Mth 08 Lecture 2: Fields, Formlly Professor: Pdric Brtlett Week UCSB 203 In our first lecture, we studied R, the rel numbers. In prticulr, we exmined how the rel numbers intercted with the opertions of ddition nd multipliction, listing number of properties tht we believed the rel numbers stisfied. From there, we showed tht some of those properties were superfluous: i.e. we showed tht we could prove tht some of these properties were necessry consequences of other properties, nd therefore tht listing them ws unnecessry. Finlly, t the end of clss, we defined the mthemticl concept of fields. We restte this here, s fields will be the subject of tody s tlk: Fields: Definitions nd Exmples Definition. A field is set F long with pir of opertions +,, typiclly thought of s ddition nd multipliction, such tht the following properties hold: Closure(+):, b F, we hve +b F. Identity(+): 0 F such tht F, 0 +. Commuttivity(+):, b F, + b b +. Associtivity(+):, b, c F, ( + b) + c + (b + c). Inverses(+): F, some F such tht + ( ) 0. Closure( ):, b F, we hve b F. Identity( ): F such tht F,. Commuttivity( ):, b F, b b. Associtivity( ):, b, c F, ( b) c (b c). Inverses( ): 0 F, some F such tht. Distributivity(+, ) :, b, c F, ( + b) c ( c) + (b c) Typiclly, when we write down field, we do so in the form F, +,, so tht we understnd tht F is field with respect to those two opertions + nd. This my seem odd t first why would we need to tlk bout wht ddition nd multipliction re? but this comes in hndy lter, when we wnt to work on sets where wht ddition nd multipliction even re cn be tricky to understnd. To mke n nlogy: this would be like listing set of properties of the UCSB cmpus, nd listing seprtely tht () we hve building clled Phelps Hll nd (b) tht we hve clss clled Mth 08 tht is tught in Phelps Hll, room 3505. Sure, these re both true properties: but if you know the second property thn the first one flls s logicl consequence! So there s no need to list the first property, if you hve the second.
As we discussed in our lst clss, the rel numbers R long with stndrd ddition nd multipliction form field. Wht other objects re fields? Wht objects re not fields? Let s consider N, +, the nturl numbers: i.e. the nonnegtive whole numbers {0,, 2, 3,...}. Does this set form field, long with the typicl opertions of ddition nd multipliction? As it turns out: no! This set does not hve dditive inverses. In specific, is nturl number, nd yet there is no nturl number tht we cn dd to to get to 0. Now, let s consider Z, +,, the integers, i.e. the whole numbers where we llow negtive nd positive vlues: Z {..., 3, 2,, 0,, 2, 3,...}. Unlike N, this set stisfies the dditive inverse property: given ny Z, we know tht is lso n integer, nd furthermore tht + ( ) 0. However, this set is lso not field, becuse it fils the property of hving multiplictive inverses. In prticulr, 2 is n integer, nd yet there is no integer such tht 2. Now, let s try [, ], +,, the closed intervl contining ll of the rel numbers between nd, long with the stndrd opertions of + nd. This set hs dditive inverses: for ny number x [,, ], the number x is lso in [, ], becuse this intervl is symmetric. However, it lso fils to hve multiplictive inverses: 2 [, ], nd yet there is no x [, ] such tht x 2. Moreover, this set fils to be dditively closed: i.e. it is possible to dd two numbers in our set nd get number outside of our set! For exmple, 2 nd 2 3 re in our set, while 2 + 2 3 7 6 is not in our set. So: we hve lot of exmples of things tht re not fields. How bout n exmple of something tht is field? In prticulr, let s consider Q, +,, the rtionl numbers. If you hven t seen these before, here s definition: Definition. Q, the rtionl numbers, is the collection of ll numbers of the form p q such tht:. Both p nd q re integers. 2. q is nonzero. We don t wnt to ccidentlly divide by 0. We define ddition on this set s norml by declring b + c d for ny b, c d Q. Similrly, we define multipliction by for ny b, c d Q. d + bc, bd b c d c bd, 2
