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.

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).

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 + ... .)

   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.

   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.)

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

   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).

   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.