330 Problem Set 5 Comments

Problem set 5:

18.37  1*1 = 1, so 1 is a unit and in U.  As U is a subset of ring and multiplication in a ring is associative (R_2).  Because a*a^{-1}=1, if a is a unit (has an inverse), then a^{-1} does also (a itself).  Recall the inverse of ab is b^{-1}a^{-1], and so if we assume and b have inverses, then ab does also.  Be careful here, this seems logical, but … why is b^{-1}a^{-1} in U?  In fact, that’s exactly  what we’re proving!  This is subtle.  b^{-1}a^{-1}  is in R because R is closed under multiplication because multiplication is a binary operation.  We’re not showing that b^{-1}a^{-1} is in U, we’re showing that ab times something in R gives 1.  Interesting. 

18.38  This seems much easier than 37.  (a+b)(a-b) = a^2 + ba - ab -b^2.  This equals a^2-b^2 iff ba-ab=0, which is true iff ab=ba.   This is true for all a,b iff R is commutative.

18.39  I think this is also easier than 37.  You need to check the axioms. 
a(bc) = a0=0 and (ab)c = 0c = 0√
a(b+c) = 0 and ab + ac = 0+0=0√
(a+b)c = 0 and ac + bc = 0+0=0√

18.44  For the first part (ab)^2=(ab)(ab) = abab = aabb=a^2b^2=ab where the middle = uses commutativity

0^2=0 and 1^2=1 always
Notice that in Z_6
2^2 = 4 so not
3^2= 9mod6=3 √
4^2=16mod6=4 √
5^2=25mod6=1 x

And in Z_12
2^2= 4 x
3^2=9 x
4^2 = 16mod12 = 4 √
5^2 = 25mod12 = 1 x
6^2 =36 mod12 = 0 x
7^2=49mod12=1 x
8^2=64mod12 = 4 x
9^2 =81mod12= 9√
10^2 = 100 mod 12 = 4x
11^2=121mod12 =1 x

So, all ordered pairs where the first element is 0, 1, 3, or 4, and the second element is 0, 1, 4, 9. 

18.45  Fun to think of linear algebra for a change … although really this is just running through symbols.  If you write out P^2 you get a nifty cancelation in the middle of A^TA. 

18.47  I think this is best done with double contrapositive, since there is so much negativity around.  That changes the statement to show that a ring has a nonzero nilpotent element iff there is a nonzero solution to x^2=0.  One direction is short.  Suppose a is a nonzero solution to x^2=0, then a^2=0 and this satisfies the definition of nilpotent. 
    The other direction isn’t bad.  Suppose a is a nonzero nilpotent element.  This says a^n=0, for a≠0.  Let us suppose that n is the minimal power so that a^n=0.  n > 1, and because it is an integer this means n ≥ 2.  Consider x=a^{n-1}, which is not zero by n being minimal.  I claim x^2 = 0.  x^2 = (a^{n-1})^2 = a^{2n-2}=(a^n)(a^{n-2}).  Notice because n ≥ 2, that the second factor is has an exponent that is non-negative and is less than n, so, the second factor is defined and is not zero, but the first is zero, therefore x^2=0.  Be careful not to use the  zero product property - there are surely zero divisors here.

18.48 Two directions.  If S is a ring, then it contains an additive identity as an additive group, and it is closed under inverses and addition, so contains a-b.  Also it is closed under multiplication. 

If these properties hold and S is a subset, then associativity of both operations comes from being a subset, along with commutativity of addition and both distributive properties.  Remembering exercise 5.45, we know S is nonempty because it contains the additive identity for R.  ab^{-1} written additively is a-b, which is our other condition, therefore S is a subgroup additively.  By the third condition, S is closed multiplicatively. 

18.49 Intersections seem to work great for objects.  Here's another example.  We already know that intersections of subgroups is a subgroup (7.4), and this proof is much the same as there and needs to be for arbitrarily many sets.  It is abelian, associative and obeys the distributive properties because it is a subset.  The only thing we really need here is to be closed under multiplication.  Because each of the subrings are, the intersection also is. 
    For fields, we also need closure under inverses, again this follows from it being true for each of the individual subfields.  That's the big one, but we also need a unity (which all the subfields would have), and commutative multiplication which is inherited as a subset. 

