Proofs and Theorems for Exam

Here are some proofs that I may need iterate on the exam. Following that are some theorems that I'll probably need to know.

• Claim: A finite integral domain is a field.
  Pf: Let D be a finite integral domain with unity 1. Let a be any nonzero element in D. I will show that a is a unit. If a=1, it is its own inverse, so I am done. So, assume that a =! 1. Then, let D contain a, a^2, a^3, a^4.... D is finite by assumption, so there must be two positive integers i and j such that i>j and a^i=a^j. Then, a^(i-j)*a^(j)=1*a^j. By cancellation, a^(i-j)=1. Since a=!1, i-j>1. Then, a*a^(i-j-1)=1. So, a^(i-j-1) is the inverse of a. qed.
  
• Claim: The characteristic of an integral domain is 0 or prime.
  Pf: It suffices to show that if the additive order of 1 is finite, it must be prime, because, in a ring with unity, the additive order of 1 is characteristic of the ring. Suppose that 1 has order n and that n=st, where 1<=s, t<=n. Then, 0=n*1=(st)*1=(s*1)*(t*1). There are no zero-divisors in an integral domain, so either s*1=0 or t*1=0. Since n is the least positive integer with the property that n*1=0, we must have s=n or t=n. Thus, n is prime. qed.
  
• Let R be a commutative ring with unity and let A be an ideal of R.
  Claim: R/A is an integral domain iff A is prime.
  Pf: Suppose that R/A is an integral domain and ab in A. Then, (a+A)(b+A)=ab+A=A. So, either a+A=A or b+A=A; that is, either a in A or b in A. Then, A is prime. To prove the converse, observe that R/A is a commutative ring with unity for any proper ideal A. Suppose that A is prime and that (a+A)(b+A)=0+A=A. Then, ab in A. Then, since A is prime, a in A or b in A. Then, a+A=A or b+A=A. That is, there can be no zero-divisors. qed.
  

There are two more that I am encouraged to know, but I don't think I'll need to know proofs. He generally only asks for proof of a theorem found in the book, but gives us three or four different options and lists five different theorems on the study guide. So, I'll be fine if I know these three.
Here are some theorems that I'll probably need to know:

• Cancellation: Let a, b, c be in an integral domain. If a=!0 and ab=ac, then b=c. NOTE: THIS IS NOT DIVISION.
  
• A finite integral domain is a field.
  
• For every prime p, Z(p) is a field.
  
• Let R be a ring with unity 1. If 1 has infinite additive order, then the characteristic of R is 0. Else, if 1 has order n, then the characteristic of R is n.
  
• The characteristic of an integral domain is 0 or prime.
  
• R/A is an integral domain iff A is prime.
  
• R/A is a field iff A is maximal.
  
• The kernel of any homomorphism is an ideal the ring.
  
• Every ideal A of a ring R is the kernel of a ring homomorphism r-->r+A from R to R/A.
  
• A ring with unity contains Z(n) or Z.
  
• A field contains Z(p) (if char=n) or Q (if char=0).
  

Ok, that's about all I need to know. Not bad, really. Its just the first midterm. Not bad at all.

Study Guide for Ring and Field Theory

Here are some definitions for the upcoming Ring and Field Theory midterm.

• Ring: A ring R is a set with two binary operations -- generally called addition and multiplication -- such that, for all a, b, c in R, the following properties hold:
  

1. a+b=b+a (the ring is commutative under addition)
2. there exists an additive identity element called 0 such that 0+a=a+0=a
3. there exists an additive inverse of each element such that a+(-a)=0
4. a+(b+c)=(a+b)+c (the ring is associative under addition)
5. (ab)c=a(bc) (the ring is associative under multiplication)
6. a(b+c)=ab+ac and (b+c)a=ba+ca (multiplication distributes over addition.
   Examples: Z; nZ; M2(Z), or the ring of 2x2 matrices with integer entries; Z[x], ring of polynomials with integer coefficients

• Zero-divisor: In a ring R, an element a is a zero-divisor if there exists an element b such that ab=0.
  Examples: in Z(4), 2 is a zero-divisor.
  
