The Birth of Abstract Algebra

Michel 1 Bibliographical Narrative Jerod Evan Michel is planning to graduate from Southern Oregon University in June 2006 with a Bachelor of Science in mathematics. He is focusing his degree on abstract algebra, and is hoping to continue his search for new ideas as a neophyte in graduate school. Jerod says that when he tells people what his major is, they look at him in disbelief. “Why would anyone want to major in math?” is a common response as well as “I could never do math in school!” Jerod responds “Anyone can do it, but if your as strange as me and actually get excited about math, it would seem less scary.”

Jerod maintains that his goal is to teach higher math as well as engage in original research within abstract algebra. Ultimately, Jerod’s goal is to expand the frontier of algebra.

Abstract The brief period of the history of the theory of numbers and equations that led to group theory and abstract algebra took place within the seventeenth and eighteenth centuries. Although mankind has been solving polynomials for approximately 4000 years, it was during this brief period that mathematicians struggled when trying to account for the general solvability of polynomials. In the early 1800s, mathematicians turned toward abstraction when solving polynomials, this was the beginning of a whole new paradigm of the solvability of equations. It was the seemingly endless search for the solution to the quintic polynomial (polynomial whose highest power of x is five) that triggered the discovery of groups. Many scholars find it baffling that it took mathematicians approximately 3500 years to generally go from solving the quadratic (polynomial whose highest power of x is two) to solving the cubic (highest power of x is three), and only one year to go from solving the cubic to the quartic (highest power of four). Those mathematicians looking for the solution to the quintic, though, were more than baffled.

Key Terms: Polynomial, Unsolvability, Group, Permutation.

The Birth of Abstract Algebra Scholar Jerod Michel and Mentor Dr. Sherry Ettlich The purpose of this essay is expository. It is to show the brief period of evolutionary works in the theory of numbers and equations that led to the birth of group theory and what is contemporarily known as abstract algebra.

Unfortunately, to go into depth on the entire development of abstract algebra is beyond the scope of this essay, however, revealing the particular works which triggered the discovery of groups and rings is not.

Also, it is the author’s goal to briefly appraise these brilliant mathematicians’ rights to the credit for the birth of abstract algebra. Though several of these mathematicians made worthy contributions, we will presume that only a handful are the rightful midwives that safely brought abstract algebra into the world.

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Following the discussion of these individuals, we will examine their impacts by asking questions concerning their contributions such as: (1) what has this person contributed to the theory of numbers, equations, or groups? (2) does this person offer the first or most complete proof concerning their contribution? (3) is there substantial evidence that groundwork was laid by some predecessor? (4) would the successor have reached proof without the necessary foundational theorems? and (5) should predecessors or successors be credited for the contribution?

Readers with no exposure to abstract algebra should feel free to skip over the sections entitled “Discussion of the Proof of …”, whereas readers with some background in abstract algebra will find these sections provide an outline of the modern or classical proof, including reminders of key definitions, theorems, and of concepts needed for the development.

Searching for solutions to polynomials led mathematicians to abstraction and the discovery of groups. The fundamental theorem of algebra is needed to prove the unsolvability of the quintic polynomial. Since it was the search for the solution to the quintic that triggered the discovery of groups, we will begin with the discussion of the fundamental theorem.

The Fundamental Theorem The “fundamental theorem of algebra” refers to this statement: (1.1) The number of roots of a nonzero polynomial over the field C (multiplicities counted), is equal to the degree of the polynomial.

The works leading to this theorem rest heavily on Girard (1595 – 1632).

Girard’s theorem on the existence of imaginary roots was certainly necessary in proving the fundamental theorem. One should note that this version of Girard’s theorem is Gauss’s interpretation which he created to make it applicable to the fundamental theorem (James 58).

Girard’s Theorem For any nonconstant polynomial α(x) F [ x ] , there is a field T such that F T and α(x) is made up of linear factors from T[x]. i.e. α(x) = a( x - z1 ) • • • ( x - zn ) in T[x] where a, z1,..., zn

T.

Discussion of the Proof of Girard Girard’’s Theorem: We will obtain T from polynomial rings modulo principle ideals.

Before we begin, here are a few definitions and theorems on which we can build.

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Theorem (1.2) Let F be a field. The polynomial f(x) F [ x ] is irreducible if and only if whenever f(x) = g( x )h( x ) , either g(x) or h(x) is a constant polynomial.

Definition (1.3) If f(x) = an x n + an-1x n-1 + • • • + a1x + a0 , where a n ≠ 0, we say that f(x) has degree n; if an is the multiplicative identity element of R , we say that f(x) is monic (Gallian 285).

Algebraists have the tendency to want to deal with monic polynomials. Having a leading coefficient of 1 often abridges the process of factorization.

Theorem (1.4) Let F be a field, I a nonzero ideal in F[x]. Then, I = g ( x ) if and only if g(x) is a nonzero polynomial of minimum degree in I.

Definition (1.5) A subring A of a ring R is called a (two-sided) ideal of R if for every r every a A , both ar and ra are in A (Gallian 253).

Definition (1.6) The set x + A = { x + a a

R and

A } is called a left coset of A.

(1.7) Thus x + A = y + A if and only if x - y

A.

Theorem (1.8) Let R be a ring and let A be a subring of R . The set of cosets {r + A | r R} is a ring under the operations (s + A) + (t + A) = (s + t) + A and (s + A)(t + A) = st + A iff A is an ideal of R (Gallian 255).

