GALOIS; The first Memoir

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1 GALOIS; The first Memoir Trying to understand it 1 Preliminaries 1.1 Notational Conventions Results directly copied from Galois Memoir, or revisions of such results, are stated using small caps; for example, Theorem. Results that I am adding to make (I hope) the Memoir more accessible to people whose algebra background is not much above mine, are stated in italics and numbered for easy reference. 1.2 Fields The concept of a field was developed decades after Galois time; however, it helps to be able to use the language of fields. After all, nowadays, Galois Theory is a chapter of field theory. So what is a field? For us it will suffice to say that a field is any subset of the set of complex numbers that contains 0 and 1 and that is closed under addition, subtraction, multiplication, and contains the reciprocal (inverse) of all of its non-zero quantities. Nowadays one considers more general fields, but we are not in nowadays, we are in Galois days. The smallest possible field, by our definition, is Q itself. Once 0,1 are in the field, we also must have = 2, = 3, etc., 0 1 = 1, 1 1 = 2, etc.; all the integers must be in it. But then so are the reciprocals of the non-zero integers: ±1/2, ±1/3,.... Finally any number of the form a/b with a, b, integers, b 0, must be in the field since a/b = a(1/b) and fields are closed under multiplication. Other obvious fields are, of course, the field R of all real numbers, and the field C of all complex numbers. There are zillions of less obvious fields. The ones of interest to us (because they are the fields Galois considered, even though he did not call them fields) are obtained by adding a certain quantity of numbers to Q and then adding all possible products, divisions, etc. until one has a field. That is, one gets the smallest field containing these added numbers. This might seem like a terribly complicated procedure but it frequently isn t. For example, the field Q( 2) is obtained by adding 2 to Q. It has a simple description: Q( 2) = {a + b 2 : a, b Q}. It should be clear that any field containing 2 will have to contain all numbers of the from a + b 2, with a, b Q; but why doesn t it contain anything else? Is the set Q( 2) of the displayed description a field? It is very easy to see that two numbers of the form a + b 2, c + d 2 when added, subtracted or multiplied yield another number of the same type. Moreover 0 = and 1 = are of this type. Division is also easy, once we see that because 2 is irrational, the only way we can have a + b 2 = 0 is if both a and b equal 0. a + b 2 c + d 2 = (a + b 2)(c d 2) (c + d 2)(c d ac ad 2 bc 2 2bd = 2) c 2 2d 2 = Thus the numbers of the form a + b 2, a, b Q, do form a field. ac 2bd ad bc c 2 + 2d2 2. c 2 2d 2 If K is a field and a is a complex number, we denote by K(a) the smallest field containing a and all the elements of K. Of course, if a K, then K(a) = K. Before we continue with fields, we should know a bit about polynomials and irreducible polynomials. 1.3 Polynomials Suppose K is a field; so a subset of C containing 0 and 1 and closed under the four basic arithmetic operations. One usually writes K[X] to denote the set of all polynomials with coefficients in K, so K[X] contains all expressions a 0 + a 1 x + + a n x n with a 0, a 1,..., a n K. I will assume we know a little bit about polynomials but just in case I ll remind you that if p(x) = a 0 + a 1 x + + a n x n and a n 0, then one says that p(x) has degree n. A polynomial of degree 0 is a non-zero

2 1 PRELIMINARIES 2 constant polynomial; one of the form p(x) = a 0, a 0 0. The 0 polynomial is said to have degree. Operating with infinity in the usual way, one has deg(p(x)q(x)) = deg p(x) + deg q(x) for all polynomials p9x), q(x). A very basic and important result is the division algorithm: Theorem. (Division Algorithm) Let p(x), q(x) be polynomials in K[X], with q(x) not the zero polynomial. There exist polynomials h(x), r(x), with the degree of r < degree of q such that p(x) = q(x)f(x) + r(x). These polynomials are unique. The easiest proof of the existence part is by induction on the degree of p. If the degree of p is less than that of q, we take f = 0 and r = p, and we are done. So assume that the degree of p degree of q. Our base case is then deg p = deg q, say p(x) = a m x m + + a 0, q(x) = b m x m + + b 0, a m 0 b m. Let f(x) = a m /b m (constant polynomial); then setting r = p fq it is clear that r is of degree < m and we are done. Assume the result now proved when deg p n, for some n m = deg q. Assume deg p = n + 1. Let p(x) = a n+1 x n a 0, q(x) = b m x m + + b 0, a n+1 0 b m. Then p(x) (a n+1 /b m )x n+1 m q(x) is of degree n; by the induction hypothesis there exist f, r, r of degree < m such that p(x) (a n+1 /b m )x n+1 m q(x) = q(x)f(x) + r(x), hence p(x) = q(x) ( (a n+1 /b m )x n+1 m + f(x) ) + r(x), and we are done in what concerns existence. For uniqueness, assume we have qf + r = qf + r where r, r are of degree less than q. Then r r = q(f f) and this implies deg q+deg (f f) = deg (r r) < deg q, only possible if deg (f f) = ; i.e., f = f, and then also r = r. One consequence of this algorithm is that K[X] is what one calls a principal ideal domain. We don t need this result in full, so I ll skip it; we only need a consequence of this result that can easily be established independently. Theorem. (Greatest Common Divisor) Let p(x), q(x) K[X]. Assume at least one of p(x), q(x) is not the zero polynomial. There exists a polynomial f(x) K[X] such that: 1. f divides both p and q. 2. If h K[X] and h divides both p and q, then h divides f The polynomial f is unique up to multiplication by a non-zero constant. Here is a non-constructive proof of this theorem. Consider the set S of all polynomials in K of the form g(x)p(x) + h(x)q(x), where g(x), h(x) K[X]. This set must contain a non-zero polynomial f(x) of minimum degree. (Because not both p(x), q(x) are zero, the set of degrees of elements of S contains non-negative integers; one of these must be smallest.) Say f(x) = g(x)p(x) + h(x)q(x). By the division algorithm, there are polynomials a(x), r(x), with deg r < deg f, such that p(x) = a(x)f(x) + r(x). Then r(x) = p(x) a(x)f(x) = p(x) a(x)(g(x)p(x) + h(x)q(x)) = (1 a(x)g(x))p(x) + (a(x)h(x))q(x) S. By the definition of f(x), there is no nonzero polynomial in S with degree < deg f, thus r is the zero polynomial and f divides p. Similarly, f divides q. Because f = g(x)p(x) + h(x)q(x), it is clear that any polynomial dividing p and q will divide f. Finally, if f is another polynomial dividing p, q and being divided by any common divisor of p, q, then f divides f and f divides f, from which f = cf for some c 0 follows (by degree comparison, for example). It is useful to know that the greatest common divisor f(x) of polynomials p(x), q(x) in K[X] satisfies for some g(x), h(x) K[X]. This is sometimes known as Bezout s equation. 1.4 Fields, again f(x) = g(x)p(x) + h(x)q(x) (1) We return to fields obtained by adjoining quantities to Q. One example plays somewhat of a role, and it is the following. Suppose p(x) = a m x m + a m 1 x m a 1 x + a 0