• Integral domain: An integral domain is a ring R such that R has unity (a multiplicative inverse), R is commutative, and there are no zero-divisors.
  Examples: Z; M(2)Z; Z(p), where p is prime; NOT Z(n), where n is NOT prime.
  
• Field: A field F is a ring such that F has unity, F is commutative, and every nonzero element in F has a multiplicative inverse.
  Examples: R; Q; any finite integral domain; Z3(i).
  
• Characteristic of a ring: The characteristic of a ring R is the smallest positive integer n such that, for all a in R, na=0. If no such positive integer exists, then the characteristic is 0.
  Examples: In Z(n), the characteristic is n; in Z, the characteristic is 0; in Z[x], the characteristic ic 0.
  
• Ideal: An ideal I is a subring of a ring R such that, for all a in I and for all r in R, ar and ra are in I.
  Examples: In Z, nZ is an ideal; in Z[x], nZ[x] is an idea; in a commutative ring R with 1 and a, the set , where 'a'={ra| r in R} is called the principle ideal generated by a; similarly, the set 'a,b'=(r1a+r2b| r1, r2 in R}, and so on.
  
• Prime ideal: An ideal I is prime if it is a proper ideal and if ab in I implies that a is in I or b is in I.
  Examples: In Z, nZ is a prime ideal iff n is prime.
  
• Maximal idea: A proper ideal I is maximal to R if, given an ideal I', I subsets I' implies that I=I' or that I'=R.
  Examples: In Z, nZ is maximal iff n is prime; is maximal in Z[x].
  
• Ring Homomorphism: A ring homomorphism P is a mapping from a ring R to a ring S such that, for all a,b in R, P(a+b)=P(a)+P(b) and P(ab)=P(a)P(b), that is, if P preserves both operations.
  Examples: if P maps all elements in R to 0, then P is a homomorphism; there is no homomorphism from 2Z to 3Z (or to 4Z); for any pos int n, the mapping from k to kmodn is a ring homomorphism.
  

Proofs to come soon.
Bubye.

Nasty little problem,,,

This is a problem that's been stumping me for a few hours now.
Let R be a commutative ring with unity. Let x and y be in R, such that x is a unit and y is nilpotent.
Claim: (x+y) is a unit.
Side Bar: Ok, here's what that means: x is unit implies that there exists b in R such that xb=1, or, that x has a multiplicative inverse. y is nilpotent implies that there is a positive integer n such that y^n=0. I want to show that there exists c in R such that (x+y)c=1, or that (x+y) has a multiplicative inverse. Observe that, if we set c to y^(n-1), then we get x*y^(n-1). If I can show that that has a multiplicative inverse, then I'll be done. But I'm not sure that that is the way to go. How about this instead: (x+y)(a+b)=1. Hmmm, that has potential. Then, xa+xb+ya+yb=1. Now, let a=x and let b=-y. Then, xa+xb+ya+yb=x^2-y^2. Wait, multiply this by x^2+y^2. Then, we have x^4-y^4. Do this enough times, and eventually, you'll get to y^n or y to some number that is greater than n, which amounts to the same thing, namely, y goes to 0. Then, we are left with x^m. Does x^m have a multiplicative inverse? Well, of course, it does. x^m=x*x^(m-1), and x is a unit. So, multiply x^m by x^-m and you get 1. Pretty facile, I know, but it's worth explaining these things. So, now for the actual proof.
Pf: Observe that multiplying x+y by (x^-m)(x-y)*II(x^i+y^i), i=2, m, where m is sufficiently large (n is fine), you have (x^-m)(x^m-y^m). But, y^m=0, so, you have x^-m*x^m, which is 1. So, x+y is a unit; specifically, its multiplicative inverse is (x^-m)(x-y)*II(x^i+y^i), i=2, m, where m is a number such that the product of the nonzero even numbers preceding m is greater than or equal to n. QED.