Then we can call {r + A | r R} the factor ring (or quotient) of R by the ideal A and denote it by R/A. Suppose R = F[x] and A = α ( x ) . So α ( x ) = {α(x)β(x) | β(x) F[x]}, a set of multiples of the polynomial α(x).

It is easily shown that α ( x )

ideal

≤ F[x] (since α ( x ) ≤ R [ x ] ).

Therefore we can build the quotient ring F[x]/ α ( x ) = { β( x ) + α ( x ) β ( x )

F [ x ]} .

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Lemma (1.9) If α ( x ) F [ x ] is irreducible, then F[x]/ α ( x ) is a field such that F

F[x]/ α ( x )

and a root of α(x) is in F[x]/ α ( x ) as well.

Proof: Suppose α( x ) F [ x ] is irreducible.

WTS If δ ( x ) + α ( x )

F [ x ] / α ( x ) then ( δ ( x ) + α ( x ) )-1

F [ x ] / α( x )

whenever δ ( x ) + α ( x ) ≠ 0 + α( x ) , which is the additive identity in

F[x]/ α ( x ) .

Suppose δ ( x ) + α( x ) ≠ 0 + α( x ) .

We get, from (1.7), that δ(x) is not in α ( x ) and it is obvious that δ(x) is not divisible by α(x).

Then α(x) and δ(x) are relatively prime because α(x) is irreducible, by assumption.

Definition (1.10) Let F be a field and let f(x) and g(x) belong to F[x]. If there is no polynomial of positive degree in F[x] that divides both f(x) and g(x) where f(x) and g(x) are said to be relatively prime, then there exist polynomials h(x) and k(x) in F[x] such that f(x)h(x) + g(x)k(x) = 1 (Gallian 292).

We get from (1.10) that δ(x)d(x) + α(x)a(x) = 1 for some d(x), a(x) F [ x ] .

Then it is obvious that α(x) divides δ(x)d(x) – 1 and δ ( x )d ( x ) -1

α( x ) .

We get from (1.7) that δ(x)d(x) + α( x ) = 1+ α( x ) .

Therefore from (1.8) we have ( δ( x ) + α ( x ) )( d ( x ) + α ( x ) ) = 1+ α( x ) which is unity in F[x]/ α ( x ) .

Hence ( δ( x ) + α ( x ) )-1 = d ( x ) + α( x ) exists in F[x]/ α ( x ) .

WTS F

F [ x ] / α( x ) .

Consider φ : F → F [ x ] / α ( x ) defined by φ( z ) = z + α( x ) .

Michel 5 No nonzero element of F is divisible by α since deg α ≥ 1.

So if φ(x) = 0 then x = 0 and φ is 1:1.

Therefore φ is a monomorphism.

Hence F is isomorphic to a subfield of F [ x ] / α ( x ) thus proving (1.9).

WTS F[x]/ α ( x ) contains a root of α(x).

From treating α like a function (since it is a polynomial), we get α(x+ α ( x ) ) = α( x ) + α( x ) from (1.8), which is the zero element in the factor ring F[x]/ β( x ) by (1.7).

(1.11) Hence x + α ( x ) is a root of α(x) in F[x]/ α ( x ) and we are done proving lemma (1.9).

To complete Girard’s Theorem we need to show α(x) is the product of linear factors in T[x] for some field T containing F. Suppose m ∈ Z + ∪{0} is the number of linear factors of α(x). So m is also the number of roots of α(x) in F[x] (multiplicities counted). i.e.

α ( x ) = ( x - a1 ) • • • ( x - am )β0 ( x ) where a1,..., am ∈ F and β0 ( x ) Suppose deg β0 = 0. Then α ( x ) = b( x - a1 ) • • • ( x - am ) , for some b

F [ x ].

F.

Then obviously T = F since all the roots of α(x) are in F and we are done.

Suppose deg β0 > 0. Let α 1 be an irreducible factor of β0 ( x ) in F [ x ].

Since β0 ( x ) has no linear factors, deg α1( x ) ≥ 2 . Let F 1= F [ x ] / α1( x ) . Since

α1( x ) has an irreducible factor of β0 ( x ), α1 has a root in F 1 by (1.11) . i.e. β0 ( x ) = ( x - am +1 )β1( x )

F1 [ x ] because ( x - am +1 ) α1( x ) and α1( x ) β0 ( x ).

Theorem (1.12) Let F be a field, α F , and f(x) F [ x ] . Then a is a zero of f(x) if and only if x - a is a factor of f(x) (Gallian 288).

Michel 6 Then α( x ) has at least one more linear factor in F1[ x ] than it had in F [ x ] , specifically am +1 , by (1.12) .

Let m1 be the number of linear factors of α ( x ) in F1 [ x ] and we get m1 ≥ m + 1.

Then α ( x ) = ( x - a1 )( x - a2 ) • • • ( x - am1 ) β1( x ). As stated earlier, if deg β1 = 0, we are done. If not, repeat the process above to obtain m2 factors and β2 ( x ) F2 [ x ].

Hence it follows that by repeating this process; there is a finite chain of fields F = F0 < F1 < ... < Fk = T where α(x) breaks down into linear factors in T [ x ] .

Q.E.D. This concludes the discussion of Girard’s Proof.