3 1 PRELIMINARIES 3 is an irreducible polynomial of degree m 2 over the rational numbers. That means it cannot be obtained as the product of two polynomials with rational coefficients both of degree < m. In particular, all of its zeros are irrational; otherwise it would decompose into a factor of degree 1 times one of degree m 1. Suppose ξ is one of its zeros. Then one writes Q(ξ) to denote the smallest field containing ξ. It is not to hard to show that Equivalently, Q(ξ) = {b 0 + b 1 ξ + + b m 1 ξ m 1 : b 0, b 1,..., b m 1 Q}. Q(ξ) = {q(ξ) : q a polynomial with rational coefficients of degree m 1}. It is sort of obvious that as defined Q(ξ) is closed under addition and subtraction. Concerning products, suppose α, β Q(ξ). Then there are polynomials with rational coefficients q 1, q 2 of degree m 1 such that α = q 1 (ξ), β = q 2 (ξ). As polynomials, the product q 1 (X)q 2 (X) can be of degree m. But, we divide by p; by the division algorithm there exist polynomials g(x), r(x), with r(x) of degree < m (i.e., m 1) such that Evaluating at ξ and remembering that p(ξ) = 0, we get q 1 (X)q 2 (X) = g(x)p(x) + r(x). αβ = q 1 (ξ)q 2 (ξ) = g(ξ)p(ξ) + r(ξ) = 0 + r(ξ) = r(ξ) Q(ξ). For inverses, we appeal to our friend Bezout. Let α Q(ξ) and α 0. Then α = q(ξ) for some non-zero polynomial q of degree m 1. Because p is irreducible, there exist polynomials g(x), h(x) with rational coefficients such that g(x)p(x) + h(x)q(x) = 1. We may assume h is of degree m 1; otherwise apply the division algorithm to get h(x) = p(x)f(x) + r(x), where r is of degree m 1. Then 1 = g(x)p(x) + (p(x)f(x) + r(x))q(x) = (g(x) + f(x))p(x) + r(x)q(x), and we replace g by g + f and h by r. Evaluating at ξ: thus 1/α = h(ξ) Q(ξ). 1 = g(ξ)p(ξ) + h(ξ)q(ξ) = 0 + h(ξ)α, It is an easily proved fact that the representation of elements of Q(ξ) as polynomials of degree m 1 with coefficients in Q is unique: If q 1 (ξ) = q 2 (ξ) and q 1 (X), q 2 (X) Q[X] are of degree m 1, then q 1 (X) = q 2 (X). In fact, otherwise q = q 1 q 2 is a polynomial of degree m 1 with q(ξ) = 0. We prove in the section dedicated to Galois first lemma that (as that first lemma states), p must divide q, an impossibility since the degree of q is strictly less than the degree of p, except if q(x) = 0. In the language of vector spaces, this implies that Q(ξ) is an m-dimensional vector space over Q, with basis {1, ξ,..., ξ m 1 }, In the construction above, Q can be replaced by any other field; in particular by a field coming from Q by adjoining some numbers. If K is a field, if p(x) is a polynomial of degree m that is irreducible over K (can t be expressed as the product of two polynomials with coefficients in K of degree < m), if ξ C is a zero of p, then the field K(ξ) is the smallest field containing all the elements of K and ξ, and the proof above, replacing all mentions of Q by K shows that K(ξ) = {b 0 + b 1 ξ + + b m 1 ξ m 1 : b 0, b 1,..., b m 1 K}. Then, similarly as for Q, K(ξ) is an m-dimensional vector space over K, with basis {1, ξ,..., ξ m 1 }. An important consequence of this result, surely not unknown to Galois, is the following Lemma 1 Let K be a field, let p(x) K[X] be irreducible of degree m. Let ξ be a root of p(x) = 0. Every element of K(ξ) satisfies a polynomial equation of degree at most m.

4 2 RATIONAL QUANTITIES 4 Proof. Because {1, ξ,..., ξ m 1 } is a basis of K(ξ) as a vector space over the field K, if α K(ξ), the m elements {1, α,..., α m } cannot be linearly independent; thus there exist c 0, c 1,..., c m K, not all of these elements equal to 0, such that c 0 + c 1 α + + c m α m = 0. In general, if K is a field, one denotes by K(b 1,..., b n ) the smallest field containing all the elements of K and b 1,..., b n. One can show that this can be done step by step. That is K(b 1, b 2 ) = K(b 1 )(b 2 ); K(b 1, b 2, b 3 ) = K(b 1, b 2 )(b 3 ), etc. And now, let s move on to the actual Memoir. 2 Rational Quantities For Galois, a rational quantity is defined in terms of a given algebraic equation, and some finite number of other quantities that may have been adjoined. Assume an equation a m x m + a 1 x m a 1 x + a 0 = 0 (2) is given, where the coefficients a 0,..., a m are complex numbers and a m 0, m 1. equation; the one we are trying to solve (or show it can t be solved) by radicals. We then have the following situations. We ll call this equation the main Case 1. The coefficients a 0,..., a m are rational and no other numbers have been adjoined. Then rational means rational in the usual, current sense. Case 2. Some coefficients are not rational, or there are adjoined quantities. In that case a rational quantity is any quantity that, as Galois puts it, is a rational function of the coefficients and the adjoined numbers; i.e., can be obtained from these, and rational numbers, by the operations of addition, subtraction, multiplication and division. In modern parlance we d say that a Galois rational is an element of the smallest field containing the coefficients and the adjoined quantities. An example Consider the equation x = 0 (Actually, at this point any equation with rational coefficients will do). If we say nothing further, rational means rational; a number that can be expressed as the quotient of two integers. But if we consider the equation and adjoin 2 (in modern parlance, consider it as an equation in Q( 2), numbers like 2, , 2 become rational. If I adjoin 2 and 3, then 6 is rational. Or should one say g rational for Galois rational? Here is where the field language can help. Originally, we let K be the smallest field containing all the coefficients of the main equation; K = Q if all coefficients are rational. In general K = Q(a 0,..., a m ). As the arguments progress one may have to adjoin further numbers to K, but at each point it should be clear with what field we are working. A g rational quantity is thus simply an element of K. 3 Irreducible equations Having defined what he means by rational quantity, Galois defines what he means by an irreducible equation. The equation a m x m + a 1 x m a 1 x + a 0 = 0, a m 0 is irreducible iff the polynomial p(x) = a m x m +a 1 x m 1 + +a 1 x+a 0 cannot be expressed as the product of two polynomials of degree < m with rational coefficients. Rational has to be interpreted in the Galoisian sense. Here are three examples. 1. Suppose the equation is x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 = 0 (this example is given by Galois). Consider the polynomial p(x) = x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 = x7 1 x 1. It is not terribly hard to prove it is irreducible (the easiest way is to use the Eisenstein criterion-with a standard translation trick about which we don t have to worry, but it can also be done by hand). That is, there are no two

