If S is a polynomial of degree n having roots al, This is the first time that roots, including impossible ones, appear within formulas exactly like ordinary numbers. Excerpt from the book Invention nouvelle We conclude by identifying the homogeneous components of this last equality. Let F and G be two polynomials in K[X], of degree m and n respectively. We will write Resx F, G whenever it is necessary to distinguish the variable with respect to which the resultant is taken. Symmetric Polynomials 2 The resultant of two polynomials in K[X] is zero if and only if the two polynomials have a greatest common divisor which is neither a constant nor zero in K [X].

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The converse is obvious. Now part 1 suffices to conclude. Formula 4 : this follows from formula 1. Historically, the resultant was introduced as a determinant see Exercise 3. This approach is very natural and actually renders the result of Proposition 3. But to compute this determinant quickly, in general, one needs to revert to the method presented above.

By Proposition 3. Real roots and sign of the discriminant Exercises for Chapter 3 39 Exercises for Chapter 3 Exercise 3. Show that there exists an elementary symmetric polynomial Sk for some 1 ; k ; n such that Sk XI, Exercise 3. The reader who quails before the rather lengthy computations involved in this exercise may employ the following shortcut: plug the results given in the solution to this exercise back into the exercise and check that they work.

By induction, this method was supposed to give solutions of polynomial equations of all degrees. But it came to be understood latS! The resultant as a determinant Let A be an integral domain. Recall how to compute the Vandermonde determinant V TI , This determinant is exactly the one which we introduced above; of course, Euler did not call it a determinant this terminology was introduced by Sylvester in , and he only wrote the explicit formula for polynomials of small degree.

Symmetric Polynomials Exercise 3. Descartes' Lemma Let n ; 1 be an integer, and let xo, Any classical algebra text contains results of Sturm generalizing this lemma. II X - Xi and l::;i::;n X - Yi are distinct since l::;i::;n they do not have the same roots; thus their coefficients are not all equal. The result follows from formula 3. Solution to Exercise 3. Solution to Exer. This gives V Tl, Once we have factored out these quantities, we are left with a determinant which can be computed as a product of the 46 3. Symmetric Polynomials determinants of two blocks, each of which is a Vandermonde determinant.

The second formula is obtained by a simple computation. Similarly, because D X Now, if a, b, e denote the roots of X 3 from Exercise 3. These computations can also be done using the method of Euclidean division or by computing a determinant. If P does not have only real roots, then it has an even number of non-real roots, say k pairs of complex conjugates. Let us reason by regrouping the factors occurring in the formula of part 2. The rest follows immediately. Abel and Galois defined the elements of a generated extension, but they did not envision these elements as forming a set.

The concept of a field and the word did not appear until the work of Dedekind between and The abstract definition of a field was given about 20 years later by Weber and Moore. One hundred years ago, the language of linear algebra did not exist and results were formulated very differently from the way they are today, as can be seen, for example, in Weber's book, listed in the bibliography. We will usually represent field extensions in one of the forms shown in Figure 4. An extension of degree 2 is called a quadratic extension. The general study of extensions of infinite degree needs topology.

C Kr; such a sequence is called a tower of fields. The formula gives the degree of Q[ -v2, j] over Q, using the intermediate extension Q[ -v2] and Proposition 4. The np numbers limj for 1 ::; i ::; n, 1 ::; j ::; p, form a basis of Mover K. Field Extensions and for each of the Xj there exist Xlj, Then L is a finite-degree extension of K and M is a finite-degree extension of L. Thus np::; [M : KJ, which bounds nand p. This family is non-empty since L belongs to it, and the intersection of all fields in the family satisfies the defining property of K A.

An element a of L is said to be algebraic over K if there exists a non-zero polynomial in K[X] having a as a root. A complex number transcendental over '01 such a number is simply called a transcendental number is thus a number that is not a root of any polynomial with coefficients in Z. Joseph Liouville was the first to prove, in , that certain real numbers are transcendental over '01, for instance lO-n!