Girard’s Theorem is, essentially, the piece that Lagrange left out of his proof of the fundamental theorem. Lagrange (1736 – 1813) had progressed, in 1772, in proving the fundamental theorem of algebra. Unfortunately Lagrange neglected to show that n imaginary roots of any equation of degree n indeed existed. What Lagrange did prove, however, was that the form of these imaginary roots is α + β - 1 with α , β R. It wasn’t until 1815 that Gauss filled this gap by interpreting Girard’s Theorem (Shenitzer and Stillwell 99).

According to Tignol the following fundamental theorem statements are equivalent:

(1.1) The number of roots of a nonzero polynomial over the field C (multiplicities counted), is equal to the degree of the polynomial.

(1.13) Each nonconstant polynomial in R[x] has one or more roots in C. (1.14) Each nonconstant polynomial in R[x] is the product of polynomials in R[x] of degree 1 or 2.

It is one thing to prove these three statements are equivalent (i.e. if one is true, all must be true), and another to prove any one is actually valid. For the sake of brevity we omit the proof of equivalence. e.g. (1.1) (1.13) (1.14) (1.1) and prove (1.13) is valid. With the proof of their equivalence, knowing (1.13) is true tells us all three statements are valid theorems.

Discussion of the Proof of the Fundamental Theorem: WTS Each nonconstant polynomial in R[x] has one or more roots in C.

Michel 7

WLOG We may assume α(x) is monic since α(x) may be divided by the leading coefficient referred to in (1.3) .

We will need the following theorems in the steps that follow:

Intermediate Value Theorem (1.16) If f ( x )

R[x] and deg f is odd then f(x) has a real root.

Theorem (1.17) If f ( x ) = x 2 + ax + b

C[x] then f(x) has two complex roots.

Proof From the quadratic formula (completing the square) we have

x =

b ± b 2 - 4 ac 2a

.

We know square roots exist in C since we can represent z θ z = re iθ = r (cos θ + i sin θ ) . So z = r e i ( 2 ) . Q.E.D. Suppose α ( x )

C as

R[x] and deg α = m .

Let m = 2e n where n is odd.

If e = 0 then by (1.16) α has a root in R and we are done.

Suppose e ≥ 1 and let a1 ,a2 ,..., am be the roots of α ( x ) . Then α ( x ) C[x], and

by Girard’s Theorem, α1,..., α m K , some field containing C. To continue, we will need the following definition concerning automorphisms and Galois extensions.

Definition (1.18) Let E be an extension field F. An automoprphism of E is a ring isomorphism from E to E. The Galois group of E over F, Gal(E/F), is the set of all automorphisms of E that take every element of F to itself. If H is a subgroup of Gal(E/F), the set EH = { x ∈E φ ( x ) = x for all x ∈H } is called the fixed field of H (Gallian 548).

Michel 8

(1.19) Here is another theorem we need: H = Gal ( E / F ) if and only if EH = F .

Now let us call the Galois extension of R containing i and all the roots of α ( x ) , L = R ( a1 ,a2 ,..., am , i ) . It is the smallest subfield of K containing R, I, and all the roots of α ( x ).

Let w be an arbitrary real number. Consider U w =

∏[ x - (a

p

+ aq + wa p aq )].

1≤ p<q ≤ m

R) permutes the terms of U w ; the By (1.18) above, an automorphism in Gal(L/R coefficients of U w will be held fixed.

Therefore the polynomial Uw has real coefficients by (1.19).

Choosing any natural number m, it can easily be shown that the number of terms m( m-1) in the product U w is 2 .

Hence the degree of U w is

m( m-1) 2

= 2e-1n( 2e n - 1) obviously.

But n( 2e n - 1) is odd, and the power of 2 in the degree of U w is less than e.

Therefore, by the induction hypothesis, U w has a root in C. Then for all w

R, there exists p,q

Z + such that ap + aq + wapaq

C, such that

1≤ p < q ≤ m .

Thus there are distinct real numbers w and s where, ap + aq + wa paq and

ap + aq + sapaq are both in C because R is infinite and roots of U w are finite.

Suppose p = 1 and q = 2, since the order of the permutations of U w is insignificant.

Let c = a1 + a2 + wa1a2 and a1 + a2 + sa1a2 .

Since w ≠ s , we get a1a2 = wc --ds

C and a1 + a2 = c - wa1a2

C.

But a1 and a2 are the roots of the quadratic equation x 2 - ax + b with coefficients in C, and, by (1.17) , are elements of C and we are done.

Q.E.D. This concludes the discussion of the Fundamental Theorem.

Michel 9

Impacts Lagrange contributed some of the most useful tools in the theory of both numbers and equations. His paper on the theory of equations “Reflexions sur la resolution algebrique des equations” must have been the most influential of its period as it was the groundwork for several of Lagrange’s successors such as Gauss, Abel, Cauchy and Galois (O’conner 2). He contributed the very bridge adjoining the theory of equations and its contemporary destination by reviewing the general solutions to the cubic and the quartic and then venturing into the frontier of equations of higher degrees. Also, based on the work of Euler and Bezout, he was able to construct the “Lagrange resolutes.” This, according to Tignol, played one of the key roles in the development of Galois theory.

Girard’s contribution, Girard’s Theorem, to the theory of equations made it possible to prove the fundamental theorem. He was indeed the first to give a complete proof of it, however, as Gauss mended Lagrange’s proof of the fundamental theorem, he had to translate (re-prove) Girard’s theorem to be able to use it (Hollingdale 312). Because Girard’s proof was completed in the early 1600s, Gauss’s credit is limited to its elegant translation into what was contemporarily useful.