Finlly, tke ny element p q Q, nd suppose tht both p nd q shre common fctor r Z. In other words, suppose tht we cn write p r nd q r b, for some set of integers r,, b Z. Then we consider p q nd b to be the sme frction. Essentilly, this is like sying tht p q r r b b. In other words, we cn cncel out common fctors. This defines Q. Is this field? Well: let s check! Does closure(+) hold: i.e. is it dditively closed? Well: let s tke ny pir of frctions b, c d. We know, becuse of how ddition is defined on frctions, tht b + c d d+bc bd. Both d+cb nd bd re integers, becuse, b, c, d re ll integers; furthermore, becuse neither b or d re zero, we know tht bd is lso nonzero. Therefore, we stisfy the properties required to be rtionl number listed bove. Consequently, we hve shown tht dding ny two rtionl numbers together yields nother rtionl number: in other words, we ve proven tht ddition is closed! How bout closure( )? Agin: let s tke ny pir of frctions b, c d. We know, becuse of how ddition is defined on frctions, tht b c d c bd. Agin, both c nd bd re integers, becuse, b, c, d re ll integers; furthermore, bd is still not zero becuse neither b nor d re zero. Therefore, we ve shown tht the product of ny two rtionl numbers is still rtionl: i.e. tht our field is multiplictively closed. Commuttivity(+) is reltively simple. All you hve to do is notice tht for ny, we cn use the fct tht multipliction in the integers is commuttive to see tht b, c d b + c d + bc d + cb c d bd db d + b. Commuttivity( ) is pretty much the sme rgument: for ny b, c d, we gin use the fct tht multipliction in the integers is commuttive to see tht b c d c bd c db c d b. Identity(+): The frction 0 is the identity. To see this, tke ny b tht 0 + b 0 b + b b. We will often write 0 insted of 0, for convenience s ske. Identity( ): The frction is the identity. To see this, tke ny b tht b b b. We will often write insted of, for convenience s ske. Q, nd notice Q, nd notice 3
Associtivity(+) is pin to write down, but not ctully ny more complicted thn the bove rguments: ll we hve to notice is tht for ny b, c d, e f Q, we hve ( b d) + c ( c b + d + e f + e f ) b + e (d + bc)f + (bd)e df + bcf + bde, nd f (bd)f bdf cf + de (df) + b(cf + de) df + bcf + bde +. df b(df) bdf d + bc bd Associtivity( ) is similr: for ny b, c d, e f Q, we hve ( b c ) e d f c bd e f (c)e (bd)f ce bdf, nd ( c b d e ) f b ce df (ce) b(df) ce bdf. Inverses(+) is pretty esy, lso. For ny p p q Q, consider the frction q. Becuse p is n integer nd q is nonzero integer, this is well-defined frction. Furthermore, we cn esily see tht p q + p q pq + ( p)q q 2 pq pq q 2 0 q 2 0 q2 q 2 0. Becuse 0 is the dditive identity, we hve just shown tht every element hs n dditive inverse. Inverses( ) is similrly simple. Tke ny p q Q tht is not equl to 0. Notice tht this mens tht p 0, becuse if p 0, we would hve 0 q 0 q q 0. Now consider the frction q p. Becuse p nd q re nonzero integers, this is welldefined frction. Then, we hve p q q p pq qp pq pq. Becuse is the multiplictive identity, we hve just shown tht every element hs multiplictive inverse. Lst one! Distributivity(+, ) is proven just like the rest: : ll we hve to notice is tht for ny b, c d, e f Q, we hve ( b d) + c e f d + bc bd ( b e ) ( c + f d e ) e f bf + ce df e f (d + bc)e (bd)f (e)(df) + (bf)(ce) (bf)(df) de + bce, nd bdf def + bfce bdf 2 de + bce. bdf 4
This should persude you of three things:. Q is field! 2. It cn be relly tedious to check if something is field. 3. Despite the fct tht it s tedious, it s not ctully tht hrd to check if something is field: every single property we tested ws pretty trivil to prove, there were just lot of them to check. So: we ve discovered two fields, R nd Q. Are there more? 2 Complex Numbers A commonly-sked question in mthemtics is the following: Given some polynomil P (x), wht re its roots? Depending on the polynomil, we ve seen severl techniques for finding these roots (Rolle s theorem, qudrtic/cubic formuls, fctoriztion.) However, t times we hve encountered polynomils tht hve no roots t ll, like x 2 +. Yet, despite the observtion tht this polynomil s grph never crossed the x-xis, we cn still use the qudrtic formul to find tht this polynomil hd the forml roots 0 ± 4 2 ±. The number, unfortuntely, isn t rel number (becuse x 2 0, for ny rel x) so this polynomil hs no roots over R. This ws rther frustrting block to run into; often, we like to fctor polynomils entirely into their roots, nd it would be quite nice if we could lwys do so, s opposed to hving to worry bout irreducible functions like x 2 +. So: wht if we just dd in into the rel numbers? Formlly, define the set of complex numbers, C, s follows: Definition. The set of complex numbers, denoted C, is the set of ll objects of the form + bi, where, b re rel numbers nd i. We cll the -piece of complex number the rel prt of the complex number, nd the bi -piece of complex number the imginry prt of the complex number. Given two complex numbers + bi, c + di, we define their sum s the object ( + bi) + (c + di) ( + c) + (b + d)i. Becuse +c, b+d re both rel numbers, this result is in the form tht we request complex numbers to be in, nd therefore is complex number. Similrly, given two complex numbers +bi, c+di, we define their product s the object ( + bi) (c + di) c + (di) + (bi)c + (bi)(di) c + (d + bc)i + (bd)i 2 c + (d + bc)i + bd( ) 2 c + (d + bc)i bd (c bd) + (d + bc)i. 5
