Still in Chapter 2 - field extensions. Then Chapter 3, for the idea of dimension (and a review of some linear algebra).

1. Sethuraman 54/4. We've actually discussed this in class. I'm asking you to write out the proof so you get more practice writing. I expect to see a large ratio of words to formulas - more like the text than the chicken scratches I put on the board.

   There are lots of parts to that problem - pretty much one per sentence. I'll put each part of the answer in a separate bullet below.

   Let p be a prime. Since it's not easy to put a sqrt sign on this web page, I will write S for sqrt(p).

   • To show that the ring R generated by Q and S is the set of numbers of the form a+bS for rational a and b all I need to show is that when you add or multiply two numbers of that form the answer is of that form. That's obvious for sums and almost obvious for products.

     	(a + bS)*(a' + b'S) = aa'+ bb'p + (ab' + a'b)S
     

     If a, b, a' and b' are rational then so are (aa'+bb'p) and (ab'+a'b) so this product is in Q[S].

   • Next I'm asked to show that if a+bS=0 then a=b=0. I'll do this considering two cases. If b=0 then 0=a+bS=a, so a=0 too. If b != 0 (I use the computer programming symbols "!=" for "is not =") then a+bS=0 implies S = a/b, which is rational since a and b are. But S is the square root of a prime, so it's not rational. That disposes of the assumption b=0.

   • The preceding theorem easily implies that if

     	a + bS = a' + b'S   (*)
     

     then a=a' and b=b'. Here's the proof. Equation (*) implies that

     	(a-a') + (b-b')S = 0
     

     which we know forces a-a'=0 and b-b'=0, hence a=a' and b=b'.

   • Finally, I'm going to show that R=Q[S] is actually a field, not just a mere ring. To do that I need to show that whenever a+bS in R is not 0 then it has a multiplicative inverse in R. Well suppose z = a+bS != 0. Then I know two things about it. First, I know a and b are not both 0, since I proved it above. Second, I know that z has a multiplicative inverse 1/z in the real numbers, since the real numbers do form a field. So I will play with that: in the real number ring

                1         1  (a-bS)    (a-bS)
     	------  = -----------  =  ----- 
     	(a+bS)    (a+bS)(a-bS)    a2-b2p
     

     Now I know what I need to know to write the proof going forwards. Given z = a + bS, let t = a²-b²p. Then t is not 0 (if it were then the prime p would = (a/b)², which is impossible). So let

     	w = (a/t) - (b/t)S
     

     which is clearly an element of R. I claim w is a multiplicative inverse of z. To prove the claim, just carry out the multiplication to see that wz=1. (I won't type it over again here. It has to work because of the preliminary algebra I did figuring out what w had to be.)

2. Sethuraman 88/2. This is very much like the problem two homeworks ago in which I asked you to prove that some things in the arithmetic of a ring that seemed obvious in fact could and should be proved to be consequences of the ring axioms. Almost everyone did a terrible job. Here's your chance to show that you learned something from the exercise and my comments on your paper. You are to prove that the vector space properties in Remark 3.3 (page 69) follow from the vector space axioms (Definition 3.1 on page 65).

   These are all short proofs in which every step must be justified. It's important to keep the notation clean, and to be sure to reason from what you are given to what you want. If you end your proof with something like "0=0" it isn't right. Be sure to use "=" only where you mean it, not as a synonym for "therefore".

   1. Theorem: For any scalar f, f*0 = 0.

      Proof: Since 0+0=0 in V (the vector space) I know (from the meaning of "=") that

      	f*(0+0) = f*0
      

      Then

      	f*0 + f*0 = f*0  (Axiom 1)
      

      Now I know that the vector f*0 has an additive inverse in V; call it w. Then adding w to both sides of the previous equation and using the associative law for + on the left tells me

      	f*0 = 0
      

      which is what I wanted to show.

   2. Theorem: For any vector v, 0*v = 0.

      Proof. This is similar to the last one. Start with 0+0=0 in F. Then

      	0*v = (0+0)*v = 0*v + 0*v   (Distributive law.)
      

      Let w be the additive inverse of 0*v in V and add it to both sides to reach the desired concusion.

   3. For any scalar f and vector V, (-f)*v = -(f*v).

      Proof. This seems obvious and is easy to prove, once you understand what's tricky about it. Remember that there is really no such thing as subtraction in either F or V. (-f) is just the way we write the additive inverse of f in F, and -(f*V) is the way we write the additive inverse of (f*V) in V. To show that the vector on the left in the theorem is the additive inverse of f*V we need to show that when we add it to f*V we get 0:

      	(-f)*v + f*V = (-f + f)*v   (distributive law)
      	             = 0*v          (meaning of -f in field F)
      	             = 0            (previously proved) 
      

   4. If v is a nonzero vector then f*v=0 implies f=0.

      Proof. Suppose f*v=0 but f !=0. Then since the scalars form a field, f has a multiplicative inverse g. Then v = 1*v = (g*f)*v = g*(f*v) = g*0 = 0, which is a contradiction. (In the preceding chain of equalities I used Axiom 4 for vector spaces, the associative law, and a fact I just proved.)