I hope... I bet there is an easier way to SHOW that there must be an inverse, other than actually FINDING the inverse... I'll keep thinking on this.

Welcom Back to Blogdom

Ok, ok, I'm back. Keep your yappers shut, I know you all missed me.
So I'm taking Ring and Field Theory this term, which is a branch of group theory.
Here is a proof that has been stumping me for a bit now. I think it's clever more than anything...

Suppose that R is a commutative ring and |R|=30, with an ideal I, where |I|=10.
Claim: I is a maximal ideal of R.
Sidebar: The order of R is 30, and, by Lagrangre's Theorem, the order of any subset divides the order of the group. Then, there are supgroups of order 10 and order 15 (call them I and J). But we aren't talking about groups, we're talking about rings. We know that there is a subring of order 10 (I). So, I need to show that I is not a subset of J or that J is not a subring (although it is a subgroup). A third possibility is to show that all groups of order 30 are isomorphic to Z(30), at which point I can fairly easily prove that I is maximal. No, I just Googled it: not every group of order 30 is isomorphic to Z(30). Wait. Wait. Wait. In I, there are elements of order 2 (because |I|=10=5*2), whereas in J, there are no elements of order 2 (because |J|=15=5*3). So, I is not a subset of J. And, by Lagrange's Theorem, there can be no other subgroups. So, I is maximal. Incidentally, J is maximal, too. Now, the formal proof.
Pf: Let R be a commutative ring, with |R|=30, and two ideas, I and J, where |I|=10 and |J|=15. By Lagrange's Theorem, there can be subgroups (and, so, subrings) of orders that divide the order of the group (or, rather, ring). Then, there can be no proper subring of order greater than 15. So, either I is maximal or J is maximal (or both). If I is not maximal, then I is a subset of J. However, I contains elements of order 2 (by Lagrange's Theorem and because 10=2*5), whereas J does not (by Lagrange's Theorem again, and because 15=3*5). So, I cannot be a subset of J. So, I is maximal. Incidentally, if there even exists a subring of order 15, then J is also maximal. QED.

Here's another that's been stumping me:
Question: Is the ring 2Z isomorphic to the ring 3Z?
Answer: Assume that 2Z is isomporphic to 3Z, where P is such an isomorphism. Then, by principal, 3Z is isomorphic to 2Z, where P(inv) is such an isomorphism. The ring 3Z has unity 1, so P(inv)(1)=1*, where 1* represents the unity of 2Z. However, no such unity exists. Then, 1 cannot map to 1*, so there is no such isomorphism P(inv). So there is no such isomorphism P, which contradicts our initial assumption.
Question: Is the ring 2Z isomorphic to the ring 4Z?
Answer: This one is a little more tricky. It feels obvious to me, frankly, that they are. They are both infinitely countable, so any mapping can be one-to-one and onto. Hmmm, let's see. Let P be such an isomorphism. Then, P(2)=4n, n in Z. Then, P(2*2)=P(2+2), implies that P(2)*P(2)=P(2)+P(2), which implies that 4n*4n=4n+4n, which implies that 16n^2=8n, which implies that 2n=1, which is impossible. So, we have a contradiction, so P cannot exist. Well, well, well, my intuition was wrong. The two aren't isomorphic. Incidentally, I can use this same method for the previous problem.

That all took me way too long. Time to get it formally written up, then off to read more ethics.

Good Morning, Proof!

Here is the third proof (or, rather, set of proofs) I'll be expected to know.
Properties of Elements and Subgroups Under Homomorphisms
Let G and G' be groups, where P maps G into G', and H is a subgroup of G.