5 4 THE FIRST LEMMA 5 polynomials of degree < 6 with rational coefficients which multiply to p(x). But now suppose we adjoin r = e 2πi/7. Then p(x) = (X r)(x r 2 )(X r 3 )(X r 4 )(X r 5 )(X r 6 ) and since all the numbers r, r 2,..., r 6 are g rational, the polynomial has ceased to be irreducible. 2. Consider x = 0. This equation is irreducible, but if we adjoin 2, it becomes reducible since x = (X x + 1)(X 2 2 x + 1). 3. Consider the equation x 2 + (1 + 7)x 7 = 0. This equation is reducible; in fact and 7 = (1 + 7) 1 is g rational. 4 The first Lemma x 2 + (1 + 7)x + 7 = (X + 1)(X + 7) Lemma I. An irreducible equation cannot have any root in common with a rational equation without dividing it. One has to remember that here rational means g rational and that what Galois means an equation dividing an equation is more precisely expressed by replacing the word equation by the word polynomial. That is, in more precise terms the lemma should be stated as: A polynomial corresponding to an irreducible equation cannot have any zero in common with a g-rational polynomial without dividing it. In other words, we have our equation, p(x) is the corresponding polynomial and the equation is irreducible. There may or may not be adjoined quantities; that q(x) is g-rational means that its coefficients are in the field K. By the Euclidean algorithm, p(x), q(x) have a greatest common divisor f(x) K[X]. Because p(x) is irreducible, there are only two possibilities for f(x); either f(x) is a non-zero constant, or a nonzero constant times p(x). But by Bezout s formula there exist two other polynomials g(x), h(x), also with coefficients in K, such that g(x)p(x) + h(x)q(x) = f(x). (3) That p(x), q(x) have a common zero means that there exists a number ξ (and it can t be in K!; otherwise p(x) is not irreducible) such that p(ξ) = q(ξ) = 0. But then, by (3), f(ξ) = 0, so f(x) is not a non-zero constant. Thus f(x) = p(x) (or a constant times p(x)) and p(x) divides q(x). 5 The second Lemma Galois considers the first lemma obvious; the second one is more interesting. He also considers it obvious. Permutations appear for the first time. I will assume we know what is meant by a permutation and use the symbol S m for the set of permutations of m objects. These objects are best described as being the integers from 1 to m. An element σ of S m is a rearrangement of the integers 1, 2,..., n; more precisely it is a one to one function from {1, 2,..., n} onto itself. One convenient way to describe such rearrangements/functions is to associate with a permutation σ a 2 n matrix in which the first row consists of the integers in their natural order, the second row are the rearranged integers. For example, if n = 3, one element of S 3 is ( ) σ = As a function, σ(1) = 3, σ(2) = 2, σ(3) = 1. Another element of S 3 is ( ) τ = ; that is τ(1) = 2, τ(2) = 3, τ(3) = 1. The compositions στ and τσ are, respectively ( ) ( ) στ =, τσ =

6 5 THE SECOND LEMMA 6 In general, if σ, τ S m, then στ is the usual composition of function; τ first, then σ: στ(i) = σ (τ(i)). I won t go into permutations per se in any detail. I ll just mention that S m is a group under the operation of composition and that there are m! permutations in S m. I think it is the granddaddy of all groups; the first one to be considered. But I could be wrong. We come now to the second lemma. Here is how Galois formulates it. Lemma 2. Given an arbitrary equation which has no equal roots, of which the roots are a, b, c,..., one can always form a function V of the roots such that none of the values that are obtained by permuting the roots in this function in all possible ways will be equal. The only proof he provides is: For example one can take V = Aa + Bb + Cc +, A, B, C,... being suitably chosen whole numbers. I will phrase the lemma in a slightly different form (using subindices for extra precision), going directly to the one can take part. It should perhaps also be mentioned that what Galois means by a function of the roots is, most likely, a polynomial function of the roots with coefficients in K. Lemma II. Assume a m x m + a 1 x m a 1 x + a 0 = 0, a m 0 is an equation with m distinct roots r 1,..., r m. There exist integers A 1,..., A m such that setting V = V (r 1,..., r m ) = A 1 r 1 + A 2 r A m r m, then no two of the values obtained by permuting the roots in all possible ways will be equal. In an even more precise way, there exist integers A 1,..., A m such that A 1 r σ(1) + A 2 r σ(2) + + A m r σ(m) A 1 r τ(1) + A 2 r τ(2) + + A m r τ(m) if σ, τ S m, σ τ. The following notation might be convenient.if and if σ S m, then I ll define V σ by V = A 1 r 1 + A 2 r A m r m, V σ = A 1 r σ(1) + A 2 r σ(2) + + A m r σ(m). Galois does not provide a proof. The proof I will give is from the article by Harold Edwards [2]. It is a constructive proof, but in practice a method of trial and error might work just as well, assuming the degree of our main equation is not too large. And assuming we know the roots. Before continuing, I want to do a couple of examples that I hope will clear this up, and that will be useful later on. It might be a good idea to try to follow all the examples. The V appearing there will be used in further examples. E1. Let s begin with something fairly simple, the equation x 3 + x 2 + x + 1 = 0. The field is Q, and the roots are i, 1, i. Almost any choice of A 1, A 2, A 3 will work. Let us try A 1 = 2, A 2 = 1, A 3 = 1. Then V = 2i + 1 i = 1 + i. Here are the six values one gets by permuting the roots. ( ) σ = identity =, V σ = V = 1 + i, ( ) σ =, V σ = 2i + i 1 = 1 + 3i, ( ) σ =, V σ = 2( i) i = 1 i, ( ) σ =, V σ = 2( 1) i + ( i) = 2 2i, 5. σ = ( ), V σ = 2( 1) ( i) + i = 2 + 2i,