Carl negonly of a 56 4. For example, Alan Baker, a Fields Medalist, extended these results considerably at the end of the s. Transcendental number theory is at present a rapidly developing subject. There exists a unique monic polynomial P in K[X] having a as a root and of minimal degree among all the non-zero polynomials in K[X] having a as a root. Now, consider the set I of polynomials of K[X] which vanish at a. Dividing T by its leading coefficient, we obtain a monic polynomial P which generates I.

The uniqueness of P is a consequence of the fact that if T is a monic polynomial in K[X] satisfying the same conditions as P, then P - T is an element of I which vanishes at a, so because its degree is 4. By the preceding section, there exists a unique monic polynomial of minimal degree in K[X] which vanishes at a; it is called the minimal polynomial of a over K. The minimal polynomial of an element a of K is X-a. If neither S nor T is a constant, then deg S 1 of P, i. We discuss methods for factoring by hand. Let us first recall a classical result. Z[X], whose content, which is the greatest common divisor of its coefficients, is equal to 1, is irreducible in Q[X] if and only if it is irreducible in Z[X].

This criterion remains valid if we replace Z by any factorial ring A and the field Q by the fraction field of A, and if we assume the existence of a prime i. Then qn P pjq is an integer, but it is equal to zero, so it follows that q divides an and p divides ao, which makes it possible to obtain the set of rational candidates for roots of P. If this set does not have too many elements, we can test the candidates one by one.

Indeed, if there exists a prime number p such that the image of P in ZjpZ [X] is an irreducible polynomial of the same degree, then P is itself irreducible assuming that its content is equal to 1. We show later in Exercise The irreducibility of Pin ZjpZ [X] can be proved, for example: 1 for polynomials of degree 2 or 3, by systematically testing if the elements of ZjpZ are roots of P; 2 for polynomials of degree 4, by systematically testing if the elements of ZjpZ are roots of P and showing that a decomposition into a product of two quadratic polynomials is impossible; 4.

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Let K[a] denote the smallest subring containing K and a, i. Thus, the expressions K[a] and K a refer respectively to images of the ring of polynomials K[X] and of the field of rational functions K X. If a is algebraic over K, these two images coincide. By Proposition 4. Thus S a jT a belongs to K[a]. As deg S 3 The map f defined above has image K[a], and kernel the set of polynomials vanishing at a, i.

This gives the desired fac- torization. Every extension L of K of finite degree n is algebraic over K, and every element of L is algebraic of degree ; n over K. Proposition 3. Write K[al, It is the K-algebra generated by al, A more general statement is that the complex numbers which are algebraic over a subfield K of C form a field called the algebraic closure of K in C. Fields that are subfields of C correspond to this situation; it is the situation which Dedekind considered in , when he gave the first definition of a field.

Almost all the exercises in Chapters concern this situation. However, it is also possible to consider the following more general situation: K is an arbitrary field and P is a polynomial K[X] that has no roots in K. Then, even without knowing any extension of K beforehand see, for example, Chapter 14 , we can construct an extension L of K in which P has at least one root. This more general situation was studied by Leopold Kronecker and Henri Weber in the years to Moreover, P is the minimal polynomial of x over K.

Since i is a homomorphism of rings with unit, it is injective, so L is an extension of K. Consequently, P is the minimal polynomial of x over K. The extension L of K contains a root of p, so it contains a root of P. P has a universal property which is an immediate consequence of the universal property of quotients. Field Extensions which gives the factorization of of quotients. Toward Chapters 5 and 6 In order to prove the impossibility of certain geometric constructions that were sought for over years, it suffices to use just a part of the results of this chapter. These problems and the proofs of their impossibility form the subject of Chapter 5, which should be considered as a joyous digression.

We will return to Galois theory proper in Chapter 6. Exercises for Chapter 4 We found it natural, in a chapter concerning algebraic numbers, to present some of the famous and little taught examples of transcendental numbers. We follow the proofs given in Alan Baker's book listed in the bibliography. Exercise 4. An example ofa transcendental number: L a n n! Suppose that e is algebraic over Q, a root of the irreducible polyqk Xk in Z[X]. Show that IJ t 1 ::; ItleltlPl 1tl. Field Extensions Exercise 4. The degree of an algebraic extension Determine the degrees of the following extensions by finding bases for them, and answer the questions.