18.50 Do we want to use 18.48?  It seems tempting.  0 is in I_a because a0=0.  If ax = 0 and ay = 0, then a(x-y) = ax - ay = 0 - 0 = 0, so (x-y) is in I_a.  And a(xy) = (ax)y = 0y = 0, so xy is in I_a.  Looks clean. 

19.24 Intersection again.  We have from 18.49 that it is a ring.  Commutativity comes from subsethood.  If each of the subrings are integral domains, they all have identity, so the intersection does.  Finally, if there are no zero divisors in all of D, there will not be in any subset. 

19.26 a.  Suppose ax=0, for neither a nor x equal to zero.  Through an idea that I wouldn't have thought of, but which was suggested in office hours, using the special b for a, consider a(b+x)a = aba+axa (using both distributive properties) = a + 0a = a + 0 = a.  But, then b+x which is not b because x is not zero is another element satisfying b's property, but b was claimed to be unique.  This is a contradiction, so there are no zero divisors.

b. The ring has no zero divisors, so we have cancellation.  aba=a implies baba=ba by left multiplication by b, and we may cancel a from the right since a was assumed to begin with to be nonzero, hence bab=b.
c. Pick a nonzero a.  Consider ab.  I claim this is a unity.  We must show for any x, xab=x and abx=x.  We know that aba=a.  We can multiply on the left by xto get xaba=xa and then use cancellation to find xab=x, as desired.  For the other direction, start with bab=b.  Multiply on the right by x to get babx=bx and this time cancel b from the left (it is worth noticing b is not zero because otherwise aba=0, not a) to get abx=x.  It's worth noting that ab is our identity for any nonzero a and it's special b.

d.  To show a division ring, we need a ring with unity (we have that so far).  Our unity is not zero because there are no zero divisors, and a is not zero, so ab is not zero.  We need to show every nonzero element has an element that we can multiply it by to get the identity, ab.  Again, this is for any a and it's special b.  It's pretty apparent that ab=ab, so that direction is good.  We only  need to show that ba = ab.  We know that aba=a.  We also know that ab is the identity, so a(ab) = a.  From transitivity we see aba = aab and by canceling the nonzero a on the left we get ba=ab, as desired. 

19.27  Integral domains contain unity.  So, the subdomain contains 1 (it's unique), and we may use the test of 1 to check the characteristic.  The operation is inherited from D so, 1+1+ ... + 1 (n 1's) is the same in D as it is in any subdomain. Therefore, the number of copies necessary to reach zero remains the same. 

20.27  If x is its own inverse, x^2=1.  Equivalently x^2-1=0, i.e. (x-1)(x+1)=0.  Z_p is a field, so it has no zero divisors.  Therefore x-1 =0 mod p or x+1=0 mod p.  The first equation says x=1 modp.  The second says x=-1mod p, but p+(-1)=p-1, which is the other solution. 

20.29 Follow Example 20.5.  Note:  36=37-1, 36=(19-1)2, 36=(13-1)3, 36=(7-1)6, 36=(3-1)18, 36=(2-1)36.  That should give you all the cases. 

20.30 36=(5-1)9 also, so we can multiply 383838 by 5, but Fraleigh probably thought that number isn't as cute. I will give two extra points on PS5 to the first person who sends me a proof for a larger number than 1919190.

Now, what about: 

The map x |-> mx from Z_p to Z_q is a homomorphism iff (something about m, p, and q).  On tempting attempt was made with q|mp.  This is true for our example with p=4, q=10, and m=5, 10|4(5).  But, it is also true for p=q=5, m=anything, in particular m=2, which is not a homomorphism as 1*1=1, but 2(1)*2(1) is not 2(1) in Z_5. I'll keep this one on offer also.  Same deal - two extra points for the first correct proof.