7 5 THE SECOND LEMMA 7 6. σ = ( ), V σ = 2( i) i + ( 1) = 1 3i, E2. Lets look now at x 3 2 = 0. If a cubic equation reduces it must have a linear factor, hence a root. Clearly, the equation has no rational roots, so it is irreducible. With ω = e 2πi/3 = i, its roots are 3 2, 3 2 ω, 3 2 ω 2. Because 1, ω, ω 2 are independent over Q, it suffices to choose A 1, A 2, A 3 distinct to get that A 1 r 1 + A 2 r 2 + A 3 r 3 A 1 r 1 + A 2 r 2 + A 3 r 3 if r 1, r 2, r 3 and r 1, r 2, r 3 are two different permutations of the roots of the equation. I ll select A 1 = 1, A 2 = 2, A 3 = 0. Then V = i. E3. Consider x 4 + x 3 + x 2 + x + 1 = 0. This is the type of equation Galois writes as x5 1 x 1 = 0. If r = e2πi/5, its roots are r, r 2, r 3, r 4 and because the polynomial x 4 + x 3 + x 2 + x + 1 = 0 is irreducible over Q it is easy to see that any choice of integers A 1, A 2, A 3, A 4 will work, as long as they are distinct. A not too bad choice of coefficients is A 1 = 1, A 2 = 2, A 3 = 2, A 4 = 1. Then ( V = r + 2r 2 2r 3 r 4 = 2 sin 2π sin π ) i. 5 These examples will be taken up again later on. Now it is time to prove the Lemma. We begin proving a simple result that will be needed in the proof. Auxiliary Lemma 1. Let p(x 1,..., X n ) be a non-zero polynomial in n variables with coefficients in some field. There exist integers A 1,..., A n such that p(a 1,..., A n ) 0. Proof. (of the Auxiliary Lemma 1) The proof is by induction on n. If n = 1, the result is immediate; non-zero polynomials in one variable have a finite number of zeros; the integers constitute an infinite set. Assume the result proved for polynomials of n 1 variables, n 1 1. Let p(x 1,..., X n ) be a non-zero polynomial in n variables. We can then write p in the form p(x 1,..., X n ) = q 0 (X 1,..., X ) + q 1 (X 1,..., X )X n + + q N (X 1,..., X )X N n where q 0,..., q N are polynomials in the n 1 variables X 1,..., X. This is a case sensitive document, there is no reason to assume any relation between N and n. At least one of the q i s must be a non-zero polynomial; that is, there is i such that q i is a non-zero polynomial. By the induction hypothesis, there exist integers A 1,..., A such that q i (A 1,..., A ) 0. But then p(a 1,..., A, X n ) = q 0 (A 1,..., A ) + q 1 (A 1,..., A )X n + + q N (A 1,..., A )X N n is a non-zero polynomial in one variable, thus there is A n such that p(a 1,..., A n ) 0. And now we can prove Lemma II. Let r 1,..., r m be the roots of our equation and for σ S m consider the degree one polynomial q σ (X 1,..., X m ) = r σ(1) X r σ(m) X m. That is, we start with the polynomial q(x 1,..., X m ) = r 1 X r m X m, and consider all polynomials that come from permuting the coefficients (the roots). Because the roots are all distinct, no two of these polynomials are equal, thus the difference of any two of them will be a non-zero polynomial (of degree one). Now we form a polynomial of a possibly very high degree as follows. We have m! of these first degree polynomials, this gives m!(m! 1) differences q σ q τ with σ τ. Multiply all these together to get a polynomial Q(X 1,..., X m ) of degree m!(m! 1). Because each of its factors was a non-zero polynomial, Q is a non-zero polynomial. Since Q is not the zero polynomial, by the Auxiliary Lemma 1, there is a choice of integers A 1,..., A m such that Q(A 1,..., A m ) 0, and that is equivalent to q σ (A 1,..., A m ) q τ (A 1,..., A m ) if σ τ.

8 6 SYMMETRIC POLYNOMIALS 8 But we should not say goodby yet to the polynomial Q we just constructed. It has an additional property that is important. The coefficients of this polynomial Q will be polynomials in the roots r 1,..., r n, and more importantly, they will be symmetric polynomials in the roots. We ll expand on this in the next section but, in case you decide to skip it, the main result applied to our (i.e., Galois ) situation is that because the coefficients of Q are symmetric polynomials in the roots of (2), they can be rewritten as polynomials in the coefficients of this equation; in other words they are in K even if the roots of (2) are not. That is Q K[X] and since Q(V ) = 0 (because q id (V ) = 0), we have the important result that V is the zero of a polynomial with coefficients in K. 6 Symmetric Polynomials In this section we have to consider polynomials in m variables, with coefficients in the field K. The simplest polynomials in m-variables are expressions of the form a X k 1 1 Xk 2 2 Xkn n, where a K and k 1,..., k n are non-negative integers. Such expressions are called monomial; the degree of the shown monomial is k = k k n. Lets call this, just to invent a convenient name, a (k 1,..., k n )-monomial, and lets call (k 1,..., k n ) the type of the monomial. For example, if n = 3, then 5X 1 X2 2 X 3 is a (1, 2, 1)-monomial, 7X3 4 is a (0, 0, 4)- monomial. A general polynomial in m variables is any linear combination of monomials using coefficients in K. The standard notation for the set of all polynomials in mn variables with coefficients in K is K[X 1,..., X n ]. (You may have noticed that round parentheses are for fields, square brackets for polynomials.) The role of the variables X 1,..., X n is just as placeholders. We add, subtract, and multiply polynomials in n-variables in the obvious way, finally collecting terms. For example, if n = 2, (5 3X 1 + 7X 2 1 X 2 + X 1 X 2 2 )(1 + X 1 X 2 + X 2 1 ) = 5 + 5X 1 X 2 + 5X 2 1 3X 1 3X 2 1 X 2 3X X 2 1 X 2 + 7X 3 1 X X 3 1 X 2 + X 1 X X 2 1 X X 3 1 X 2 2 = 5 3X 1 + 5X 1 X 2 + 5X X 2 1 X 2 + X 1 X 2 2 3X X 3 1 X 2 + 8X 3 1 X X 2 1 X 3 2 Two polynomials are equal if and only if all corresponding monomials have the same coefficients. That is, any polynomial can be written as a finite sum of terms, each term being a monomial, and any two monomials appearing in the sum having different types. Then two polynomials are equal if and only if the coefficients of all the corresponding monomials are the same. Maybe its easier to just to say that they are the same if they are the same. The degree of a polynomial is the maximum of the degrees of its constituting monomials. Special notation has been developed to be able to handle these polynomials in a compact, efficient manner. For example, if n = 3, a general polynomial in 3 variables of degree 4 looks like a a 100 X 1 + a 010 X 2 + a 001 X 3 + a 200 X a 020 X a 002 X a 110 X 1 X 2 + a 101 X 1 X 3 + a 011 X 2 X 3 + a 300 X 3 1 +a 030 X a 003 X a 210 X 2 1 X 2 + a 201 X 2 1 X 3 + a 120 X 1 X a 021 X 2 2 X 3 + a 102 X 1 X a 012 X 2 X a 111 X 1 X 2 X 3 +a 400 X a 040 X a 004 X a 310 X 3 1 X 2 + a 301 X 3 1 X 3 + a 130 X 1 3X a 031 X 3 2 X 3 + a 103 X 1 X a 013 X 2 X 3 3 +a 220 X 2 1 X a 202 X 2 1 X a 022 X 2 2 X a 211 X 2 1 X 2 X 3 + a 121 X 1 X 2 2 X 3 + a 112 X 1 X 2 X 2 3. One needs a better way of expressing these objects, and there is a better way, but we ll simply try not to write any of these polynomials out in full again. In the theory of equations, an important set of polynomials in K[X 1,..., X m ] is the set of symmetric polynomials. These are polynomials left invariant by permutations of the variables. The formal definition is: p(x 1,..., X m ) is a symmetric polynomial iff for every σ S m we have p(x 1,..., X m ) = p(x σ(1),... X σ(m) ). In other words, the variables have to appear in a symmetric way. Constant polynomials are trivially symmetric. It is easy to see that the only symmetric polynomials of degree 1 are polynomials of the form a + b(x X m ),