1. Claim: P maps the identity of G to the identity of G'.
   Pf: This proof is the same as the proof for property 1 of the previous set of proofs. But I'll do it again for practice. P(e)=P(ee)=P(e)P(e), and e'P(e)=P(e). So, e'P(e)=P(e)P(e), implying that e'=P(e). qed
2. Claim: P(g^n)=[P(g)]^n.
   Pf: Again this is the same as property 2 of the previous set of properties. If n is positive, then the claim follows by mathematical induction and by the definition of homomorphisms. If n=0, then the claim follows by the previous property. If n is negative, then -n is positive. Then, P(e)=P((g^n)(g^-n))=P(g^n)P(g^-n)=P(g^n)[P(g)]^-n. Then, P(g^n)=[P(g)]^n. qed
3. Claim: If |g| is finite, then |P(g)| divides |g|.
   Pf: Let |g|=n. Then, g^n=e. Then, e=P(e)=P(g^n)=[P(g)]^n. Then, by Thm 4.2, |P(g)| divides n=|g|. qed
4. Claim: KerP is a subgroup of G.
   Pf: By property 1, KerP is nonempty. Let a,b be in KerP. Then, P(a)=P(b)=e. Then, P(ab^-1)=P(a)P(b^-1)=P(a)[P(b)]^-1=ee^-1=e. Then, ab^-1 is in KerP, so KerP is a subgroup of G. qed
5. Claim: P(a)=P(b) iff aKerP=bKerP.
   Pf: I will prove the statement "If P(a)=P(b), then aKerP=bKerP" first. Assume that P(a)=P(b). Then, e=P(a)[P(b)]^-1=P(a)P(b^-1)=P(ab^-1), which is in KerP. By property 5 of the Lemma of Cosets, iff ab^-1 is an element of a subgroup H, then aH=bH. Then, the claim follows. Now, conversely, assume that aKerP=bKerP. Then, by the same property of the Lemma of Cosets, P(ab^-1) is in KerP, and so on backwards to P(a)=P(b). qed

There is a bunch more, but I think I get the general idea. Besides, I'll be asked to prove one of several, and I doubt I'd choose this one. I'll probably choose the set of properties in the last posting. Ok, back to PHL.

Another Group Proof

Need a breather from PHL, so here's another group theory proof.
Properties of Isomorphisms Acting on Elements and Subgroups
Let G and G' be groups, and let P map G to G'. Let H be a subgroup G.

1. Claim: P maps the identity of G to the identity of G'
   Pf: Because e=ee, P(e)=P(ee)=P(e)P(e). Also, because P(e) is in G', P(e)=e'P(e). Then, e'P(e)=P(e)P(e) --> e'=P(e). qed
2. Claim: For every integer n and for every group element a in G, P(a^n)=[P(a)]^n.
   Pf: By definition of isomorphisms and by mathematical induction, if n is positive, then the claim is true. If n=0, P(a^n)=P(a^0)=P(e) and [P(a)]^n=[P(a)]^0=e. If n is negative, then -n is positive, and we have from the first claim and mathematical induction that e=P(e)=P([g^n][g^-n])=P(g^n)P(g^-n)=P(g^n)[P(g)]^-n =e. Then, P(g^n)=[P(g)]^-(-n)=P(g)^n. qed
3. Claim: For any elements a and b in G, a and b commute iff P(a) and P(b) commute.
   Pf: I will prove the claim "P(a) and P(b) commute if a and b commute" first. Assume that a and b commute. Then, ab=ba. P(ab)=P(a)P(b) and P(ab)=P(ba)=P(b)P(a), so P(a)P(b)=P(b)P(a). Now I will prove the converse. Assume that P(a) and P(b) commute. Then, P(a)P(b)=P(b)P(a). Then, P(a)P(b)=P(ab)=P(b)P(a)=P(ba), so P(ab)=P(ba). Because P is one-to-one, ab=ba. qed
4. Claim: G='a' iff G'='P(a)'.
   Pf: I will prove the claim "G'='P(a)'" first. Let G='a'; then, by closure, 'P(a)' is a subset of G'. Because P is onto, for any element b in G', there is an element a^k in G such that P(a^k)=b. Thus, b=[P(a)]^k, so b is in 'P(a)'. Then, G'='P(a)'. Now I will prove the converse. Suppose that G'='P(a)'. 'a' is a subset of G (by closure, etc.). For any b in G, we have P(b) is an element of 'P(a)'. Then, for some integer k, we have P(b)=[P(a)]^k=P(a^k). Because P is one-to-one, b=a^k. Then, G='a'. qed
5. Claim: Isomorphisms preserve orders.
   Pf: Let g be in G, where |g| is n. Then, P(g^n)=P(e)=e and P(g^n)=[P(g)]^n. So, [P(g)]^n=e. So, |P(g)|=|g|. qed
6. Claim: If G is finite, then G and G' have exactly the same number of elements of every order.
   Pf: By property 5, elements of order n map to elements of order n. Then, G and G' must have the same number of elements of each order. qed
7. Claim: G is Abelian iff G' is Abelian.
   Pf: By property 3, this property follows.
   
