is the summatory function of the arithmetic finction f , we have F (n) =
d|n
f (d).
Now,
∑
d|n
μ(d)F ( n d
d|n
μ(d)
k|(n/d)
f (k)
dk|n
μ(d)f (k)
k|n
f (k)
d|(n/k)
μ(d)
= f (n)
as for n/k > (^1) ∑
d|(n/k)
μ(d) = 0,
so n/k = 1 ⇒ n = k.
Corollary 2.8.1. For the arithmetic function, Euler′s φ-f unction we have
φ(n) = n
d|n
μ(d) d
Theorem 2.9. In any finite field Fq, F∗ q forms a cyclic multiplicative group.
Proof. We know that cardinality of F∗ q is q − 1. Now if d|q −1, then xd^ −1 divides xq−^1 −1 and so it divides xq^ −x. Therefore xd^ − 1 has d roots, which are distinct. Let G ⊆ F∗ q defined by G := {g ∈ G : gd^ = 1F}. We claim that G is a subgroup of F∗ q with cardinality d. It is clear that G contains d distinct elements. Now 1F satisfies the deginition of G, so 1F ∈ G. Let g 1 , g 2 ∈ G, then (g 1 g 2 − 1 )d^ = gd 1 (g 2 d )−^1 = 1F ⇒ g 1 g 2 − 1 ∈ G ⇒ G is a subgroup of F∗ q. ψ(d) be the number of elements of order d in the multiplicative group F∗ q. Then (^) ∑
t|d
ψ(t) = d.
So by M¨obius Inversion Formula we have
ψ(d) =
t|d
d μ(t) t
= φ(d).
Using the corollary of the previous theorem. So ψ(q − 1) = φ(q − 1) ≥ 1, hence the proof follows.
We already defined an element and a field to be algebraic over some field in Chapter 1. Now we define the minimal polynomial or the def ining polynomial or the irreducible polynomial more generally.
Definition 2.2. Minimal polynomial : Let α ∈ L be algebraic over the field K, then the uniquely determined monic polynomial m ∈ K[x], that generates the ideal {h ∈ K[x] : h(α) = 0} of K[x] is called the minimal polynomial.
Note : Let α ∈ L be algebraic over the field K, then it’s minimal polynomial m over K has the following properties:
As ζ is a root of f , by above argument ζq, ζq 2 ,... , ζq m− 1 are also roots of f. Now we have to show this roots are simple roots. Suppose to the contrary, ζq i = ζq j for some i, j ∈ Z and without loss of generality let, 0 ≤ i < j ≤ m − 1, then
ζ = ζq
m
= ζq
m−(j−i) , using ζq
i = ζq
j .
So by Lemma 2.10 we have f divides xq m−(j−i) − x and by Lemma 2. m|m − (j − i), which is a contradiction as (j −i) > 0 ⇒ m−(j −i) < m.
Corollary 2.12.1. If f ∈ Fq[x] be an irreducible polynomial such that deg(f ) = m. Then the splitting field of f over Fq is given by Fqm.
Corollary 2.12.2. Let f, g ∈ Fq[x] be two irreducible polynomials with deg(f ) = deg(g) < ∞, then the splitting fields corresponding to f and g are isomorphic to each other.
By Theorem 2.12 we have some specific roots related to a field and it’s splitting field. From now on we will identify these roots with a special terminology given below.
Definition 2.3. Conjugates of ζ : Let Fqm be a field extension of the field Fq and ζ ∈ Fqm^. Then the conjugates of ζ with respect to the field Fq are ζ, ζq, ζq 2 ,... , ζq m− 1 . Note 1 : The conjugates of ζ with respect to Fq are distinct if and only if the minimal polynomial of ζ is of degree m. In case, the conjugates are not distinct, then the minimal polynomial is of degree r < m. In fact r is a proper divisor of m and each distinct roots of the minimal polynomial is repeated m d times.
Note 2 : All the conjugates of ζ( 6 = 0) ∈ Fqm^ with respect to the field Fq have the same order in the multiplicative group F× qm.
Our primary objective lies in Gauss Sums and it’s application, so we will restrict our discussion topic as much it is needed to reach the goal. For our convenience from now onwards we will denote F = Fqm and K = Fq, i.e., F is the finite field extension of K(ζ).
Definition 2.4. Trace of ζ : For any ζ ∈ F, the trace of ζ, denoted by T rF/K(ζ), is defined by
T rF/K(ζ) = ζ + ζq^ + ζq 2
m∑− 1
j=
ζq j .
Now if K is a prime-subfield of F, then T rF/K(ζ) is called the Absolute T race of ζ and then it is denoted by T rF(ζ).
From the above definition it is clear that, trace of ζ over K is the sum of all conjugates of ζ over K.
Theorem 2.13. The trace function T rF/K satisfies the following properties:
Proof.
(ζq^ + ζq 2
Since ζq m = ζ, we have {T rF/K(ζ)}q^ = T rF/K(ζ). Then by Corollary 2.5.2 it is clear that T rF/K(ζ) ∈ K, ∀ζ ∈ F.
T rF/K(ζ 1 + ζ 2 ) = (ζ 1 + ζ 2 ) + (ζ 1 + ζ 2 )q
2
m− 1
= (ζ 1 + ζ 2 ) + (ζq
2 1 +^ ζ
q^2 2 ) +^ · · ·^ + (ζ
qm−^1 1 +^ ζ
qm−^1 2 ) = (ζ 1 + ζq
2 1 +^ · · ·^ +^ ζ
qm−^1 1 ) + (ζ^2 +^ ζ
q^2 2 +^ · · ·^ +^ ζ
qm−^1 2 ) = T rF/K(ζ 1 ) + T rF/K(ζ 2 ).
T rF/K(cζ) = (cζ)q^ + (cζ)q 2
= cζq^ + cq
2 ζq
2
m− 1 ζq
m− 1
= cζq^ + cζq 2
= c(ζq^ + ζq 2