9 7 THE THIRD LEMMA 9 where a, b K. But then things get more complicated. However, there are certain symmetric polynomials that form a sort of basis for all of them, namely the following polynomials: s 0 (X 1,..., X m ) = 1, s 1 (X 1,..., X m ) = X X m, s 2 (X 1,..., X m ) = X 1 X 2 + X 1 X X 1 X m + X 2 X X m 1 X m,. s m (X 1,..., X m ) = X 1 X 2 X m. That is s k (X 1,..., X m ) is the sum of all possible products of k of the variables. One has: Lemma 2 Let f(x 1,..., X m ) K[X 1,..., X m ] be invariant under the action of S m, meaning that f(x σ (1),..., X σ(m) ) = f(x 1,..., X m ) for all σ S m. Then there exists a polynomial h K[X 1,..., X m ] such that f(x 1,..., X m ) = h(s 0 (X 1,..., X m ), s 1 (X 1,..., X m ),..., s m (X 1,..., X m )). In particular, if r 1,..., r m are the roots of the main equation, then f(r 1,..., r m ) K. I omit the proof. 7 The third lemma Here one sees the raison d être of the V of the preceding Lemma. At least, one begins to see it. This time I won t modify Galois version. Lemma III. The function V being chosen as is indicated in the preceding article, it will enjoy the property that all the roots of the proposed equation will be rationally expressible as a function of V. In lengthier sentences, if our main equation has roots r 1,..., r m and r i r j if i j; if V is as in Lemma II, then all the roots of the equation are of the form φ(v ), where φ is a polynomial function with coefficients in K. We had V = A 1 r 1 + A 2 r A m r m. We will show that r 1 is a polynomial in V with coefficients in K; the same argument works for the remaining roots. For the proof, and also because it is needed later on, it will be convenient to analyze a bit more the polynomial Q(X) of the second lemma. To recall the assumptions, we assume that our main equation has m distinct roots and we formed V = A 1 r 1 + +A m r m, where A 1,..., A m Z are selected so that any permutation of the roots in the expression for V gives a different value. We defined Q(X) as the product of all degree one polynomials of the form x (A 1 r σ(1) + + A m r σ(m) ) as σ ranges over all permutations of {1,..., m}. Let us recall that Q(X) K[X]. Let G be the set of all permutations of 1,..., m leaving 1 fixed and form the product of all expressions of the form x (A 1 r 1 + A 2 r σ(2) + + A m r σ(m) ); in other words the product of all factors of Q(X) in which the root r 1 stays in the first place. Using modern symbols, we form ( X (A1 r 1 + A 2 r σ(2) + + A m r σ(m) ) ). σ G This is a polynomial of degree (m 1)! in x (because G S m 1 has m 1 elements). Its dependence on r 2,..., r m is symmetric. A permutation of the roots r 2,..., r n is the same as a reordering of the factors, thus the end product is the same. By the auxiliary lemma of this section, a symmetric polynomial in r 2,..., r m is a polynomial with coefficients in K of a; that is we can write ( X (A1 r 1 + A 2 r σ(2) + + A m r σ(m) ) ) = F (X, r 1 ), σ G where F (X, Y ) is a polynomial in two variables x, y with coefficients in K. Notice that because the identity permutation is (of course) in G, one of the factors of the product above is x (A 1 r 1 + A m r m ) = x V ; thus this product is 0 if we set x = V.

10 8 THE FOURTH LEMMA 10 Just to make sure we are on the same page, suppose m = 3, the main equation is a 3 x 3 + a 2 x 2 + a 1 x + a 0 = 0, the roots are r 1, r 2, r 3, and V = A 1 r 1 + A 2 r 2 + A 3 r 3 = q(r 1, r 2, r 3 ). The permutations leaving 1 fixed are ( ) ( ) id = identity =, and σ = Thus in this case (X A 1 r 1 A 2 r 2 A 3 r 3 )(X A 1 r 1 A 2 r 3 A 3 r 2 ) = x 2 (2A 1 r 1 + (A 2 + A 3 )(r 2 + r 3 ))x + A 2 1r A 1 (A 2 + A 3 )r 1 (r 2 + r 3 ) + (A A 2 3)(r 2 r 3 ) + A 2 A 3 (r r 2 3). Now r 1 + r 2 + r 3 = a 1, so that r 2 + r 3 = a 1 r 1 ; and r 1 r 2 + r 1 r 3 + r 2 r 3 = a 2, thus r 2 r 3 = a 2 r 1 (r 2 + r 3 ) = a 2 + r 1 (a 1 + r 1 ) = a 2 + a 1 r 1 + r 2 1, r 2 + r 2 3 = (r 2 + r 3 ) 2 2r 2 r 3 = a 2 1 2a 2 4a 1 r 1 r 2 1. Returning to our argument, for the reasons mentioned above, we will have F (V, r 1 ) = 0. At the same time, we will have F (X, r k ) 0 for k > 1. This is because, for example, F (X, r 2 ) = ( X (A1 r 2 + A 2 r σ(1) + A 3 r σ(3) + + A m r σ(m) ) ) σ G 2 where G 2 is the set (subgroup) of all permutations in S m leaving 2 fixed, and F (X, Y ) is the same polynomial in the variables x, y as before. That is because in this new product, the roots r 1, r 3,..., r m appear in exactly the same places as the roots r 2, r 3,..., r m did; thus the expression of each symmetric appearance of these roots as a polynomial in r 2 will be identical to the expression of the corresponding appearance of the roots r 2,..., r m as a polynomial in r 1. Because of the choice of the integers A 1,..., A m, V A 1 r 2 + A 2 r σ(1) + A 3 r σ(3) + + A m r σ(m) for all σ G 2, thus F (V, r 2 ) 0. Similarly, F (V, r k ) 0 for all k > 1. We also notice here, for future reference that because we can organize the factors of Q(X) into m groups, the k-th group being all those that have the root r k multiplying A 1, we have that Q(X) = F (X, r 1 )F (X, r 2 ) F (X, r m ). (4) Let us turn the tables a bit, set x = V, and consider the polynomial f(y ) = F (V, Y ) (as a polynomial in y). Its coefficients are in the smallest field containing K and V ; that is, in K(V ). As mentioned, f(r 1 ) = F (V, r 1 ) = 0. If we denote by p(y ) the polynomial corresponding to our main equation, just for a moment writing the variable as y; that is, p(y ) = a m y m + + a 0, then we also have, of course, p(r 1 ) = 0. But p, f cannot have any other common zeros. In fact, the other zeros of p are r 2,..., r m and, as we saw, these are not 0 s of f(y ) = F (V, Y ). Let q be the greatest common divisor of f and p; that is q is a polynomial of highest degree that divides both f and p. Such a polynomial exists, with coefficients in K(V ) by the Euclidean algorithm. But any zero of this polynomial would be a zero of both f and p, so that q is a polynomial having r 1 and only r 1 as a zero. But a polynomial with only one (non-repeated) zero has to have degree 1, so that q(x) = ax + b, where a, b K(V ), a 0. Then 0 = ar 1 + b implies r 1 = b/a K(V ). This shows r 1 K(V ); the proof for the other roots is identical. 8 The fourth lemma Things are beginning to take shape. We continue assuming (2) has m distinct roots; form V as before. We saw at the end of the section on the second lemma that V was a zero of a polynomial Q(X) with coefficients in K (with g-rational coefficients). There is, of course, no reason why this polynomial of degree m! should be irreducible over K. Nor why it shouldn t. However, Q decomposes into the product of polynomial irreducible over K, one of which has V as a zero. That is, we can write Q(X) = Q 0 (X)Q 1 (X),

