12-240/Classnotes for Tuesday September 18

DBN: Classes: 2012-13: 12-240 / Navigation Additions to this web site no longer count towards good deed points. # Week of... Notes and Links 1 Sep 10 About This Class, Tuesday, Thursday 2 Sep 17 HW1, Tuesday, Thursday, HW1 Solutions 3 Sep 24 HW2, Tuesday, Class Photo, Thursday 4 Oct 1 HW3, Tuesday, Thursday 5 Oct 8 HW4, Tuesday, Thursday 6 Oct 15 Tuesday, Thursday 7 Oct 22 HW5, Tuesday, Term Test was on Thursday. HW5 Solutions 8 Oct 29 Why LinAlg?, HW6, Tuesday, Thursday, Nov 4 is the last day to drop this class 9 Nov 5 Tuesday, Thursday 10 Nov 12 Monday-Tuesday is UofT November break, HW7, Thursday 11 Nov 19 HW8, Tuesday,Thursday 12 Nov 26 HW9, Tuesday , Thursday 13 Dec 3 Tuesday UofT Fall Semester ends Wednesday F Dec 10 The Final Exam (time, place, style, office hours times) Register of Good Deeds Add your name / see who's in!

Various properties of fields

Thrm 1: In a field F: 1. a+b = c+b ⇒ a=c

2. b≠0, a∙b=c∙b ⇒ a=c

3. 0 is unique.

4. 1 is unique.

5. -a is unique.

6. a^-1 is unique (a≠0)

7. -(-a)=a

8. (a^-1)^-1 =a

9. a∙0=0 **Surprisingly difficult, required distributivity.

10. ∄ 0^-1, aka, ∄ b∈F s.t 0∙b=1

11. (-a)∙(-b)=a∙b

12. a∙b=0 iff a=0 or b=0

. . .

16. (a+b)∙(a-b)= a^2 - b^2 [Define a^2 = a∙a] Hint: Use distributive law

Thrm 2: Given a field F, there exists a map Ɩ: Z → F with the properties (∀ m,n ∈ Z):

1) Ɩ(0) =0, Ɩ(1)=1

2) Ɩ(m+n) = Ɩ(m) +Ɩ(n)

3) Ɩ(mn) = Ɩ(m)∙Ɩ(n)

Furthermore, Ɩ is unique.

Rough proof:

Test somes cases:

Ɩ(2) = Ɩ(1+1) = Ɩ(1) + Ɩ(1) = 1 + 1 ≠ 2

Ɩ(3) = Ɩ(2 +1)= Ɩ(2) + Ɩ(1) = 1+ 1+ 1 ≠ 3

. . .

Ɩ(n) = 1 + ... + 1 (n times)

Ɩ(-3) = ?

Ɩ(-3 + 3) = Ɩ(-3) + Ɩ(3) ⇒ Ɩ(-3) = -Ɩ(3) = -(1+1+1)

What about uniqueness? Simply put, we had not choice in the definition of Ɩ. All followed from the given properties.

At this point, we will be lazy and simply denote Ɩ(3) = 3_f [3 with subscript f]

∃ m≠0, m ∈ N, Ɩ(m) =0

In which case, there is a smallest m<0, for which Ɩ(m)=0. 'm' is the characteristic of F. Denoted char(F). Examples: char(F_2)=2, char(F_3)=3... but NOTE: char(R)=0

Thrm: If F is a field and char(F) >0, then char(F) is a prime number.

Proof: Suppoer char(F) =m, m>0. Suppose also m is not prime: m=ts, t,s ∈ N.

Then, Ɩ(m) = 0 = Ɩ(t)∙Ɩ(s) ⇒ Ɩ(t)=0 or Ɩ(s)=0 by P12.

If Ɩ(t)=0 ⇒ t≧m ⇒ m=t, s=1 or likewise for Ɩ(s)=0, and m=s, t=1

In any factorization of m, one of the factors is m and the other is 1. So m is prime. ∎

Lecture 3, scanned notes upload by Starash

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