Lagrange’s proof of the fundamental theorem, though technically incomplete, was still phenomenal. It relied on several arguments first made by Euler, which is why it is sometimes referred to as the Euler-Lagrange proof.

Lagrange is, however, the first to link these arguments together to form what was almost the whole of the fundamental theorem (Hollingdale 307). We might say it is ironic that the first and most complete proof came from Gauss in 1815 due only to his recognition that Girard’s theorem could be used. It seems fair to say that Girard, Lagrange and Gauss each deserve some large ration of praise for proving the fundamental theorem.

Lagrange This section we will devote to the period in which Lagrange’s work was applied to group theory after the times of Abel and Galois. In the late 1800s, algebra was growing into the subject we now call pure mathematics. By now, the advances made in solving the polynomials, and all that was gained through Cardano and Ferrari (whose methods for solving third and fourth degree polynomials are in the appendix), were now commonly known (Hollingdale 297).

Also, the facts obtained from the work of De Moivre on the roots of complex numbers gave mathematicians such as Lagrange and his contemporaries great insight into the theory of equations. We have already discussed the contribution made by Lagrange known as the Fundamental Theorem of Algebra. The following section looks at some of the work Lagrange did which is directly related to the theory of groups.

When Lagrange first obtained this result, he did not create new notation, which makes it hard to follow. Hopefully, using modern notation will help.

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For any integer q ≥ 1, call Sq the symmetric group on { 1,2,..., q } . The set of all permutations of the numbers 1,2,...,q. We will apply φ

Sq to some arbitrary

rational function g in q indeterminates x1, x 2 ,..., x q , to permute the indeterminates as follows: φ( g ( x1, x 2 ,..., xq )) = g ( xφ( 1) , xφ( 2 ) ,..., x φ( q ) ) . We will call the permutation subgroup { φ

Sq φ( g ) = g } , ε( g ) (the permutations that hold g invariant). Let

us also note that the number of distinct permutations in Sq is q!.

Theorem (2.0) Let g = g( x1 , x 2 ,..., x q ) be the rational fraction [function] in q indeterminates. The number n of different values that g takes under the permutations of x1, x2 ,..., x q is equal to the quotient of q! by the number of permutations that leave g invariant: q! n = ε ( g ) (Tignol 139).

Discussion of the Proof of Lagrange Lagrange’’s Theorem: Let G = { g ( x φ( 1) , x φ( 2 ) ,..., x φ( q ) ) φ

Sq } .

(2.1) Call g = g1 ,g 2 ,..., g n the distinct elements of G and also recall ε ( g i ) = {φ ∈ S q φ ( g ) = g i } for i = 1,2,…,n.

Therefore ε( g1 ) = ε( g ) = { φ

Sq φ( g ) = g1 } .

Now let some i { 1,2,…,n} and φ ∈ ε( g i ) .

Suppose δ

ε( g ) .

So φ � δ ( g ) = φ( δ ( g )) = φ( g ) = g i .

Then φ � δ

{φ

Also suppose κ

Sq φ( g ) = g i } = ε( g i ). ε( g i ) so κ ( g ) = g i .

So φ -1( κ ( g )) = φ -1( g i ) = g .

(2.2) Then φ -1 � κ

ε( g ).

Michel 11 Define T : ε( g ) → ε( g i ) by T ( δ ) = φ � δ (notice from our work above that T is well-defined).

WTS T is one to one.

Let α , β

ε( g ) such that T ( α ) = T ( β ).

Then φ � α = φ � β .

Therefore α = β , since φ is a permutation and the cancellation property holds.

WTS T is onto.

So κ

ε( g i ). Let δ = φ -1 � κ . Then δ

ε( g ) by (2.2).

Also T ( δ ) = φ � δ = φ � ( φ -1 � κ ) = κ , hence T is onto.

Therefore ε( g ) = ε( g i ) .

Since i was arbitrary, we know ε( g ) = ε ( g i ) for all i , 1 ≤ i ≤ n.

Also, since g1 ,..., g n are distinct, ε( g i ) ∩ ε ( g ) is empty for all i ≠ 1 (g = g1 ). n

Therefore Sq = ∪ ε ( g i ) and we also get Sq = ∑i =1 ε( g i ) = ∑i =1 ε( g ) . n

n

i =1

Then q! = n ε( g ) .

Q.E.D. This concludes the discussion of Lagrange’s Theorem.

Let us notice the similarities when comparing the above theorem from Article 97 in Lagrange’s “Reflexions sur la resolution algebrique des equations” to its more contemporary meaning in group theory.

Theorem (2.3) If G is a finite group and H is a subgroup of G, then H divides G . Moreover, the number of distinct (left or right) cosets of H in G is

H G

(Gallian 137).

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Discussion of the Proof of the Contemporary Version of Lagrange Lagrange’’s Theorem: Lemma (2.3) Let H be a subgroup of G, and let a and b belong to G. Then (1) a aH , (2) aH = bH or aH ∩ bH is empty , (3) aH = bH .

Proof of (1): It is obvious that a = a • e

aH .

Proof of (2): Suppose aH ∩ bH is not empty .

WTS aH = bH .

Then there exists some y1 , y 2 and bH ).

H such that x = ay 1 and x = by 2 (definition of aH

Hence a = xy1-1 = by 2 y 1-1 by substitution.

: Let ah

aH .

