Survey of Modern Algebra

MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 33 Straightedge and compass constructions In classical geometry, the tools of a straightedge (a ruler without markings) and a compass (which makes circles) are used to construct new shapes out of old ones. Many old problems in geometry ask whether certain operations or quantities can be constructed in this fashion. In this lecture we solve some of these problems by relating them to field theory. A (real) number α is constructible if we can construct a line segment of length |α| in the plane R2 starting from a line segment of length 1 and using a straightedge and compass. 1 Let F Ă R be the set of constructible numbers. At the beginning, we know only that t´1, 0, 1u Ă F . Indeed, we are allowed to start with a segment of unit length; and we can draw a point in R2 which is a line segment of length 0. What is allowed? Using the straightedge, we can draw straight lines as we like through points we have already drawn, but the only arcs we can draw are the ones whose lengths are already in F . Similarly, with the compass we can draw circles of radius which is either specified by two points we have already drawn, or is the length of something in F . So what other numbers are constructible, i.e. are contained in F ? Using our straightedge we can draw unit length arcs in sequence in a straightline: 1 1 This shows that Z Ă F , i.e. the integers are constructible numbers. § If α, β P F then α ` β and α ´ β are also in F . The idea is the same as above: given arcs of length α and β we can draw them side by side, parallel and adjacent, using our straightedge. Then we have an arc of length α ` β. α β Similarly, if we draw the arcs as starting from the same point, overlapping, then we can extract an arc whose length is the difference α ´ β, assuming for simplicity that α ą β. α β In the picture, the arcs should really be superimposed. The portion of the arc of length α, in red, which does not overlap the arc of length β, in blue, is a segment of length α ´ β. 1 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 33 What about multiplication and division of constructible numbers? This is a bit more work. Suppose α P F , α ą 0. Then we can draw an arc of length α. We can then draw two circles each of radius α, centered at the two points of the arc, as in the right picture below. α Then using the straightedge we can draw a line passing between the two points of intersection of the circles. We have then bisected the arc! Thus we now know how to draw an arc of length α{2. Thus if α is constructible, then α{2 is constructible. We should also record a consequence of our construction: given an arc, we can draw a line perpendicular to that arc passing through the midpoint of the arc. Now consider a point lying above a straight line. We claim that using our tools we can draw a line parallel to the given line, and passing through the given point P . We assume the the distance between the point and the line is constructible. P To do this, first draw a circle centered at the point and of radius large enough so that it intersects the line twice. An arc is formed between the intersection points (in red below). From our previous construction we can draw a line perpendicular to this arc which bisects it. This perpendicular also passes through P . Next, consider the arc from the intersection of the lines, call this point O, to the point P . As the length of OP is constructible, we can draw an arc of the same length starting from P upwards, ending at a point Q. Q P O Finally, we know from above that we can draw a line perpendicular to the segment OQ and which passes through its midpoint. This is the desired parallel line. 2 MTH 461: Survey of Modern Algebra, Spring 2021 § If α, β P F Lecture 33 then αβ and α{β (β ‰ 0) are also in F . Proof. Assume β ą 1. Draw an arc of length α, an arc of length 1, and an arc of length β all emanating from the same starting point. The 1 and β arcs should be parallel, and the angle between these and the α arc is some arbitrary angle between 0 and 90˝ . β β 1 1 α α The situation is depicted above on the left. Then use the straightedge to draw a segment from the endpoint