Does v3 lie in Q[j]?

## Groupes, algèbres et représentations

Does i lie in Q[j]? Does j lie in Q[i]?

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Computing in algebraic extensions Consider the polynomial P X one of its roots. Algebraic extensions 1 Let L be an extension of finite degree n of a field K. What happens if L' is another extension of K, contained in L and of degree n over K? Let b be an algebraic element over K whose degree m is relatively prime to n. Determine the degree of K[a, b] over K and show that P is irreducible over K[b].

What is the intersection K[a] n K[b]? Field Extensions 3 Let x be an algebraic element of odd degree over a field K. Show that deg P divides n. Give an example of an infinite degree extension of Co 7 Let K be a field contained in C, and let L be an algebraic but not necessarily finite extension of K. Let a be an element of C algebraic over L. Show that a is an algebraic element over K. We see that 5 is a root, divide by it and complete the factorization.

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Solution to Exercise 4. It follows that p is irreducible. This classical result is generalizable to all cyclotomic polynomials see Chapter 9. As 3 is irreducible in the factorial ring Z[i], it is prime in Z[i]. Every other choice of leads to the same result. Consider the situation mod 2. Solutions to Some of the Exercises 71 Solution to Exercise 4.

As the minimal polynomial of. The tower rule gives [1Q[. Finally, [1Q[. Consequently, [1Q[. A basis of IQ[. This last remark leads to the equality Q[. We could have proved this equality directly by showing two easy converse inclusions. In what follows, note that two quadratic extensions Q[a] and Q[b] are equal whenever a E Q[b]. Field Extensions Later Exercise 8. Alternatively, we can compute using a table of the an expressed in terms of basis vectors Table 4. Solutions to Some of the Exercises 75 3 As u is not in Q, it is of degree 3 over Q, and we obtain its minimal polynomial over Q by computing the powers of u in the basis 1, a, a 2 , using Table 4.

As x is a root of the polynomial X2 - x 2 with coefficients in K[x 2], x is algebraic of degree at most 2 over K[X2]. The minimal polynomial of x over K gives an expression for x in terms of x 2 by regrouping the terms of even order and the terms of odd order.

The field C X of rational functions with complex coefficients is an example of an extension of C that is not algebraic, and has infinite degree. This solves the problem. To find a polynomial that vanishes at these numbers, we can a look for a relation of linear dependence between the k-th powers of the numbers for 0 ; k ; n, in a suitable extension of Q; b use resultants as described above; c proceed more directly if the expression of the number allows this.

There are several ways to prove that it is of degree 6.

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The corresponding polynomial is the minimal polynomial of z over Q, which is irreducible by Eisenstein's criterion. This is the minimal polynomial of t over Q, by a reasoning analogous to the one used for y. This is a minimal polynomial for u since b lies in a cubic extension of Q and not in Q one can also use Eisenstein's criterion. In this chapter, we consider planar problems, in the sense of elementary geometry. The verb "construct" means construct with straightedge and compass, according to the procedures described more precisely below. Write V E for the set of lines in the plane passing through two distinct points of E, and let CE denote the set of circles in the plane whose center is a point of E and whose radius is a distance between two distinct points of E.

Constructions with Straightedge and Compass A point P in the plane is said to be constructible from E if there exists an integer n such that P is constructible in n steps from E. The number of steps depends on the construction procedure, though there exists a minimal such number. Construction of a projection 5.

Construction of a line parallel to a given line The desired line is then M N. Constructions with Straightedge and Compass 5. Moreover, if a circle of CE has center xo, Yo in n and radius equal to the distance between two points of E with coordinates Xl, YI and X2' Y2 in n, then its equation is given by which has the desired form.