11 9 PROPOSITION 1, AND THE GALOIS GROUP OF THE EQUATION. 11 where Q 0, Q 1,... are irreducible polynomials of K[X] and Q 0 (V ) = 0. If Q was irreducible, then Q = Q 0 and there are no other polynomials. Since V was a simple 0 of Q(X) (if you don t see why, review how V was defined and Q was constructed), if there are any further polynomials Q 1, Q 2, etc., V is not a zero of any of them. More importantly perhaps, the construction of V and Q shows that all zeros of Q are simple thus the same is true of Q 0. Here is Galois fourth lemma. Lemma IV. Let us suppose that the equation in V has been formed and that one of its irreducible factors has been taken so that V is the root of an irreducible equation. Let V, V, V,... be the roots of this irreducible equation. If a = f(v ) is one of the roots of the proposed equation, f(v ) will also be one of the roots of the proposed equation. In our notation, let n be the degree of the irreducible polynomial Q 0 (X) (Galois also uses the symbol n for its degree) and (using the same notation as Galois), let V, V,..., V be the roots of Q 0. As was proved in Lemma III, all the roots of (2) are of the form φ(v ), where φ(x) is a polynomial with coefficients in K. Lemma IV states that if φ(v ) is a root of (2), then so are φ(v ),..., φ(v ). No claim is made that for a given fixed φ all the roots of (2) obtained this way are distinct; in fact, since it is quite possible that n > m, there definitely can be repetitions. The proof goes as follows. Because V, V,... will be roots of Q(X) = 0, V must be of the form A 1 r σ(1) + A m r σ(m) for some permutation σ S m. If k = σ(1), then V also satisfies F (V, r k ) = 0 (see formula (4)) and the same argument showing that r 1 = φ(v ) now shows r k = φ(v ). Notice that in this argument, k = 1 is quite possible. Until further notice we will keep the assumptions that our main equation has m distinct roots r 1,..., r m, V is defined as above; let us denote by Q 0 (X) the irreducible factor of Q(X) with V as a zero, and let V,..., V be the other zeros of Q 0 (X). The degree of Q 0 (X) is denoted (thus) by n. Looking at things for a moment from a more modern point of view, recalling the construction of a field obtained from a field K by adjoining the zero of an irreducible polynomial (see 1.4), the fact that Q 0 has degree n means that every element of K(V ) can be expressed uniquely in the form b 0 +b 1 V +...+b V, where b 0,..., b K. Remark. For later reference let us notice that every one of the permutations V σ = m A lr σ(l) is then the root of an irreducible polynomial of degree n. In fact, because each one of these permutations is a combination of roots of the main equation, it is in K(V ). This implies that it is the root of an irreducible polynomial of degree n. But we can reverse the roles of V and V σ and conclude that the irreducible polynomial of V has to have a degree n the degree of the polynomial of V σ. At this point, Galois is done with Lemmas. The serious stuff begins. 9 Proposition 1, and the Galois group of the equation. Let us recall that all roots of our main equation are polynomials (that can be assumed of degree n 1) in V. Galois introduces the notation r 1 = φv, r 2 = φ 1 V,..., r m = φ m V for this purpose. In other words, φ(x), φ 1 (X,..., φ m 1 (X) are m polynomials with coefficients in K, which we may assume to be all of degree n 1, such that r 1 = φ(v ), r 2 = φ 1 (V ),..., r m = φ m 1 (V ). By Lemma IV, if we replace V by V (or V, or... ), we again obtain roots of the equation. Because V, V play symmetric roles, both are roots of the irreducible polynomial of degree n, this new list of m-roots can t contain repetitions; because all the polynomials φ j are of degree n 1, φ j (V ) = φ l (V ), will imply j = l. In other words, the list φ(v ), φ 1 (V ),..., φ m 1 (V ) is a permutation of the roots of the main equation. Galois omits the parentheses, and for a while I ll follow his example and write, for example, φv instead of φ(v ) The set of all the permutations obtained this way forms a group and is what we nowadays call the Galois group. Galois used the word group in his writing, but probably not in a technical sense; a group of permutations where just a bunch of permutations. Of course, as we ll see below, at some point he notes that if S and T (his notation) are in this group, so is ST. I will state Proposition I in Galois words; he also calls it a theorem (that is, the first proposition is not a lemma; it is a theorem). Theorem. Let an equation be given of which the m roots are a, b, c,.... There will always be a group of permutations of the letters a, b, c,... that will enjoy the following property: 1. That every function of the roots invariant under the substitutions of this group will be rationally known; 2. Conversely, that every function of the roots that is rationally determinable will be invariant under the substitutions.

