Wilkins. Introduction to Galois theory(45s).pdf: eKnigu

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Wilkins. Introduction to Galois theory(45s).pdf

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We denote by (f1 , f2 , . . . , fk ) the ideal of R generated by any ﬁnite subset {f1 , f2 , . . . , fk } of R. We say that an ideal I of the ring R is ﬁnitely generated if there exists a ﬁnite subset of I which generates the ideal I . Lemma 3.5 Let R be a unital commutative ring, and let X be a subset of R. Then the ideal generated by X coincides with the set of al l elements of R that can be expressed as a ﬁnite sum of the form r1 x1 + r2 x2 + · · · + rk xk , where x1 , x2 , . . . , xk ∈ X and r1 , r2 , . . . , rk ∈ R. Pro of Let I be the subset of R consisting of all these ﬁnite sums. If J is any ideal of R which contains the set X then J must contain each of these ﬁnite sums, and thus I ⊂ J . Let a and b be elements of I . It follows immediately from the deﬁnition of I that 0 ∈ I , a + b ∈ I , −a ∈ I , and ra ∈ I for all r ∈ R. Also ar = ra, since R is commutative, and thus ar ∈ I . Thus I is an ideal of R. Moreover X ⊂ I , since the ring R is unital and x = 1x for all x ∈ X . Thus I is the smallest ideal of R containing the set X , as required. Each integer n generates an ideal nZ of the ring Z of integers. This ideal consists of those integers that are divisible by n. Lemma 3.6 Every ideal of the ring Z of integers is generated by some nonnegative integer n. Pro of The zero ideal is of the required form with n = 0. Let I be some non-zero ideal of Z. Then I contains at least one strictly positive integer (since −m ∈ I for all m ∈ I ). Let n be the smallest strictly positive integer belonging to I . If j ∈ I then we can write j = q n + r for some integers q and r with 0 ≤ r < n. Now r ∈ I , since r = j − q n, j ∈ I and q n ∈ I . But 0 ≤ r < n, and n is by deﬁnition the smallest strictly positive integer belonging to I . We conclude therefore that r = 0, and thus j = q n. This shows that I = nZ, as required....

for all I + x ∈ R/I and I + y ∈ R/I . One can readily verify that R/I is a ring with respect to these operations. We refer to the ring R/I as the quotient of the ring R by the ideal I . Example Let n be an integer satisfying n > 1. The quotient Z/nZ of the ring Z of integers by the ideal nZ generated by n is the ring of congruence classes of integers modulo n. This ring has n elements, and is a ﬁeld if and only if n is a prime number. Deﬁnition A function ϕ: R → S from a ring R to a ring S is said to be a homomorphism (or ring homomorphism ) if and only if ϕ(x + y ) = ϕ(x) + ϕ(y ) and ϕ(xy ) = ϕ(x)ϕ(y ) for all x, y ∈ R. If in addition the rings R and S are unital then a homomorphism ϕ: R → S is said to be unital if ϕ(1) = 1 (i.e., ϕ maps the identity element of R onto that of S ). Let R and S be rings, and let ϕ: R → S be a ring homomorphism. Then the kernel ker ϕ of the homomorphism ϕ is an ideal of R, where ker ϕ = {x ∈ R : ϕ(x) = 0}. The image ϕ(R) of the homomorphism is a subring of S ; however it is not in general an ideal of S . An ideal I of a ring R is the kernel of the quotient homomorphism that sends x ∈ R to the coset I + x. Deﬁnition An isomorphism ϕ: R → S between rings R and S is a homomorphism that is also a bijection between R and S . The inverse of an isomorphism is itself an isomorphism. Two rings are said to be isomorphic if there is an isomorphism between them. 6...

bj c0 is not divisible by p, since p is prime and neither bj nor c0 is divisible by p. Therefore aj is not divisible by p, and hence j = n and deg g ≥ n = deg f . Thus deg g = deg f and deg h = 0. Thus the polynomial f does not factor as a product of polynomials of lower degree with integer coeﬃcients, and therefore f is irreducible over Q (Proposition 3.16)....

Deﬁnition Let L: K be a ﬁeld extension, and let α be an element of L. If there exists some non-zero polynomial f ∈ K [x] with coeﬃcients in K such that f (α) = 0, then α is said to be algebraic over K ; otherwise α is said to be transcendental over K . A ﬁeld extension L: K is said to be algebraic if every element of L is algebraic over K . Lemma 3.19 A ﬁnite ﬁeld extension is algebraic. Pro of Let L: K be a ﬁnite ﬁeld extension, and let n = [L: K ]. Let α ∈ L. Then either the elements 1, α, α2 , . . . , αn are not all distinct, or else these elements are linearly dependent over the ﬁeld K (since a linearly independent subset of L can have at most n elements.) Therefore there exist c0 , c1 , c2 , . . . , cn ∈ K , not all zero, such that c0 + c1 α + c2 α2 + · · · + cn αn = 0. Thus α is algebraic over K . This shows that the ﬁeld extension L: K is algebraic, as required. Deﬁnition A polynomial f with coeﬃcients in some ﬁeld or unital ring is said to be monic if its leading coeﬃcient (i.e., the coeﬃcient of the highest power of x occurring in f (x) with a non-zero coeﬃcient) is equal to 1....

