Your SlideShare is downloading. ×
Denitions you should know:
• Ring
• homomorphism (and isomorphism, automorphism)
• R[X]
• integral domain, eld
• unit, zer...
2
Example. Show R × R × R × R and M2(R) are not isomorphic.
Idea: R × R × R × R is commutative, but M2(R) is not.
Suppose ...
3
Example. In this way, you can think of Z being in M2(R) by thinking of
1 0
0 1
+ · · · +
1 0
0 1
z times
=
z 0
0 z
as z ...
Upcoming SlideShare
Loading in...5
×

Math 110 review_session_2_notes

32

Published on

Logic and proof review

Published in: Science
0 Comments
0 Likes
Statistics
Notes
  • Be the first to comment

  • Be the first to like this

No Downloads
Views
Total Views
32
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
1
Comments
0
Likes
0
Embeds 0
No embeds

No notes for slide

Transcript of "Math 110 review_session_2_notes"

  1. 1. Denitions you should know: • Ring • homomorphism (and isomorphism, automorphism) • R[X] • integral domain, eld • unit, zero divisor, nilpotent • characteristic Theorems you should know: • all the nice polynomial stu (divisibility, gcds, roots, degree, etc.) • subring criterion • basic ring facts (0R · a = 0R for all a ∈ R, etc.) How to show a map f : R → S is an isomorphism: • Show it's a homomorphism: f(x + y) = f(x) + f(y) for all x, y ∈ R f(xy) = f(x)f(y) for all x, y ∈ R • Show it's injective: If f(x) = f(y), then x = y. • Show it's surjective: For all s ∈ S, there is x ∈ R such that f(x) = s. Example. Show that complex conjugation f : C → C, a + bi → a − bi is an automorphism. This is a homomorphism: For all a1 + b1i, a2 + b2i ∈ C: f((a1 + b1i) + (a2 + b2i)) = f((a1 + a2) + (b1 + b2)i) = (a1 + a2) − (b1 + b2)i = (a1 − b1i) + (a1 − b2i) = f(a1 + b1i) + f(a2 + b2i) f((a1 + b1i)(a2 + b2i) = f((a1a2 − b1b2) + (a1b2 + b1a2)i) = (a1a2 − b1b2) − (a1b2 + b1a2)i = (a1 − b1i)(a2 − b2i) = f(a1 + b1i)f(a2 + b2i) This is injective: If f(a1 + b1i) = f(a2 + b2i), a1 − b1i = a2 − b2i, so a1 = a2 and (−b1) = (−b2), so b1 = b2, and a1 + b1i = a2 + b2i. So f is injective. This is surjective: If a + bi ∈ C, f(a + (−b)i) = a − (−b)i = a + bi, so f is surjective. Thus f is an isomorphism How to show two rings are not isomorphic: assume they are, and get a contradiction somehow (look for some way the ring structures are dierent. 1
  2. 2. 2 Example. Show R × R × R × R and M2(R) are not isomorphic. Idea: R × R × R × R is commutative, but M2(R) is not. Suppose there is an isomorphism f : M2(R) → R × R × R × R. Consider a = 1 0 0 0 and b = 0 1 0 0 . f(ab − ba) = f(a)f(b) − f(b)f(a) = f(a)f(b) − f(a)f(b) = 0 (since R × R × R × R is commutative). But ab − ba = 0 1 0 0 , so then f 0 1 0 0 = 0 0 0 0 = f 0 0 0 0 , so f is not injective. Example. Show Z × Z2 not isomorphic to Z. Several possible ways: (0, 1) is a nontrivial idempotent in Z × Z2, but Z has no nontrivial idempotent. (1, 0) · (0, 1) = (0, 0), but there are no zero divisors in Z. (0, 1) + (0, 1) = (0, 1), but there is no such nonzero element in Z. (Assume there is an isomorphism, and use one of these (or something else), to get a contradiction) To show a polynomial is irreducible: • If degree is 2 or 3: show it does not have a root (rational roots theorem is one way in Q[x]) • If in Q[x] or Z[x]: Eisenstein. To show a polynomial is reducible: • Factor it. • Show it has a root. (long division of polynomials in Zp[x]) Characteristic of a ring. If R is a commutative ring with identity 1R, and z ∈ Z, even though z is not necessarily in R, we can kind of see z as being in R by taking 1R + · · · + 1R z times , which we write z·1R for short. If there is a positive integer n such that 1R + · · · + 1R n times = 0, we can actually view the elements of Zn as being contained in R in the same way: identify [z] with 1R + · · · + 1R z times . We call the smallest such n the characteristic of R. If there is no such n, we say that R has characteristic 0. On the rst exam, you showed that Z has characteristic 0 and Zn has characteristic n.
  3. 3. 3 Example. In this way, you can think of Z being in M2(R) by thinking of 1 0 0 1 + · · · + 1 0 0 1 z times = z 0 0 z as z ∈ Z. You can think of Z2 being in M2(Z2) by thinking of [1] [0] [0] [1] + · · · + [1] [0] [0] [1] z times = [z] [0] [0] [z] as [z] ∈ Z.

×