Solutions to Math 103B Test #1,  January 28, 2009

(1A) Z[x] is a subring of Q[x], 
but it is not an ideal, as it is not closed under
multplication by Q[x].  To see this, multiply 1/2 from Q[x] times
1 from Z[x] to get an element 1/2, which is not in Z[x].

(1B)  Let A be the 2 by 2 zero matrix whose upper left entry is 1, and whose
other three entries are 0.  Let B be the 2 by 2 matrix whose 
lower right entry is 1, and whose other three entries are 0.  
Then AB is zero but neither factor is zero in the matrix ring, 
so the matrix ring is not an integral domain.

(1C)  Z/4Z is not an integral domain, since (2+4Z)*(2+4Z) is zero,
but neither coset is zero in the quotient ring.

(2) Let J be any ideal of Q[x] strictly containing the ideal <x>.  It suffices
to prove that 1 is in J, for this will force J to be the whole ring, so that
<x> must be maximal.  Since J is not <x>, J must contain a polynomial of
the form xg(x)+c, where c is a NONZERO rational. Since xg(x) is in I and hence
in J,  c= (xg(x) +c) - xg(x) must also be in J.  Multiply by 1/c to see that
1 = (1/c)*c  is also in J.  

(3)  x is in Z[x] and 3 is in S.  Since the product 3x is not in S,
it follows that S cannot be an ideal.


(4) The ideal <I, 2>  of Z[x} strictly contains I, since 2 is not in I.
It remains to show that 1 is not in <I, 2>, for then <I, 2> will not be
the entire ring.  Assume that 1 *is*  in <I, 2>.  Then 1=f(x)+2g(x).
where f(x) is in I and g(x) is in Z[x].  Plugging in x=i, we get
1 = 0 + 2(a+bi) for some integers a, b, which is impossible.


(5A)  Suppose two of these cosets were equal. 
Then c would be in I for some
c in {1,2,3,4,5}.  Then c=(3+3i)(a+bi)  for some integers a,b.
Multiply this equation by its conjugate equation to get
c^2 = 18(a^2 + b^2), which is impossible since c^2 is not a multiple of 18.  

(5B)  Add either i or 2i to each of the coset representatives 
listed in problem (5A).