Let z ∈ K (α). Then z = g (α) for some g ∈ K [x]. But then there exist polynomials l and f belonging to K [x] such that g = lm + f and either f = 0 or deg f < deg m (Lemma 3.10). But then z = f (α) since m(α) = 0. Suppose that z = h(α) for some polynomial h ∈ K [x], where either h = 0 or deg h < deg m. Then m divides h − f , since α is a zero of h − f . But if h − f were non-zero then its degree would be less than that of m, and thus h − f would not be divisible by m. We therefore conclude that h = f . Thus any element z of K (α) can be expressed in the form z = f (α) for some uniquely determined polynomial f ∈ K [x] satisfying either f = 0 or deg f < deg m. Thus if n = deg m then 1, α, α2 . . . , αn−1 is a basis of K (α) over K . It follows that the extension K (α): K is ﬁnite and [K (α): K ] = deg m, as required. Corollary 3.22 A ﬁeld extension L: K is ﬁnite if and only if there exists a ﬁnite subset {α1 , α2 , . . . , αk } of L such that αi is algebraic over K for i = 1, 2, . . . , k and L = K (α1 , α2 , . . . , αk ). Pro of Suppose that the ﬁeld extension L: K is a ﬁnite. Then it is algebraic (Lemma 3.19). Thus if {α1 , α2 , . . . , αk } is a basis for L, considered as a vector space over K , then each αi is algebraic and L = K (α1 , α2 , . . . , αk ). Conversely suppose that L = K (α1 , α2 , . . . , αk ), where αi is algebraic over K for i = 1, 2, . . . , k . Let Ki = K (α1 , α2 , . . . , αi ) for i = 1, 2, . . . , k . Clearly Ki−1 (αi ) ⊂ Ki for all i > 1, since Ki−1 ⊂ Ki and αi ∈ Ki . Also Ki ⊂ Ki−1 (αi ), since Ki−1 (αi ) is a subﬁeld of L containing K ∪ {α1 , α2 , . . . , αi } We deduce that Ki = Ki−1 (αi ) for i = 2, 3, . . . , k . Moreover αi is clearly algebraic over Ki−1 since it is algebraic over K , and K ⊂ Ki−1 . It follows from Theorem 3.21 that the ﬁeld extension Ki : Ki−1 is ﬁnite for each i. Using the Tower Law (Proposition 3.18), we deduce that L: K is a ﬁnite extension, as required....

since it is the fourth vertex of a parallelogram which has three vertices at the constructible points (x, 0), (0, y ) and (0, 1) (Lemma 3.25). But the line which passes through the two constructible points (0, y ) and (x, y − 1) intersects the x-axis at the point (xy , 0). Therefore the point (xy , 0) is constructible, and thus xy ∈ K. Now suppose that x ∈ K, y ∈ K and y = 0. The point (x, 1 − y ) is constructible, since it is the fourth vertex of a parallelogram with vertices at the constructible points (x, 0), (0, y ) and (0, 1). The line segment joining the constructible points (0, 1) and (x, 1 − y ) intersects the x-axis at the point (xy −1 , 0). Thus xy −1 ∈ K. The above results show that K is a subﬁeld of the ﬁeld of real numbers. Moreover if x ∈ K and y ∈ K then the point (x, y ) is constructible, since it is the fourth vertex of a rectangle with vertices at the constructible points (0, 0), (x, 0) and (0, y ). Conversely, suppose that the point (x, y ) is constructible. We claim that the point (x, 0) is constructible and thus x ∈ K. This result is obviously true if y = 0. If y = 0 then the circles centred on the points (0, 0) and (1, 0) and passing through (x, y ) intersect in the two points (x, y ) and (x, −y ). The point (x, 0) is thus the point at which the line passing through the constructible points (x, y ) and (x, −y ) intersects the x-axis, and is thus itself constructible. The point (0, y ) is then the fourth vertex of a rectangle with vertices at the constructible points (0, 0), (x, 0) and (x, y ), and thus is itself constructible. The circle centred on the origin and passing though (0, y ) intersects the x-axis at (y , 0). Thus (y , 0) is constructible, and thus y ∈ K. We have thus shown that a point (x, y ) is constructible using straightedge and compasses alone if and only if x ∈ K and y ∈ K. 1 Suppose that x ∈ K and that x > 0. Then 2 (1 − x) ∈ K. Thus if 1 C = (0, 2 (1 − x)) then C is a constructible point. Let (u, 0) be the point at which the circle centred on C and passing through the constructible point (0, 1) intersects the x-axis. (The circle does intersect the x-axis since it passes through (0, 1) and (0, −x), and x > 0.) The radius of this circle is 1 (1 + x)), 2 and therefore 1 (1 − x)2 + u2 = 1 (1 + x)2 (Pythagoras’ Theorem.) But then 4 4 u2 = x. But (u, 0) is a constructible point. Thus if x ∈ K and x > 0 then √ x ∈ K, as required. The above theorems can be applied to the problem of determining whether or not it is possible to construct a regular n-sided polygon with a straightedge and compass, given its centre and one of its vertices. The impossibility of trisecting an angle of 60◦ shows that a regular 18-sided polygon is not constructible using straightedge and compass. Now if one can construct a regular n-sided polygon then one can easily construct a regular 2n-sided polygon by bisecting the angles of the n-sided polygon. Thus the problem 20...