If P is a point in the plane with coordinates p, q in n, constructible in one step from E, then K p, q is equal to K or to a quadratic extension of K. The element q lies in K or in a quadratic extension of K according to whether P is not or is irreducible over K. This formula also shows that the degrees of p and q over K must divide 2m , so they are powers of 2. This was first proved by Wantzel, who published his result in although Gauss may have known it as early as D'apres eela, pourreconnattre si 1a construction d'un probleme de Geometrie peut s'efi'ectuer avec la regIe et Ie com pas, il faut chercher s'ilest possible de faire d.

Nous traiterons seulement ici Ie cas ou requation du probleme est a1gebrique. The beginning of Wantzel's proof 5. Of course, 3 does not divide any power of 2, so Proposition 5. The ancient Greeks actually possessed other methods for trisecting angles see Exercise 5.

The same reasoning as in the preceding section then shows that the construction is impossible. Constructions with Straightedge and Compass the preceding one, with Ko constructible from E. Let us show that p,O is constructible. Then p is a root of a quadratic polynomial T with coefficients in K. Exercises for Chapter 5 0, Vc 87 I Constructions of D,ye 3 This result is easily proved by induction, using the two previous results.

This concludes the proof. They enabled him to consider the product of two lengths x and y as a length, rather than considering it as the rectangle with sides x and y, and this led him to suppress the homogeneity conditions that had made the work of his predecessors so heavy: " OU il est a. Roots of quadratic equations Let 0 AB denote an orthogonal basis of the plane.

Recover the usual algebraic condition geometrically. Exercise 5. Construction of the regular pentagon 1 Check the classical construction of the regular polygon with five sides, with one vertex labeled A, inscribed in a circle C of center 0 and radius OA. It is the climax of this part of the Elements; the propositions of books 2, 3, and 4 build methodically toward this end. The solution is entirely geometric. Exercises for Chapter 5 89 Exercise 5. Trisection of angles 1 Archimedes' method Figure 5.

Archimedes' method Consider a circle of center 0 and radius T, and an angle AOB where A and B are two points on the circumference of the circle. Draw the line D parallel to OH passing through A. Suppose that we are in possession of a straightedge marked with two points at a distance of 2a from each other, and that with this straightedge, we know how to construct a line passing through 0, intersecting the segment [AH] at B and the line D at C in such a way that the length of the segment [BC] is 2a Figure 5.

Constructions with Straightedge and Compass 3 Origami the Japanese art of paper-folding, called kami in Japanese. Trisection by origami All straightedge and compass constructions can by done by origami, and many others as well. The origami method for trisecting an angle is due to the Japanese Abe. Solutions to Some of the Exercises Solution to Exercise 5.

Construction of the roots of a quadratic equation 1 We first construct a segment of length vp see Proposition 5. The point H divides DC into segments of lengths a and b, which are the roots of the given equation. Solutions to Some of the Exercises 91 The construction is possible on the one hand if p 0, on the other hand if there exist points with ordinate vP on the circle C, i.

Solution to Exercise 5. Assume the existence of L. By Descartes' method, we know that a 2 is a root of R. We know that L contains b,e,d, therefore x, a root of one of the two factors, is of degree ::; 2 over L. We conclude that L gives the desired result. We conclude by using 1 and the constructibility theorem. However, the definitions and results all generalize directly to arbitrary fields contained in an algebraically closed field C of characteristic o for fields of characteristic p i- 0, see Chapters 14 and One surprising aspect of the theory is the very minor role played by polynomials, which appeared in previous chapters as the main subject of Galois theory.

This is due to the efforts of Dedekind at the end of the 19th century, and Emil Artin in the s and s, to clarify the linear aspects of Galois theory - in particular, the notion of K-homomorphisms, which extends the original idea of permutations of roots of a polynomial. Let a be a complex number that is algebraic over K, with minimal polynomial P over K. Each root of Pin C is called a conjugate of a over K. Two roots of an irreducible polynomial of K[X] are said to be conjugate over K. K-Homomorphisms 2 The numbers v2 and -v2 are conjugate over Q, but not over Q[v2]. Because 0- is injective, the rest of the argument works by decomposing S into a product of irreducible factors.