Becuse c bd nd d + bc re both rel numbers, the result is gin in the form tht we request complex numbers to be in, nd therefore is complex number. As specil cse, the definition of the product mens tht i 2 ; check this with the bove formul if you don t believe it! Grphiclly, we cn visulize the complex numbers s plne, where we identify one xis with the rel line R, the other xis with the imginry-rel line ir, nd mp the point + bi to (, b): ir z +bi r θ b R We ve mentioned the complex numbers in the middle of lecture bout the concept of fields. Lecture theory suggests the following question: Is C field? The nswer, s you my hve guessed, is yes! We ve lredy shown tht few of the required properties of fields re stisfied: for instnce, we ve lredy shown tht both ddition nd multipliction re closed, becuse when we defined these opertions we took specil cre to note tht the output of both opertions ws complex number. Things like ssocitivity, distributivity nd commuttivity re shown just like how we did for the rtionl numbers; we reserve few of these proofs for the homework, nd leve the others for the interested reder to pursue on their own. Similrly, you cn esily show tht the dditive identity is 0 + 0i nd tht the multiplictive identity is + 0i; we gin reserve these proofs for the reder to check on their own. It bers noting tht like with Q, we will often write 0 insted of 0 + 0i nd insted of + 0i, for brevity s ske. Inverses, however, re weirder. Well: the dditive inverse is exctly wht we d expect: given ny + bi, the number ( ) + ( b)i is its inverse, becuse ( + bi) + (( ) + ( b)i) ( + ( ) + (b + ( b))i 0 + 0i 0. But the multiplictive one is weirder! Suppose we hve ny + bi 0; wht cn we multiply it by to get? This is tricky to figure out t first. Accordingly, becuse we re mthemticins, the first response we hve to problem being tricky is to try working on simpler problem. In 6
prticulr, mybe is hrd to get. Wht if we just wnted to multiply + bi by something to get rel number? This seems more promising. In prticulr, trick tht you might remember from polynomil multipliction is the observtion tht ( + bi)( bi) 2 bi + bi b 2 i 2 2 b 2 ( ) 2 2 + b 2. This is rel number, s it doesn t hve ny imginry prt! So we ve chieved our esier gol: we cn multiply ny complex number + bi by the quntity bi, nd get something tht s rel, even if it s not necessrily. This is useful. (So useful, in fct, tht mthemticins hve nme for it! Given ny complex number z + bi, we cll the complex number bi the conjugte of z, nd write it s z.) In fct, it s pretty much our nswer! We know how to form the inverse of rel number r: s long s the rel number r in question is nonzero, then its inverse is just r. We know tht becuse + bi 0, one of or b is nonzero; therefore, we know tht 2 + b 2 is nonzero, becuse one of 2, b 2 is nonzero nd they re both nonnegtive (becuse they re squred rel numbers!) Therefore, we know tht is the inverse of 2 + b 2, nd 2 +b 2 therefore tht ( + bi)( bi) Therefore, we hve tht ( bi) 2 + b 2 ( 2 bi + bi b 2 i 2) 2 + b 2 ( 2 b 2 ( ) 2) 2 + b 2 2 + b 2 2 + b 2. 2 + b 2 2 + b 2 + b 2 + b 2 i is the inverse of + bi, for ny i. Becuse, b re rel numbers such tht one of them is b nonzero, we know tht both, re both rel numbers; therefore this inverse is 2 +b 2 2 +b 2 complex number! So we hve shown tht complex numbers hve multiplictive inverses. 3 Other fields? We hve shown tht R nd Q re fields, nd (fter doing the HW) you should be convinced tht C is field too. This covers pretty much ll of the exmples tht we re going to use in this clss. However, in our lst bit of lecture, I should mention tht there re fr strnger fields out there tht we ctively use in modern reserch 2! To understnd them, consider the following object: 2 Specificlly, the reserch re of elliptic curve cryptogrphy, one of the pieces of technology tht you use every dy in encrypting your wifi, relies hevily on the fields we study here. 7