of the unit length arc to the endpoint of the α arc. We then have the right-hand picture above. Next, we know from previous constructions that we can draw a parallel arc (in green) to the arc just drawn, and passing through the endpoint of β. β β 1 1 α α Now we have an arc starting from the same point as the others, and ending at the intersection of the green arc with the line containing the α arc. This new arc is in orange, above on the right. Call the length x. The two triangles in the picture are similar, so we have α{1 “ x{β. Thus x “ αβ. Thus we have constructed an arc of length αβ. The case β ă 1 and also the case of division are similar, and left as exercises. § The set F Ă R of constructible numbers is a field. This result just follows from the above work. Thus far it is clear that Q Ă F . However the field F is larger than just the rationals, as the following result indicates. § If α P F then ? α P F. Proof. Start with two arcs of length 1 and α side by side, forming an arc of length 1 ` α. P α 1 Bisect this line and draw a circle centered at the midpoint M and with radius p1 ` αq{2 so that the arc is a diameter of the circle. See the left picture below. 3 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 33 Q P P M M P M Draw a perpendicular line to the original line which passes through the point P , as in the picture above in the center. (We can draw this line because it is passing through the midpoint of constructible arc of length 2α.) Draw arcs from the endpoints of the original arc to the point of intersection Q, as in the picture above on the right. Then the arc P Q has some length x. Using the?similarity of triangles appearing, we have α{x “ x{1 or x2 “ α. Thus we obtain that x “ α is constructible. In fact we have essentially found all of the ways of producing constructible numbers. § A number α P R is constructible if and only if there is a sequence of fields Q “ F0 Ă F1 Ă F2 Ă ⋯ Ă Fk ? such that Fi`1 “ Fi p αi q where αi P Fi , and α P Fk . We will not go through the proof in detail, although it is not too difficult. Roughly, all of the “tricks” one can do to form new cosntructible numbers are contained in the above results. § If α is constructible, then rQpαq ∶ Qs “ 2k for some k ě 0. Proof. From the previous result, we have Qpαq “ Fk for some Fk and a sequence of fields Q “ F0 Ă F1 Ă ⋯ Ă Fk which are quadratic extensions obtained by adjoing square roots, and in particular have degree 1 or 2. Then we compute rQpαq ∶ Qs to be rQpαq ∶ Qs “ rFk ∶ Fk´1 s⋯rF2 ∶ F1 srF1 ∶ F0 s which is some power of 2. We say a subset of R2 is constructible if it can be draw using straightedge and compass operations as above, using only constructible measurements. We consider the problem of constructing a regular n-gon, i.e. a symmetric polygon with n sides, using only a straightedge and compass. It is straightforward to argue: § If the regular n-gon is constructible, then cosp2π{nq is constructible. 4 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 33 Using this we now prove: § The regular 18-gon is not constructible. Proof. We only need argue that cosp2π{18q “ cosp360˝ {18q “ cosp20˝ q is not a constructible number. Using pcospθq ` sinpθqiq3 “ cosp3θq ` sinp3θqi we have the identity cosp3θq “ 4 cos3 pθq ´ 3 cospθq Setting θ “ 2π{18 with cosp2π{18q “ α gives 1{2 “ cospπ{3q “ cosp3θq “ 4α3 ´ 3α Thus α is a root of 8×3 ´ 6x ´ 1. This is an irreducible polynomial, and is the minimal polynomial of α over the field Q. Thus rQpαq ∶ Qs “ degpppxqq “ 3. This is not a power of 2 and so α cannot be constructible. 5 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 34 Splitting fields and Galois groups Let f pxq P F rxs be a polynomial with coefficients in a field F . A field extensions E over F is called a splitting field for the polynomial f pxq if there are α1 , . . . , αn P E such that E “ F pα1 , . . . , αn q and the polynomial splits into linear factors f pxq “ apx ´ α1 q⋯px ´ αn q for some non-zero constant a P F . Examples 1. Let f pxq be the polynomial x2 ` 1 P Rrxs. As f pxq “ px ´ iqpx ` iq as a polynomial in Crxs, a splitting field for f pxq is given by Rpi, ´iq “ Rpiq “ C. 2. Let f pxq “ x2 ´ 2 P Qrxs. ? Then f pxq “ px ´ by the field extension Qp 2q over Q. ? ? 