So ah = ( by 2 y -1 )h = b(( y 2 y 1-1 )h ) Then aH

: Let bh

bH since y 2 y 1-1

H.

bH .

bH .

So bh = ( ay 1y 2-1 )h = a(( y 1y 2-1 )h ) Then bH

aH since y 1y 2-1

H.

aH .

Hence aH = bH .

Proof of (3): Let the map φ : aH → bH be defined by φ( ah ) = bh. The simple proof that φ is one to one and onto is omitted.

Therefore aH = bH .

Thus we have proved (2.3).

Let α1H , α 2H ,...., α q H denote the distinct left cosets of H in G.

G,α αH by (2.3.1). Furthermore by (2.3.1), there exists a k (1 ≤ k ≤ q ), such that αH = α k H .

For all α

Michel 13

q

Thus each element of G are in some α i H . i.e. G = ∪ α i H . i =1

So G = α1H + • • • + α q H since the cosets are all disjoint by (2.3.2) .

From (2.3.3), α k H = eH = H for all k (1 ≤ k ≤ q ).

Now we get G = q H . Q.E.D. This concludes the discussion of the contemporary version of Lagrange’s Theorem.

Notice that in the proof that precedes the one above, ε( g i ) represents the same coset that α i H does in the more contemporary version of Lagrange’s Theorem. Lagrange contributed to several branches of Mathematics including number theory, equation theory, ordinary and partial differential equations, calculus of variations, analytic geometry, fluid mechanics and celestial mechanics (Gallian 149). Lagrange’s methods for solving third and fourth degree polynomials as well as theorems similar to the preceding one are what laid the groundwork for algebraists such as Abel and Galois.

Impacts Lagrange was the first to prove (2.0), and it was complete in its first publication. It appears to both Tignol and Gallian that very little was taken from the works of any other mathematicians in order for Lagrange to prove this theorem. Lagrange and Euler both worked heavily with permutations, though, which was very likely the groundwork for Cauchy’s 1815 paper on permutations from which Abel and Galois drew inspiration. There is no doubt that Lagrange should take full credit for his theorem, and partial credit for latter proven theorems which rely on Lagrange’s theorem.

Gauss The Disquisitiones Arithmeticae is the heart of Gauss’s work. It contains research important to the theory of numbers and the theory of equations such as the Fundamental Theorem of Gauss and the Law of Quadratic Reciprocity of Legendre. According to Dunnington, Disquisitiones would have placed Gauss among the top ranking mathematicians of his day. In March 1775, by induction, he discovered that for odd primes p, -1 is a quadratic residue of prime numbers of the form 4n+1 and is a quadratic nonresidue of primes of the form 4n+3, that is, 1 ≡ x 2 (mod p ) for some x Z + if and only if p is of the form 4n+1. (38) He wrote the first proof of it at the age if nineteen and published it the following year.

Now restless about this class of research, he published six different principles concerning residues and nonresidues of prime numbers.

Michel 14

Having really achieved what he aimed for, his fascination with modular arithmetic grew exponentially. The subsequent study modular arithmetic by such masters of analysis as Euler and Lagrange could hardly fail to use the work of young Gauss, a recognition unlike that of other brilliant young men experienced.

Also, the great result of this arithmetical research being the complete solution to the binomial equation, Gauss anticipated the results of his more experienced contemporaries. He was even the first to ever use the symbol − 1 to geometrically interpret the imaginary number. Concerning the theorem on binomial equations, he gave presentations in 1815 and 1816 and published the results, but these still gave him little fame. According to Dunnington, Lagrange seemed to not have heard of the first presentation, and Cauchy actually received credit as a first discoverer of the binomial equation solution in France. (41) In 1799, Gauss published his dissertation. There were several attempts to prove this Fundamental Theorem of Algebra by predecessors such as Lagrange and Ruffini (Lagrange progressed the furthest), but Gauss claimed that these proofs were incomplete as he tendered an indisputable proof in his dissertation.

According to Tignol, Gauss received the highest praise for his proof of it. (167) Gauss’s work can naturally separate into two major topics: the fundamental theorem of algebra and the solution of cyclotomic equations (solutions to irreducible factors of x n - 1 in Q[x]), which he devoted the last section in Disquisitiones to (Tignol 166). His results on cyclotomic polynomials show how to complete Vandermonde’s arguments to provide functions by radicals inductively for the roots of unity. Gauss thoroughly reviews all possible reductions of cyclotomic polynomials of prime index to those of smaller degrees.

According to Tignol, Lagrange once envisioned solving an equation by determining, successively, certain functions of the roots. (168) But Gauss shows p the solution of xx -1-1 = 0 can be reduced to something of degree equal to the prime factors of p- 1. As briefly mentioned above, it’s the seventh and last section of Gauss’s Disquisitiones that discusses the 17th roots of unity being determined solving successively four quadratic equations, since 17 - 1 = 24 .

One of Gauss’s contributions to the theory of groups is Gauss’s Lemma:

Lemma (3.0) The product of two primitive polynomials is primitive (Gallian 297).

Michel 15

Discussion of the Proof of Gauss Gauss’’s Lemma: Definition (3.1) The content of a nonzero polynomial α1x n + α n-1x n-1 + • • • + α 0 , where α Z, is the greatest common divisor of α n , α n -1 ,..., α 0 . A primitive polynomial is an element of Z[x] with content 1 (Gallian 297).