Let us show the uniqueness of IJ. This generalization is useful for proving the corollary to Proposition 6. K-Homomorphisms L.. The following formula can also be deduced from the proposition: 6. Figure 6. The preceding proposition shows that there exist r extensions of j to L[a]. Every element generating L is called a primitive element for the extension L of K. Given an arbitrary equation, which has no equal roots, whose roots are a, b, c, K-Homomorphisms A primitive element used to be called a Galois resolvent.

The same term was also used for the minimal polynomial of such an element. We prove the result here for subfields of C; it generalizes not only to fields of characteristic 0 but even to all separable algebraic extensions of finite degree defined in Chapter It is an extension of degree 4 of Q. This theorem is a direct consequence of a theorem on characters.

The characters of G are elements of the set F of set maps from G to K; we will say that characters are linearly independent if they are linearly independent in the K-vector space F. Because n is minimal, there exist elements AI, The K-homomorphisms from L to C form a set of linearly independent vectors in the C-vector space of linear maps from L to C.

Thus, Theorem 6. Exercises for Chapter 6 Exercise 6. Solutions to Some of the Exercises Exercise 6. Give other primitive elements for this extension. Exercise 6. K -endomorphisms of an algebraic extension Let L be an algebraic extension of a field K of finite or infinite degree, and let a : L --; L be a K-homomorphism.

Solutions to Some of the Exercises Solution to Exercise 6. K-Homomorphisms of the two generators and is quadratic over Q, would belong to a cubic extension. These five fields are all distinct, since the equality of two of them would imply that one of the k, 1 ::; k ::; 4, which are of degree 4 over Q, would belong to it, even though the extensions are of degree 5.

Let us show that these three fields are all distinct. The first one is contained in IR, and the other two are not. The two last ones are of degree 6 over Q.

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But the last field is of degree 12 over Q. This result does not tell us anything about the extension, but we are in the case of an extension from which the conjugates of 0, V'2, etc. Solution to Exercise 6. The result follows. An element of Q[. The image of. The element. To visualize them, draw a regular hexagon, place each of the six numbers at a vertex and draw a line directly between two vertices if we have shown that they correspond to distinct numbers.

To conclude, every pair of vertices must be connected by a line. Let Xl, The field K[XI' Thus, in characteristic 0, the splitting field of P is the same as that of PI, since the splitting field is obtained by adjoining the roots. Thus, we can restrict our attention to polynomials with distinct roots. It is an extension of degree 6 of Q. Normal Extensions 7. The splitting field of P over K is K[a,d]. K; this suffices for the proof. In other words, all conjugates of elements of N must lie in N. We will try as much as possible to reserve the letter K for the base field, the letters L, L', In characteristic 0, the two notions are equivalent; we use the the expression "normal extension".

Consequently, u x lies in N, so that u N C N. Conversely Figure 7. Then 1 N is a normal extension of finite degree of K. This gives the result. Consequently, [N: K] divides k! It is irreducible, so all of its roots are in N since N is a normal extension of K. The splitting field of P over K is thus N. Then N is a normal extension of every intermediate extension L between K and N. If Xl, The example of Figure 7. This extension always exists; it is the intersection of the set of normal extensions of K containing L, a set which is non-empty since C is a normal extension of all its subfields.

The normal closure N of Lover K inside 7. Normal Extensions As K[A] is the splitting field of the product of the minimal polynomials of al,.. Let K be an arbitrary field, and P an irreducible polynomial of degree n in K[X]. We know that P has no roots in K. Let us now indicate how to construct an extension N of K which is a splitting field of P over K, i.

By the induction hypothesis, there exists a field N which is a splitting field of T over L.

This field is a splitting field of P over K. If the roots of P are simple, the number of extensions is equal to [N:K]. To prove the lemma, we use induction on [N: K]. If the roots of P are simple, the number of such extensions al is equal to the number of roots of a S in N', i. The induction hypothesis then gives the desired number. Toward Chapter 8 Chapters 4, 6, and 7 suffice to open the doors of the "paradise" created discovered? Exercises for Chapter 7 Exercise 7. Splitting fields 1 Determine the splitting field ofaX 2 Q a, b, c.