12 9 PROPOSITION 1, AND THE GALOIS GROUP OF THE EQUATION. 12 I will try to clarify exactly what Galois means. More precisely, I will present Harold Edwards interpretation and (in a bit more detail) his proof (see [2]). Galois has some difficulties in deciding whether to use permutation or substitution; in his manuscript; the word permutation frequently appears crossed out and replaced by substitution. He considers a permutation as being a done deal. Thus if we have the letters a, b, c, a permutation is these letters arranged in a different way, say as a, c, b. The passage from a, b, c to a, c, b he calls a substitution. As he puts it, Substitutions are the passage from one permutation to another. In other words, exactly what we nowadays call permutations. As I mentioned before, it seems likely that his use of the word group is completely non-technical; it just so happens that the permutations he has in mind form a group. Here is the way Galois presents the group of permutations, in table form. The first column simply indicates what zero of Q 0 is being used; the next columns list the roots. Now I am essentially writing what Galois wrote; as translated to English in the edition of Galois complete works [3]. What Galois calls proposed equation is what I have been calling the main equation. Let φv φ 1 V φ 2 V..... φ m 1 V be the roots of the proposed equation. Let us write down the following permutations of the roots: (V ) φv φ 1 V φ 2 V..... φ m 1 V (V ) φv φ 1 V φ 2 V..... φ m 1 V (V ) φv φ 1 V φ 2 V..... φ m 1 V (V () ) φv () φ 1 V () φ 2 V ()..... φ m 1 V () I say that this group of permutations enjoys the specified property. So there you have it; the Galois group of the equation. Galois writes a function of the variable V (or V, or... ) without using parentheses; his φv is our φ(v ). The (V ), (V ), etc. on the left label the rows. From our (i.e., [2] s) version of Proposition 1 it will follow that we do have a group of permutations that can be interpreted as a subgroup of S m, and that this group is uniquely defined, depending only on the equation and the field K, not on the choice of V. But before getting to this I want to look at some examples. Examples. I want to look again at the three examples of 5 (the second lemma section), plus two additional examples. For me these examples were fairly crucial in understanding what was going on. E1. In this example the equation was x 3 + x 2 + x + 1 = 0. The field is Q, and the roots are i, 1, i. We selected (out of an infinity of possible selections) V = 1 + i. The irreducible equation over Q satisfied by V (unique up to multiplication by a non-zero constant factor) is x 2 2x + 2 = 0; its roots are V and V = 1 i. Now i = V 1, 1 = 1 and i = V + 1 so that the function φ, φ 1, φ 2 are φ(x) = x 1, φ 1 (X) = 1, φ 2 (X) = x + 1. Since φ(v ) = i, φ 1 (V ) = 1, φ 2 (V ) = i, Galois array looks like i 1 i i 1 i so that the Galois group is made up of two permutations; the identity and the transposition switching i and i, leaving 1 fixed. ( As a subgroup ) of S 3 it can be identified with G = {e, σ} where e is the identity and σ is the transposition σ = E2. In this example we had x 3 2 = 0, with roots 3 2, ω 3 2, ω 2 3 2, where ω = ( 1 + i 3)/2. We selected V = 3 2 3i. Then V is a root of x = 0. Showing that this polynomial is irreducible over Q is easy or hard, depending on how much algebra you have had. This is for people who have had some field theory: One can see that it splits into (X 3 6i 3)(X 3 + 6i 3) in Q( 3) and V is a root of x 3 + 6i 3 = 0. Because x 3 + 6i 3 is irreducible over Q( 3) (otherwise it would have a root in Q( 3) and it is easy to see that the existence of such a root would imply 3 2 Q),

13 9 PROPOSITION 1, AND THE GALOIS GROUP OF THE EQUATION. 13 one sees that the degree of Q( 3, V ) over Q( 3) is 3; thus the degree of Q( 3, V ) over Q is 6. But Q( 3, V ) = Q(V ) (for example, 3 = V 3 /6), so Q(V ) is of degree 6 over Q, and the irreducible polynomial of V over Q must be of degree 6. For people who did not have any field theory: Take my word that x is irreducible over Q. The roots are given by V, V = V = 3 2 ( 3i, V = 3 ) ( ) i 3 2, V = V 3 3 = 2 2 i 3 2, V (4) = V = ( 3 ) i 3 2, V (5) = V (4) = ( ) i 3 2 Now: 3 2 = 1 18 V 4, 3 2 ω = 1 2 V 1 36 V 4, 3 2 ω 2 = 1 2 V 1 36 V 4. Thus, in this case φ(x) = 1 18 x4, φ 1 (X) = 1 2 x 1 36 x4, φ 2 (X) = 1 2 x 1 36 x4. So the table now looks like (V ) φ(v ) = 3 2 φ 1 (V ) = 3 2 ω φ 2 (V ) = 3 2 ω 2 (V ) φ(v ) = 3 2 φ 1 (V ) = 3 2 ω 2 φ 2 (V ) = 3 2 ω (V ) φ(v ) = 3 2 ω 2 φ 1 (V ) = 3 2 ω φ 2 (V ) = 3 2 (V ) φ(v ) = 3 2 ω 2 φ 1 (V ) = 3 2 φ 2 (V ) = 3 2 ω (V (4) ) φ(v (4) ) = 3 2 ω φ 1 (V (4) ) = 3 2 ω 2 φ 2 (V (4) ) = 3 2 (V (5) ) φ(v (5) ) = 3 2 ω φ 1 (V (5) ) = 3 2 φ 2 (V (5) ) = 3 2 ω 2 As we can see, all permutations occur. The Galois group is (isomorphic to) S 3. E3 We consider the equation x 4 + x 3 + x 2 + x + 1 = 0, with roots r, r 2, r 3, r 4, where r = e 2πi/5 = cos(2π/5) + i sin(2π/5). We had taken ( V = r + 2r 2 2r 3 r 4 = 2 sin 2π sin π ) i. 5 Verifying the computations that follow could be a good exercise in trigonometry. The irreducible equation satisfied by V is Q 0 (X) = x x = 0. The other roots of this equation can be seen to be: V = V, and with some difficulty, V = 2 ( 2 sin 2π 5 sin p i 5) i, V = V. With a bit of effort one determines the polynomials φ, φ 1, φ 2, φ 3. They are Here is how the table of permutations looks like. φ(x) = 1 25 x x2 1 2 x 3, φ(v ) = r, (5) 2 φ 1 (X) = 1 50 x x x + 1, φ 1(V ) = r 2, (6) φ 2 (X) = 1 50 x x2 1 2 x + 1, φ 2(V ) = r 3, (7) φ 3 (X) = 1 25 x x x 3 2, φ(v ) = r4. (8) (V ) φ(v ) = r φ 1 (V ) = r 2 φ 2 (V ) = r 3 φ 3 (V ) = r 4 (V ) φ(v ) = r 4 φ 1 (V ) = r 3 φ 2 (V ) = r 2 φ 3 (V ) = r (V ) φ(v ) = r 3 φ 1 (V ) = r φ 2 (V ) = r 4 φ 3 (V ) = r 2 (V ) φ(v ) = r 2 φ 1 (V ) = r 4 φ 2 (V ) = r φ 3 (V ) = r 3