3. Extra credit. Let F be a field, and R = F[[x]] the ring of power series with coefficients in F.

   The double square brackets in F[[x]] is new notation. When R is a ring we write R[[x]] for the ring of "formal power series" in the variable x. That's the set of all expressions of the form a₀ + a₁x + a₂x² + ... where the coefficients ai come from the ring R. You add and multiply those objects using the usual rules for the arithmetic of exponents (which are ordinary integers) and the arithmetic of the ring R for adding and multiplying the coefficients.

   1. Show that the power series f(x) is a unit (i.e. has a multiplicative inverse) if and only if the constant term is nonzero. (Hint. When the constant term is 1, let g(x) = f(x)-1 and think about the geometric series 1 + g(x) + (g(x))² + (g(x))³ + ... .)

      The proof one way is easy. If f(x) has a 0 constant term then it looks like thisL

      	a1x + a2x2 + ...
      

      and there's clearly no power series you can multiply this by that will make all the terms but the constant term cancel. You will always be able to factor an x out of whatever you get. So there's nothing you can multiply this power series by and get just plain 1. So when f(x) has a zero constant term it does not have a multiplicative inverse.

      The proof the other way is harder.

   2. Show that x is prime in R.

   3. Show that R has just one prime, x, (up to multiplication by units) and that the fundamental theorem of arithmetic is true in R.

4. In class we discussed this sequence of solutions to the equation

   	x2 - 2y2 = +-1	(*)
   
   	n	xn	yn	check
   	1	1	1	 1 -  2*1 = -1
   	2	3	2	 9 -  2*4 =  1
   	3	7	5	49 - 2*25 = -1
   

   Lewanda noted that the pattern seems to be

   	yn+1 = xn + yn
   	xn+1 = xn + 2yn	
   

   This set of solutions to (*) isn't an accident. There's a conceptual way to see what's going on. The clue is in the arithmetic of the ring R = Z[sqrt(2)] - the set of all real numbers of the form a+b*sqrt(2) where a and b are ordinary integers. R is very far from being a field. Most of its members don't have multiplicative inverses. But some do.

   1. Before we start studying that ring, check that the table does what we thing it does: prove that the successive lines generated with those formulas do indeed provide solutions to equation (*). Remember that "prove" means "forever", not just the next few lines.

      Suppose x_n and y_n satisfy (*) and x_n+1 and y_n+1 are defined as above, and compute:

      	xn+12 - 2yn+12 
      	   = (xn + 2yn)2 - 2(xn + yn)2
                 = tedious formal algebra you all got right
                 = -(xn2 - 2yn2)
                 = -+1.
      

      so x_n+1 and y_n+1 satisfy (*) too - with the opposite sign. I know the first lines work, so this proof shows all the succeeding lines work as well. It's very important to note here that there's no +-1 until the end of the chain of equalities. If I put it in the first line then all I am doing is asserting what I am trying to prove!

   2. Now let's see how thinking about the ring R helps explain what's happening. First show that for positive integers x and y, z = x+y*sqrt(2) has a multiplicative inverse in R if and only if the pair x,y solves *. (Hint: z has a real number inverse 1/z. Figure out when that 1/z is in R by rationalizing the denominator.)

      We've essentially done this in problem 1 above. The best way to see it here is to note that when (x,y) solves * we have

      	(x + y sqrt(2))*(x - y sqrt(2)) = +11
      

      so that +-(x - y sqrt(2)) is the multiplicative inverse of (x+y*sqrt(2)). The converse is true too.

   3. Show that in any ring, whenever an element r has a multiplicative inverse so do all of its powers.

      Suppose r has a multiplicative inverse - call it s. Then I will show that sⁿ is the multiplicative inverse of rⁿ. To do that I need to show that sⁿ*rⁿ = 1. But that's easy. Using the associative and commutative laws many times tells me that sup>n*rⁿ = (r*s)ⁿ, but that's just 1ⁿ which equals 1.

      Note that this proof uses no mumbo jumbo about negative exponents, nothing about 1/r = r^-1. All that mumbo jumbo is only true once you've proved it, and we haven't.

   4. Let u = 1+sqrt(2). Since 1,1 solves (*), u has a multiplicative inverse in R (you proved that in (b)). In fact, u corresponds to the first line of the table.

   5. Show that the rest of the table contains just the powers of u.

   6. Explain how the sequence of arguments (b)-(e) provides another solution to part (a).

      Proof: When you multiply

      	(a + b sqrt(2)) * ( 1 + sqrt(2) ) = (a+2b) + (a+b)sqrt(2)
      

      you see that you get exactly the equations for computing x_n+1 and y_n+1 from x_n and y_n in the table.

   7. Optional, harder. Show that all the positive solutions to (*) come from powers of u and thus appear in the table. That's tricky. Suppose someone shows you two HUGE numbers, say x=63018038201 and y=44560482149 which happen to satisfy (*). (These do - you can check if you like.) How can you be sure they appear in the table?

   8. Optional, easier. This whole discussion can be carried out for the more general equation

      	x2 - dy2 = +-1	(**)
      

      (We've just studied the special case d=2.) See how far you can get with the cases d=3 and d=5. First look for one small solution, then use the strategy above to see how to generate the table.

      For more information on this topic (probably more than you want to know), you can google "Pell's equation". Here's the first hit, from mathworld. It's pretty informative, particularly down toward the bottom of the page, at about equation (31). If you get that far, look at the smallest solution for d=61.