Normal extensions 1 Show that a quadratic extension of a field K is necessarily normal. Normal Extensions 2 Does the fact that ij2 is an element of Q[ ij2] whereas j ij2 and j2 ij2 are not imply that Q[ ij2] is not a normal extension of itself? Is it a normal extension of every sub field K of L? Show that N' is a normal extension of L. Exercise 7. Proof of d' Alembert's theorem fundamental theorem of algebra In this problem, we will show that every non-constant polynomial P in qX] has at least one root in C, following an idea due to Pierre Samuel see the bibliography.

Take a splitting field of P. Solutions to Some of the Exercises Solution to Exercise 7. This is an extension of degree 8 over Q. Solution to Exercise 7. Normal Extensions Let y be the second root of the minimal polynomial of x over K. A false argument: let P be an irreducible polynomial of K[X] having a root in N. As a polynomial in L[X], it has a root in N, so all its roots must lie in N.

Thus N is a normal extension of K. The error in this reasoning is that in fact, P may no longer be irreducible over L. To every polynomial with coefficients in a field K, with splitting field N over K, we associate a group G called its Galois group. We show that the subgroups of G are in bijective correspondence with the intermediate extensions between Nand K. This correspondence makes it possible to solve problems about polynomials and their splitting fields algebraically, by computing groups.

Over the following chapters, we sketch out this dictionary between the properties of an equation and the algebraic properties of its associated group. The set of K-automorphisms of L is equipped with a group structure whose group law is the composition of K -automorphisms.

Galois Groups 8. They define injective homomorphisms Sn by 8. For every a in Sn, set a. Obviously, this was not the language used by Lagrange; he reasoned on. I case. Who would dare to write this way nowadays? And yet, his argument is the basis of the argument used to prove that every equivalence class of elements in a group modulo a subgroup contains the same number of elements.

Galois was the first to actually use the word "group", but he meant it as a subgroup of the permutation group of the set of roots of a polynomial, as in Proposition 8. Galois did not use the notation for permutations 8. Galois Groups that we use now; he wrote successive arrangements of the roots which he called permutations: abcde bcdea cdeab emphasizing the fundamental notion of what he calls substitution, which makes it possible to pass from one permutation to another.

Galois did not use associativity, identity elements, or inverses, only the internal group law, writing: "Thus, if in such a group we have the substitutions Sand T, we are sure to have the substitution ST. The abstract form of the definition of a group, which we use today, was built up slowly over the course of the 19th century, with suggested definitions by Cayley , Kronecker , Weber , Burnside , and Pierpont The axioms of associativity, identity element and inverses were first stated in their present form by Pierpont. Let h' : L[X] The coefficients of S are thus invariant under the elements of H and belong to I H.

Thus L is a normal extension of I H. Let us now show that L is an extension of finite degree r of I H. Let y be an element of degree s. This gives the six possibilities listed in Table 8. Note that 0"1 is the identity on N and the identity element of G, and that 0"2 is induced by complex conjugation. To determine the structure of G, we note that it has three elements of order two: 0"2,0"4,0"6, and two elements of order 3: 0"3,0"5. Let us determine the field of invariants of a subgroup H of G. It is summarized in Figure 8.

We also assume that L is a normal extension of K Figure 8. Let us explain this. A trellis is an ordered set in which the sup and the inf of any two elements exists. The fact that I and G are decreasing is clear. N is a normal extension of L and L'. Field Theory. Steven Roman. About the first edition: " Mordeson, Zentralblatt "The book is written in a clear and explanatory style. Albu, MathSciNet. Therefore, a book devoted to field theory is desirable for us as a text.

While there are a number of field theory books around, most of these were less complete than I wanted. JavaScript is currently disabled, this site works much better if you enable JavaScript in your browser. Mathematics Algebra. Graduate Texts in Mathematics Free Preview.

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