14 9 PROPOSITION 1, AND THE GALOIS GROUP OF THE EQUATION. 14 From a permutation point of view, identifying the permutation taking r, r 2, r 3, r 4 with the corresponding permutation of the exponents, the group consists of the identity permutation e and the permutations ( ) ( ) ( ) σ =, τ =, and µ = We see that τ 2 = σ, τ 3 = µ, τ 4 = e, so that G is the cyclic group {e, τ, τ 2, τ 3 }. Later on I will be proving that the group of any cyclotomic polynomial of degree prime minus one; i.e., every polynomial of the form X p X + 1, where p is a prime, is a cyclic group of order p 1. E4. I want to look at x 3 2 = 0 again, but now in the field K = Q( 3 2). The polynomial x 3 2 factors, x 3 2 = (X 3 2)(X x + 3 4). (Notice that 3 4 = ( 3 2) 2 K). The roots are the same, we can use the same V = i But this time V 2 = K = Q( 3 2), so that the irreducible polynomial satisfied by V is x The second zero of this polynomial is V = V = i This time, to get the roots, we have to replace φ, φ 1, φ 2 by ψ(x) = 3 2 (constant polynomial), ψ 1 (X) = 1 2 V 1 3 2, then ψ1 (V ) = ω 3 2, 2 ψ 2 (X) = 1 2 V 1 3 2, then ψ2 (V ) = ω where ω = e 2πi/3 = ( i). The Galois style permutation table is now much reduced, it consists of (V ) ψ(v ) = 3 2 ψ 1 (V ) = 3 2 ω ψ 2 (V ) = 3 2 ω 2 (V ) ψ(v ) = 3 2 ψ 1 (V ) = 3 2 ω 2 ψ 2 (V ) = 3 2 ω The Galois group consists of two elements; the identity and the transposition 2 3, 3 2. E5. I want to look once again at x 3 2 = 0, but this time after adjoining 3 i to the rationals; that is, in Q( 3 i). The element V is the same as before, but now the irreducible equation it satisfies is R 0 (X) = X i = 0. ( ) The other roots of R 0 (X) = 0, using the notation of example E2, are V (3) = i 3 ( ) 2, and V (4) = i 3 2. Either by direct computation, or by taking the remainders of dividing the polynomials φ, φ 1, φ 2 of E2 by R 0 (X), we see that the corresponding polynomials when our base field is Q( 3 i) are ψ(x) = 1 ( 1 3 ix, ψ1 (x) = ) ( 3 i x, ψ 2 (x) = ) 3 i x. 6 The Galois style table of permutations works out to: (V ) ψ(v ) = 3 2 ψ 1 (V ) = 3 2 ω ψ 2 (V ) = 3 2 ω 2 (V (3) ) ψ(v (3) ) = 3 2 ω 2 ψ 1 (V (3) ) = 3 2 ψ 2 (V (3) ) = 3 2 ω (V (4) ) ψ(v (4) ) = 3 2 ω ψ 1 (V (4) ) = 3 2 ω 2 ψ 2 (V (4) ) = 3 2 If we set r 1 = 3 2, r 2 = 3 2 ω, and r 3 = 3 2 ω 2, and identify a permutation of the roots with a permutation of the indices of the roots, then the group is now the cyclic group G = {e, τ, τ 2 }, where e is the identity permutation, and ( ) τ = ; notice that τ 3 = e.

15 9 PROPOSITION 1, AND THE GALOIS GROUP OF THE EQUATION. 15 It is time to prove the first proposition, or the modernized version in [2], which reads Proposition I, Version 2. Let G be the set of permutations of the roots of the main equation described by the table above. Then G has the following properties: 1. If F is a polynomial in m variables with coefficients in K and F (r 1,..., r m ) = c K, then F applied to any permutation in G of the roots r 1,..., r m is also equal to c. 2. Conversely, if F is a polynomial in m variables with coefficients in K that gives the same value to any permutation by an element of G of the roots of the main equation, then F (r 1,..., r m ) K. Let me repeat this proposition with a bit more notation. I want to identify G with a subset of S m. For this purpose, I ll denote by τ k the element of S m such that the permutation of the roots given by the k-th row of Galois table can be described as r τk (1),..., r τk (m) (with τ 0 = identity). Then we can state the proposition as follows: Proposition I, Version 2. Let G = {τ 0,..., τ }. Then 1. If F K[X 1,..., X m ] and F (r 1,..., r m ) K, then F (r τk (1),..., r τk (m)) = F (r 1,..., r m ) for k = 0, 1,..., m Conversely, if F (r τk (1),..., r τk (m)) = F (r 1,..., r m ) for k = 0, 1,..., n 1, then F (r 1,..., r m ) K. Proof of the proposition, version 2 (or 2 ) It will be convenient to write V k for what Galois calls V k ; the roots of Q 0 (X). That is, V 0 = V, V 1 = V, etc. Otherwise we might get too confused when having to take powers of these quantities. If F K[X 1,..., X n ], consider F (φ(x), φ 1 (X),..., φ m 1 (X)). This is a polynomial in one variable with coefficients in K; by the division algorithm F (φ(x), φ 1 (X),..., φ m 1 (X)) = Q 0 (X)q(X) + R(X) (9) where the degree of R is n 1, and q, R K[X]. Since V, V,..., V = V 0, V 1,..., V are roots of Q 0 we see that F (X τk (1),..., X τk (m)) = F (φ(v k ), φ 1 (V k ),..., φ m 1 (V k )) = r(v k ). If F (r 1,..., r m ) K, then R(V ) K. But the only polynomial of degree n 1 of V that can be in K is a constant polynomial; that is R(X) = b K. But then R(V k ) = b for all k. This proves the first part. For the second part, assume F K[X 1,..., X m ] and F (r τk (1),..., r τk (m)) = F (r 1,..., r m ) for k = 0, 1,..., n 1. With R as in 9, we then have R(V k ) = R(V ) for k = 1,..., n 1. Now R(X) = c 0 + c 1 x + c x ; I claim c 1 = = c = 0. In fact, the equations R(V k ) = R(V ) for k = 1,..., n 1 can be written in the form c 1 (V 1 V ) + c 2 (V1 2 V 2 ) + + c (V1 V ) = 0 c 1 (V 2 V ) + c 2 (V2 2 V 2 ) + + c (V2 V ) = 0. c 1 (V V ) + c 2 (V 2 V 2 ) + + c (V ) = 0 We can think of this as a system of () homogeneous linear equations in the unknowns c 1,..., c. The determinant of the system is V 1 V V1 2 V 2 V1 V 1 V V 2 V V 2 V V2 2 V 2 V2 V 1 V 1 V1 2 V 1 det. = det 1 V 2 V2 2 V 2 0 V V V 2 V 2 V V. 1 V V 2 V because the last determinant is a Vandermonde determinant; it equals ± the product of all differences V j V k for j k, thus is different from 0 since all these differences are non-zero. It follows that the only solution of the homogeneous system in question is the zero solution; i.e., c 1 = = c as claimed. It follows that R(X) = c 0 K, hence F (r 1,..., r m ) = c 0 K.