2qpx ` 2q. Thus a splitting field is given 2 3. Let f pxq “ px2?´ 2qpx ? ´ 3q P Qrxs. A splitting field for this polynomial is given by the field extension Qp 2, 3q over Q. 4. Let f pxq “ x3 ´ 2 P Qrxs. The complex roots of this polynomial are as follows: ? ? ? 3 3 3 2, 2 ¨ e2πi{3 , 2 ¨ e´2πi{3 ? ? ? Thus a splitting field for f pxq is given by E “ Qp 3 2, 3 2e2πi{3 , 3 2e´2πi{3 q. Note that ´ ? ¯2 ´ ? ¯´1 ? 3 3 3 2 ¨ e´2πi{3 “ 2 2 ¨ e2πi{3 so the third root is already in the field which is?generated by the first two roots. In ? ? other words, our splitting field can be written E?“ Qp 3 2, 3 2e2πi{3 q. Furthermore, clearly 3 2e2πi{3 can be replaced by e2πi{3 so that E “ Qp 3 2, e2πi{3 q. Finally, ? 1 3 e2πi{3 “ ´ ` i 2 2 ? and so e2π{3 can be replaced by ´3. In summary, we have the identification ? ? 3 E “ Qp 2, ´3q ? We leave as an exercise that also E “ Qp 3 2 ` e2πi{3 q. 5. Let f pxq “ x4 ´ 2 P Qrxs. The roots ? complex ? ? of this polynomial are We have the splitting field Qp 4 2, 4 2iq “ Qp 4 2, iq. ? ? ? ? 4 2, ´ 4 2, 4 2i, ´ 4 2i. It is a basic result, which we omit, that for a given non-constant polynomial f pxq P F rxs, a splitting field exists for f pxq; and moreover, any two such splitting fields are isomorphic. Thus given f pxq P F rxs we may speak of the splitting field of f pxq. 1 MTH 461: Survey of Modern Algebra, Spring 2021 § Given a field extension E Lecture 34 over F , we define AutpE{F q “ tσ ∶ E Ñ E ∶ σ is an isomorphism, σpaq “ a for all a P F u This is naturally a group, called the automorphism group of the field extension. Proof. Let σ, σ 1 P AutpE{F q so that σ, σ 1 ∶ E Ñ E are isomorphisms each fixing F , i.e. σpaq “ σ 1 paq “ a for all a P F . Then the composition σ ˝ σ 1 is an isomorphism of E and pσ ˝ σ 1 qpaq “ σpσ 1 paqq “ σpaq “ a. Thus σ ˝ σ 1 P AutpE{F q. Further, a “ idE paq “ pσ ´1 ˝σqpaq “ σ ´1 pσpaqq “ σ ´1 paq and thus σ ´1 P AutpE{F q. Thus AutpE{F q is a group. A field extension E over a subfield of C is called Galois if it is a splitting field of some polynomial f pxq P Qrxs. Thus all of the examples above are Galois fields. More generally, an extension E over an arbitrary field F is Galois if it is the splitting field of a polynomial f pxq P F rxs which has no repeated roots in E. We mainly focus on extensions of Q and other subfields of C, in which case Galois fields can be identified with splitting fields. § If E is a Galois extension of a field F we write GalpE{F q “ AutpE{F q and call this the Galois group of the field extension. As a Galois extension E over F is the splitting field of some polynomial f pxq P F rxs, we also call GalpE{F q the Galois group of the polynomial f pxq. § Let E “ F pα1, . . . , αnq be the splitting field of f pxq P F rxs whose roots in E are α1 , . . . , αn . Then there is a 1-1 homomorphism φ ∶ GalpE{F q ÐÑ Sn to the symmetric group Sn . If f pxq P F rxs is irreducible, then the image subgroup impφq Ă Sn is a transitive subgroup of Sn . We remark that a transitive subgroup G Ă Sn is a subgroup of the symmetric group Sn such that for all i, j P t1, . . . , nu there is a permutation f P G such that f piq “ j. Proof. First, write f pxq “ an xn ` ⋯ ` a0 where each ai P F . Consider a root αi P E of f pxq, and some σ P GalpE{F q. Using that σ ∶ E Ñ E is a homomorphism fixing F we compute: f pσpαi qq “ an pσpαi qqn ` an´1 pσpαi qqn´1 ` ⋯ ` a1 σpαi q ` a0 “ an pσpαin qq ` an´1 pσpαin´1 qq ` ⋯ ` a1 σpαi q ` a0 “ σpan pαin qq ` σpan´1 pαin´1 qq ` ⋯ ` σpa1 αi q ` σpa0 q “ σpan αin ` an´1 αin´1 ⋯ ` a1 αi ` a0 q “ σp0q “ 0 2 MTH 461: Survey of Modern Algebra, Spring 2021 Lecture 34 Thus σpαi q is also a root of f pxq, and therefore σpαi q “ αj for some j P t1, . . . , nu. Since E is Galois over F , the roots are distinct, so the index j is uniquely determined. Next, define φ ∶ GalpE{F q Ñ Sn by φpσqpiq “ j if and only if σpαi q “ αj . In other words, we look at how the automorphism σ ∶ E Ñ E permutes the roots α1 , . . . , αn . It is straightforward from this construction that φ is a homomorphism. Next, recall E “ F pα1 , . . . , αn q. As σ P GalpE{F q fixes F , it is entirely determined by where it sends the elements αi . This shows that φ is 