8. Claim: G is cyclic iff G' is cyclic.
   Pf: By property 4, this property follows.
   
9. If K is a subgroup of G, then P(K) is a subgroup of G'.
   Pf: Let a,b be in K, and P(a)=c and P(b)=d. I need to show that cd^-1 is in P(K). By definition, c and d are in P(K). P(ab^-1)=P(a)P(b^-1)=P(a)[P(b)]^-1=cd^-1, which is in P(K) by closure.
   

That was fun. Now, back to PHL.

Here's a proof to kick off my monster study sess

So I'm supposed to be studying crazy hard for this PHL final, but I'm not really feelin' it. So I'm going to spend a little time doing a proof or two for Group first, to get me motivated. Here goes:
The Fundamental Theorem of Cyclic Subgroups
Claim: Every subgruop of a cyclic group is cyclic. Moreover, if |'a'|=n, then the order of any subgroup of 'a' is a divisor of n; and, for each positive divisor k of n, the group 'a' has exactly one subgroup of order k -- namely, 'a^(n/k)'.
Pf: Let G='a' and suppose that H a subgroup of G.
To prove the first claim (that every subgroup of a cyclic group is itself cyclic), we need to show that H is cyclic. Assume that H does not equal just the identity e. I claim that H contains an element of the form a^t, where t is positive. Since G='a', every element of H has the form a^t; and, when a^t belongs to H with t<0, h="'a^m'." g="'a'," b="a^k" k="mq+r," k="a^(mq+r)="(a^mq)(a^r)-">a^r=(a^-mq)(a^k); since a^k and a^-mq are both in H, a^r is in H. However, m is the least positive integer such that a^m is in H, but r is less than m and greater than or equal to 0. So, r=0, implying that b=a^k=a^-mq=(a^m)^q which is in 'a^m'. Then, the arbitrary element b is in the cyclic subgroup of H 'a^m', so H='a^m', so H is cyclic. This proves the first claim.
To prove the second claim (if |'a'|=n, then the order of any of 'a' will be a divisor of n), suppose that |'a'|=n and H is a subgroup of 'a'. H is cyclic, so let H='a^m', where m is the least positive integer such that a^m is in H. Let b=a^mq again, such that a^mq=e. Then, e=a^mq=b=a^n, implying that mq=n. So, the order of the subgroups of a cyclic group divide the order of the group.
To prove the third and final claim (for each positive divisor k of n, the group 'a' has exactly one subgroup of order k, 'a^(n/k)'), let k be any positive divisor of n. I need to show that 'a^(n/k)' is the one and only subgroup of 'a' of order k. |'a^(n/k)'|=n/gcd(n,n/k)=k. Now, let H be any subgroup of 'a' of order k. I know that H='a^m', where m divides n. Then, m=gcd(n, m) and k=|a^m|=|a^gcd(n,m)|=n/gcd(n,m)=n/m. Thus, m=n/k, and H='a^(n/k)'.
qed

That was pretty confusing. I need to work on that one.
Ok, PHL time. I'll publish another one in a bit.