Suppose φ( x )ε( x ) is not primitive and assume the content β Z is greater than 1.

Z such that p is prime and p | β .

Then there exists p

Let the coefficients of φ( x ) and ε( x ) be ω1 ,..., ωn and κ1 ,..., κ m respectively.

Also let these coefficients be reduced (mod p) and call these ω1• ,..., ωn• and

κ1• ,..., κ m• respectively.

Call the polynomials with the reduced coefficients φ • ( x ) and ε • ( x ) .

Then φ • ( x ),ε • ( x ) Z p [x].

(3.2) We get φ • ( x )ε • ( x ) = 0 and hence p (

Z p [x] since p | β .

∑ ωl κ t ) for all s, 0 ≤

s ≤ n + m.

l +t =s

Suppose φ( x ) ≠ 0

Z p [x]. So there exists a q such that p does not divide ωq but

p does divide ωc for all c < q.

WTS ε • ( x ) = 0 Z p [x].

Consider s = q for ω0κ q + • • • + ωq 1κ 1 + ωq κ 0 .

It is obvious that p ω0 κ q + • • • + ωq -1κ 1 since p ωc for all c, 0 ≤ c < q.

And p ωq κ 0 + • • • + ωq -1κ1 + ωq κ 0 we get from (3.2) .

Therefore p must divide the difference, which is ωq κ 0 , but not ωq .

Michel 16 Hence p | κ 0 .

Suppose for b > 0, p | κ c for all c < b. Now consider s = q + b.

So

q -1

q +b

∑ω κ

∑ω κ

c

q +b -c

+ ωq κ b +

c =0

q -1 c

q +b -c

is divisible by p as well as

c = q +1

∑ω κ c

q + b -c

(since

c =0 q +b

p ωc for all c, 0 ≤ c ≤ q) and

∑ω κ c

q + b- c

(since q + b - c < b and

c =q +1

p κ c for all c < b).

Then p divides the difference, which is ωq κ b , but not ωq .

Hence p divides κ b . Thus, by induction, p κ b for all b, 0 ≤ b ≤ m, and hence

ε • ( x ) = 0 in Z p .

Thus φ • ( x ) = 0 or ε • ( x ) = 0.

Therefore p | ωi• for i 1,…,n or p | κ •j for j 1,…,m.

Hence either φ • ( x ) is not primitive or ε • ( x ) is not primitive.

Q.E.D. This concludes the discussion of Gauss’s Lemma.

It is extraordinary how close Gauss comes to defining specifically the properties of groups without necessarily doing so. Unfortunately, reviewing the entire extent of the relationship between Gauss’s Disquisitiones and group theory, through modular arithmetic, is beyond the scope of this essay.

Impacts According to Gallian, in 1801, the asteroid Ceres was observed by astronomers on three occasions before they lost it. Gauss brilliantly used these three observations to calculate the orbit of Ceres. While working on this superhuman task, he found that the variation in experimentally derived data follows a bell-shaped curve now known as the Gaussian distribution (Gallian 573).

All of the above (and more) was accomplished by the genius, Carl Frederick Gauss, before the age of 25!

It is apparent as on reads Gauss’s Disquisitiones that there was much groundwork laid by both Lagrange and Euler since the number of citations are countless. However, there are few accounts of Gauss ever offering an incomplete proof concerning any theorem he ever proved, and on several occasions he was the first discoverer. There is no substantial evidence that the

Michel 17 preceding lemma, Gauss’s Lemma, was the success of anyone but Gauss himself. The foundational arguments that Euler and Lagrange contributed to Gauss’s success were in the hands of a man who, without them, still would have succeeded.

Abel and Galois Abel (1802 – 1829) was born and raised in Norway. As a boy he already was intrigued by the work of Lagrange and Cauchy’s permutation work. Once he reached college, he progressed rapidly and in 1821 began writing original papers (Hollingdale 298). Abel eventually grew frustrated with one problem in particular for which no mathematician had yet offered a proof. It was this very problem that led Abel to discovering what are now known as Abelian equations, that is, equations that are solvable if g(h(x)) = h(g(x)) .

Abel ignored Gauss’s opinion that the quintic was unsolvable, and relentlessly searched for the general solution to it. According to Pesic, it was not until Abel submitted samples of his work to his teachers that he knew this solution was much more complicated than he had thought. (90) From there he could have spent a lifetime trying to prove its solvability. Instead he intuitively turned completely around and proved its unsolvability by contradiction. That is, he assumed that the general solution existed, and showed that it led to an absurdity (impossibility).

His first step was to describe the form the solution must take if it did indeed exist. Regarding the equation of degree n, BnY n + Bn-1Y n-1 + • • • + B1Y + B0 , Abel proved this general statement:

All algebraic functions of Y can be expressed in the form 1 2 n -1 Y = q + Z n + q2Z n + • • • + qn-1Z n where q1,q 2 ,..., qn -1 are finite sums of radicals and 1

polynomials and Z n is in general an irrational function of the coefficients of the original equation (Pesic 92).

What this means is that if Y is a solution to the equation of degree n, we can say Y is equal to the sum of terms described above. i.e.

1

2

3

Suppose n = 4. Then Y = q1 + Z 4 + q2 Z 4 + q3 Z 4 .

Michel 18 However, in the case of the quintic, where n = 5, 1 2 3 4 Y = q + Z 5 + q2 Z 5 + q3 Z 5 + q 4 Z 5 . So Abel proved that the solution either has this form, or does not exist. Next he proved that if a general solution exists, it could be expressed in terms of rational functions of the equation’s roots (Pesic 93).