1-1. Finally, suppose f pxq is irreducible. Then F pαi q – F rxs{f pxq for any root αi of f pxq. In particular, for αi and αj any two roots of f pxq that lie in E we have an isomorphism F pαi q – F rxs{f pxq – F pαj q. This can be extended to an isomorphism E Ñ E (we omit this step), from which the result follows. The above result shows that the Galois group of E over F is isomorphic to a subgroup of Sn , which has order n!. We obtain: § Let E be the splitting field of f pxq P F rxs whose degree is n. Then |GalpE{F q| divides n!. Examples 1. Let f pxq be the polynomial x2 ` 1 P Rrxs, whose splitting field is C “ Rpi, ´iq. Label the roots of f pxq by α1 “ i and α2 “ ´i. Note that complex conjugation σ ∶ C Ñ C defined by σpa ` biq “ a ´ bi is an element of GalpC{Rq of order 2. Thus tidC , σu Ă GalpC{Rq. As the Galois group has order dividing 2! “ 2, this is in fact an equality of groups. Thus GalpC{Rq – Z2 . ? to the previous ex2. Let f pxq “ x2 ´ 2 P Qrxs with splitting field Qp 2q. Similar ? ample, ? there is only σ P GalpQp 2q, Qq and it is given by ? one non-trivial automorphism ? σpa ` b 2q “ a ´ b 2. We have GalpQp 2q{Qq – Z2 . ? ? 3. Let f pxq “ px2 ´ 2qpx2 ´ 3q P Qrxs with splitting field E “ Qp 2, 3q. Recall that any element in E can be written uniquely as ? ? ? ? a`b 2`c 3`d 2 3 where a, b, c, d P Q. Define automorphisms σ1 , σ2 , σ3 P GalpE{Qq by ? ? ? ? ? ? ? ? σ1 pa ` b 2 ` c 3 ` d 2 3q “ a ´ b 2 ` c 3 ´ d 2 3 ? ? ? ? ? ? ? ? σ2 pa ` b 2 ` c 3 ` d 2 3q “ a ` b 2 ´ c 3 ´ d 2 3 ? ? ? ? ? ? ? ? σ3 pa ` b 2 ` c 3 ` d 2 3q “ a ´ b 2 ´ c 3 ´ d 2 3 3 MTH 461: Survey of Modern Algebra, Spring 2021 Labelling the roots of f pxq as α1 “ Lecture 34 ? ? ? ? 2, α2 “ ´ 2, α3 “ 3, α4 “ ´ 3 we have φpσ1 q “ p12q, φpσ2 q “ p34q, φpσ3 q “ p12qp34q ? ? Because no automorphism can interchange ˘ 2 with ˘ 3, these are all of them. Thus GalpE{Qq “ tidE , σ1 , σ2 , σ3 u – te, p12q, p34q, p12qp34qu Ă S4 Note that this group is isomorphic to Z2 ˆ Z2 . Note in this example that f pxq is not irreducible, and that the corresponding permutation group is not transitive. ? ? 4. Let f pxq “ x3 ´ 2 P Qrxs with splitting field E “ Qp 3 2, ´3q. The roots of f pxq are ? ? ? 3 3 3 α1 “ 2, α2 “ 2 ¨ e2πi{3 , α3 “ 2 ¨ e´2πi{3 A direct computation shows that every permutation of these 3 roots extends to define an automorphism, and thus GalpE{Qq – S3 . ? 5. Let f pxq “ x4 ´ 2 P Qrxs with splitting field E “ Qp 4 2, iq. The roots of f pxq are ? ? ? ? 4 4 4 4 α1 “ 2, α2 “ ´ 2, α3 “ 2i, α4 “ ´ 2i ? Note an automorphism of this extension is determined by where it sends α1 “ 4 2 and i. There are automorphisms σ, τ P GalpE{Qq such that: ? ? 4 4 σpiq “ i σp 2q “ 2i, ? ? 4 4 τ p 2q “ 2, τ piq “ ´i We have φpσq “ p1324q and φpτ q “ p34q. Any other automorphism is a composition of these two automorphisms. The Galois group is isomorphic to the following subgroup of S4 : GalpE{Qq – te, p12q, p34q, p12qp34q, p1324q, p1423q, p13qp24q, p14qp23qu Ă S4 In fact this subgroup is isomorphic to the dihedral group D4 , the symmetries of a square. 4 MTH 461: Survey of Modern Algebra, Spring 2021 Homework 10 Homework 10 1. Show each number is algebraic over Q by finding its minimal polynomial. ? (a) 5 ´ 1 a ? 3 (b) 3`i 2 ? ? (c) 3 ` 5 2. Consider Z2 rxs{px3 ` x ` 1q. Show this is a field with 8 elements. Write the multiplication table for the non-zero elements of this field. 3. (a) Let α “ ? ? 4 5 and β “ 4 5 ¨ i. Find the minimal polynomials of α and β. (b) Find the degrees of the extensions Qpαq and Qpβq over Q. (c) Show that Qpαq and Qpβq are isomorphic fields. (d) Show that Qpαq and Qpβq are not the same field. 4. Find a basis for each field extension, and compute the degree of the extension. ? (a) Qp 2, iq over Q ? ? (b) Qp 3q over Qp 27q ? ? ? (c) Qp 2, 3, 5q over Q 5. Show that the regular polygon with 9 sides is not constructible. Recall this amounts to showing α “ cosp2π{9q is not a construcible number. You will need to find the minimal polynomial of α. Use the identity pcospθq ` i sinpθqqn “ cospnθq ` i sinpnθq, with θ “ 2π{9 and n “ 3. 6. Prove that the Galois group of an irreducible cubic polynomial is isomorphic to either S3 or Z3 . 1
Purchase answer to see full attachment

 

Just $7 Welcome
Order Now