Definition. The set C, of clock numbers, is defined long with n ddition opertion + nd multipliction opertion s follows: Our set is the numbers {0,, 2, 3, 4, 5, 6, 7, 8, 9, 0, }. Our ddition opertion is the opertion ddition mod 2, or clock rithmetic, defined s follows: we sy tht + b c mod 2 if the two integers + b nd c differ by multiple of 2. Another wy of thinking of this is s follows: tke clock, nd replce the 2 with 0. To find out wht the quntity + b is, tke your clock, set the hour hnd so tht it points t, nd then dvnce the clock b hours; the result is wht we cll + b. For exmple, 3 + 5 8 mod 2, nd + 3 2 mod 2. This opertion tells us how to dd things in our set. Similrly, our multipliction opertion is the opertion multipliction mod 2, written b c mod 2, nd holds whenever +b nd c differ by multiple of 2. Agin, given ny pir of numbers, b, to find the result of this clock multipliction, look t the integer b, nd dd or tke wy copies of 2 until you get number between 0 nd. For exmple, 2 3 6 mod 2, 4 4 4 mod 2, nd 6 4 0 mod 2. We often will denote this object s Z/2Z, +,, insted of s C. This is not field. To see this, first show tht is the multiplictive identity of this field: this is not very hrd, nd is good exercise to mke sure you understnd wht s going on. Now, becuse is the multiplictive identity, we know tht if this is field, ny element x Z/2Z should hve multiplictive inverse x tht we cn multiply x by to get to. However, you cn check tht no such multiplictive inverse exists for 2. This is not hrd to see: tke ny other element y in Z/2Z, nd multiply 2 by y: you get 2y, n even integer. Adding or subtrcting multiples of 2 to 2y will not chnge this property: becuse both 2 nd 2y re even, the result of these opertions will lwys be even., however, is not n even number. Therefore, there is no wy for 2y to be equl to, no mtter wht y we pick from our set. Therefore, 2 hs no multiplictive inverse! However, some smll vritions on this object re fields: Definition. The object Z/nZ, +,,i.e.s defined s follows: Your set is the numbers {0,, 2,... n }. Your ddition opertion is the opertion ddition mod n, defined s follows: we sy tht + b c mod n if the two integers + b nd c differ by multiple of n. For exmple, suppose tht n 3. Then + 2 mod 3, nd 2 + 2 mod 3. Similrly, our multipliction opertion is the opertion multipliction mod n, written b c mod n, nd holds whenever + b nd c differ by multiple of n. For exmple, if n 7, then 2 3 6 mod 7, 4 4 2 mod 7, nd 6 4 3 mod 7. 8
There re mny vlues of n for which this is lwys field! You will study some of these vlues on the homework. We run one smple clcultion here: Clim. Z/5Z, +, is field. Proof. So: we re working with the set {0,, 2, 3, 4}, with the opertions + nd tken mod 5. We first note tht the opertions +, re closed. This is becuse given ny pir of numbers in {0,, 2, 3, 4}, their sum nd product re positive integers. If we tke wy copies of 5 one-by-one from ny positive integer, we must eventully lnd in the set {0,, 2, 3, 4}, s it is impossible to subtrct 5 from positive integer tht is not in this set (i.e. n integer greter thn or equl to 5) nd get negtive integer. (If you don t see why, drw number line nd try subtrct copies of five one by one lgorithm with some smple numbers!) We lso note tht the commuttivity, ssocitivty nd distribuitivity xioms ll hold. To see this, notice tht before tking the mod opertion, our + nd opertions re precisely the sme s those for R; therefore, before pplying mods, these two opertions preserve these properties. So: ll of these properties re equlity sttements (i.e. ( + b) + c + (b + c)). These sttements ll hold before we pply the mod opertion, becuse they hold in R. Therefore, when we pply the mod opertion to both sides, we re pplying it to the sme thing on both sides! Becuse we re strting from the sme number on both sides when we perform our dd or tke wy copies of 5 lgorithm, we cn t possibly get these two sides to go to different numbers in {0,, 2, 3, 4}: we re strting from the sme plce nd performing the sme lgorithm tht lwys hs unique output! (If you don t see why our mod lgorithm of dd or tke wy multiples of 5 until you re in the set {0,, 2, 3, 4} lwys hs the sme output from ny set input, prove this to yourself.) Therefore, we preserve these xioms. To see the identity nd inverse properties for + nd, consider the following ddition nd multipliction tbles: + 0 2 3 4 0 0 2 3 4 2 3 4 0 2 2 3 4 0 3 3 4 0 2 4 4 0 2 3 0 2 3 4 0 0 0 0 0 0 0 2 3 4 2 0 2 4 3 3 0 3 4 2 4 0 4 3 2 Notice how in the 0-row of the ddition tble, dding 0 to ny element doesn t chnge tht element. Therefore 0 is n dditive identity! Similrly, notice tht there is 0 in every row nd column in the ddition tble: this mens tht given ny element in Z/5Z, there is some other element we cn dd to it to get to 0. Therefore, we hve dditive inverses. As well, notice how in the -row of the multipliction tble, multiplying by ny element doesn t chnge tht element. Therefore, is multiplictive identity. Similrly, notice tht there is in every row nd column of the multipliction tble not corresponding to 0: this mens tht given ny element in Z/5Z, there is some other element we cn multiply by it to get to. Therefore, we hve multiplictive inverses. So this is field! A field with finitely mny elements, which is weird. 9