Now all of the irrational functions of the coefficients of Y are rational functions of the roots of Y. i.e.

1

When n =5, Z 5 is a rational function of the roots of Y. This is genius! Abel discovered that the roots of Y are irrational functions of the coefficients of Y, and the coefficients will always be rational sums or products of the roots of Y. The following is the heart of Abel’s proof of the unsolvability of the quintic. Abel proved this statement:

If a rational function of five quantities takes fewer than five values when the five quantities are permuted, it can take only two different values (equal in magnitude and opposite in sign), or one value, but never three or four values (Pesic 93).

The groundwork for this portion of Abel’s proof came partly from Cauchy’s 1 work in permutations. Abel knew from Cauchy that Z 5 could take only one, two, or five values when the roots are permuted. It was also partly the Fundamental Theorem of Algebra which we have already discussed: an equation of the fifth degree must have at least one root and no more than five roots. i.e.

1

Z 5 can’t take only one value by Abel’s assumption that the five roots are not equal.

1

However, when Abel assumes that Z 5 takes five different values when the roots are permuted, it leads to an expression with 120 values, which is impossible. Subsequently, the only other possibility Abel was considering was 1 that of two different values. But Z 5 having two values when permuting the roots led to another absurdity. When he derived the equation, it gave an expression with 120 values on the left and 10 values on the right. Then, in 1828, Abel was finally able to claim: the general solution to the equation of degree five cannot be obtained by radicals.

Everiste Galois (1811 – 1832), whose work agreed entirely with Abel’s, also proved the unsolvability of the quintic in 1830. Unaware of Abel’s work, Galois gave a complete account of the solvability of algebraic equations that was more thorough than Abel’s. He did this before the age of 20. Many minds today are intrigued at the notion that two mathematicians so young could independently prove something so brilliant and then die such mysterious deaths.

Galois is responsible for the abstraction of algebraic systems. He understood solvability better than any mathematician before him (and probably

Michel 19 after), even Abel. What Galois did, that Abel didn’t, was discuss the permutations of the roots, not case by case, but by an abstraction of the algebraic system in which they are elements. He called these systems groups!

Impacts The unsolvability of the quintic was the most valuable contribution Abel made to the birth of abstract algebra. Not only because this led to the definition of Abelian equations, but its latter application to group theory. Abel was the first to prove the quintic’s unsolvability and his proof definitely was complete. As mentioned above, Abel may not have been able to do this without the Fundamental Theorem of Algebra, which we know Lagrange, Girard, and Gauss all played parts in proving. Abel also employed Cauchy’s arguments regarding permutations of roots.

Galois too can get praise as a first discoverer since his proof of the quintic’s unsolvability came before 1831. Similar groundwork from Lagrange, Girard, Gauss and Cauchy was employed by Galois. However, the emergence of group theory was, primarily, the brilliance of Galois.

Conclusion We set out to examine the roles of a handful of mathematicians in the birth of abstract algebra. After viewing theorems such as the Fundamental Theorem of Algebra, Lagrange’s Theorem, Gauss’s Lemma, and the unsolvability of the quintic, it is safe to say we caught a glimpse of Algebra’s evolution in the last three hundred years. Out of the several mathematicians we discussed, and after discussing the impacts their discoveries made on the theory of groups and rings, it seems fair to call Euler, Lagrange, Gauss, Abel and Galois the rightful midwives of abstract algebra. Their work not only stands out in an ocean of works in number theory and equation theory, but was essential for the development of group and ring theory. We must acknowledge that before these men devoted themselves, pure mathematics was only an isolated collection of results. Let us thank these midwives by continuing to discover and expand the frontier of mathematical development!

Michel 20 Appendix Cardanoâ&#x20AC;&#x2122;s Method General Cubic: x 3 + ax 2 + bx + c = 0 First change variables y = x + a3 , to get

( y - a3 )3 + a( y - a3 )2 + b( y - a3 ) + c = 0 and simplify as follows: ( y - a3 )( y - a3 )( y - a3 ) + a( y - a3 )( y - a3 ) + b( y - a3 ) + c = 0 2

2

3

2

[ y 3 - 2 a3 y 2 + a9 y - a3 y 2 + 2 a9 - a27 ] + a [ y 2 - 2 a3 y + a9 ] + b [ y - a3 ] + c = 0 2

2

3

3

y 3 + a9 y - 2 a3 y + by + 2 a3 - a27 + a9 - b a3 + c = 0 2

2

3

3

y 3 + ( a9 - 2 a3 + b )y + ( 2 a3 + a27 + a9 - b a3 + c ) = 0.

For the sake of brevity we will call the above cubic equation y 3 + uy + v = 0. The substitution allowed us to reduce the task of solving the general cubic to the special case where there is no y 2 term.

Suppose y = 3 m + 3 n .

Then y 3 = ( 3 m + 3 n )3 = ( 3 m + 3 n )( 3 m + 3 n )( 3 m + 3 n ) = ( 3 m 2 + 23 m 3 n + 3 n 2 )( 3 m + 3 n ) = m + 23 m 2 3 n + 3 m 3 n 2 + 3 m 2 3 n + 23 m + 3 n 2 + n = m + n + 3( 3 m 2 3 n + 3 m 3 n 2 ) .

So 0 = y 3 + uy + v = ( m + n + 3( 3 m 2 3 n + 3 m 3 n 2 )) + u( 3 m + 3 n ) + v = ( m + n + v ) + ( 33 mn ( m m + 3 n )) + u( 3 m + 3 n ) .

Then ( m + n + v ) + 3( 3 mn + u )( 3 m + 3 n ) = 0 .

This equation clearly holds if the part (m+n+v) and 3( 3 mn + u )( 3 m + 3 n ) both vanish, for example, if m+n = -v and mn = - ( u3 )3 .

Michel 21 Solving this system of equations, we get solutions m, n = - v2 ± ( u3 )3 + ( v2 )2 .

Hence y = 3 - ( v2 ) + ( u3 )3 + ( v2 ) 2 + 3 - ( v2 ) - ( u3 )3 + ( v2 ) 2 and by substituting our choices for u and v back into the solution we have a general solution to x 3 + ax 2 + bx + c = 0 .

Ferrari’s Method General Quartic: x 4 + ax 3 + bx 2 + cx + d = 0 (1) Again, begin by changing variables, y = x +

a 4

so x = y - a4 , to get

( y - a4 ) 4 + a( y - a4 )3 + b( y - a4 )2 + c ( y - a4 ) + d = 0 , which simplifies to a quartic without a cubic term, y 4 + py 2 + qy + r = 0 , where p = b - 6( a4 ) 2 , q = c - b( a2 ) + ( a2 )3 , r = d - c( a4 ) + b( a4 )2 - 3( a4 ) 4 .

By completing the square we get y 4 + py 2 = - qy - r , p

p

( y 2 )2 + py 2 = -qy - r , and finally ( y 2 + 2 )2 = -qy - r + ( 2 )2 .

Now add a quantity u ∈ R as follows: p

p

( y 2 + 2 + u )2 = -qy - r + ( 2 )2 + 2uy 2 + pu + u 2 , We want to determine u in such a way that the right hand side becomes a perfect square. If the right-hand side is a square, then it is the square of q 2u y - 2 2u . p

So we have - qy - r + ( 2 )2 + 2uy 2 + pu + u 2 = ( 2u y holds if and only if

q2 8u

p

= -r + pu + ( u )2 + u 2 .

(2) Clear the denominator as follows: 8u 3 + 8 pu 2 + ( 2p 2 - 8r )u - q 2 = 0 .

Since (2) is a cubic, such a u can be found. p

Thus, if u ≠ 0, ( y 2 + 2 + u ) 2 = ( 2u y -

q 2 2u

)2 .

q 2 2u

)2 . But this equation

Michel 22 First, p

p

p

p

p

p

p

( y 2 + 2 + u )( y 2 + 2 + u ) = y 4 + 2 y 2 + uy 2 + 2 y 2 + ( 2 )2 + 2 u + uy 2 + 2 u + u 2 p

= y 4 + py 2 + 2uy 2 + pu + ( 2 ) 2 + u 2 , and y 4 + py 2 = - qy - r as stated above. p

p

Therefore ( y 2 + 2 + u ) 2 = -qy - r + ( 2 )2 + 2uy 2 + pu + u 2 , and ( 2u y -

q 2 2u

)( 2u y - 2

q2

q 2u

q2

q

q2

) = 2uy 2 - 2 2 y + 8u = 2uy 2 + 8u - qy . p

So 2uy 2 + 8 u - qy = y 4 + py 2 + 2uy 2 + pu + ( 2 ) 2 + u 2 . p

(3) So returning to ( y 2 + 2 + u )2 = ( 2u y - 2

q 2 2u

) , if u ≠ 0, then y must satisfy one p

of the following quadratic equations: y 2 + 2 + u = ±( 2u y -

q 2 2u

).

If u = 0, then q = 0 and (1) is y 4 + py 2 + r = 0 , which is quadratic in y 2 and easily solved.

If u ≠ 0, then solving the two quadratic equations (3) for y is all that remains.

Michel 23 Works Cited Dunnington, Waldo. Carl Friedrick Gauss Titan of Science. Exposition Press Inc. New York New York, 1955.

Gallian, Joseph. Contemporary Abstract Algebra. Houghton Mifflin Company Boston New York, 2002.

Gauss, Carl. Disquisitiones Arithmeticae English Edition. Springer Verlag New York New York, 1986.

Hollingdale, Stewart. Makers of Mathematics. Penguin Books New York New York, 1989.

Hofmann, Joseph. The History of Mathematics to 1800. Littlefield, Adams & Co. Totowa New Jersey, 1967.

James, Ioan. Remarkable Mathematicians from Euler to Neumann. The Mathematical Association of America, Cambridge University Press. New York New York, 2002.

“O’conner, JJ.” History of Abstract Algebra. 09 July 2005 http://wwwhistory.mcs.st-andrews.ac.uk/HistTopics/Quadratic_etc_equations.html

“O’conner, JJ.” History of Abstract Algebra. 09 July 2005 http://wwwgroups.dcs.stand.ac.uk/~history/HistTopics/Development_group_theory.html Pesic, Peter. Abel’s Proof. Massachusetts Institute of Technology. Cambridge Massachusetts, 2003.

Shenitzer, Abe and Stillwell, John. Mathematical Evolutions. The Mathematical Association of America Washington DC, 2002.

Tignol, Jean-Pierre. Galois’ Theory of Algebraic Equations. World Scientific Publishing Co. River Edge New Jersey, 2001.