Solutions to 103B final exam, March 18, 2009

(1A)  Let (c) = (b).  Then c = br and b=cs for some r,s  in R. 
Thus c = csr. Thus 1 = sr,  so r is a unit, so c and b are associates.

(1B)  Say b = cu  for some unit u in R.  Then  b is in the ideal (c),
so (b) is contained in (c).  We also have c = bv, where v is the inverse
of u in R.  Thus c is in the ideal (b), so (c) is contained in (b).
Thus (b) = (c).

(2)  The zero ideal of the ED is clearly principal (generated by 0),
so consider a nonzero ideal I, and let  x  be the element of I with
smallest (positive) measure.  Let b be an arbitrary element of  I.
It suffices to show that b is a multiple of x, for that will show that
I equals the principal ideal (x).  If  b is not a multiple of x, then
b = qx + r  for a nonzero "remainder" r whose measure is less than the
measure of x.  But because  r = b - qx is an element of I 
(since b and x are in I), this contradicts the minimality of the measure
of x.

(3A)  Let b and c be elements of I.  Then both b and c are elements of
I_k  for some sufficiently large k.  Since I_k is an ideal,  b-c is also
in I_k, so b-c is in I.  The same argument shows that rc is in I
for any ring element r.  Thus I is an ideal.

(3B)  Since R is a PID,  I = (w)  for some w in R.  The element w is in I,
thus in I_k for some k, so (w) is contained in I_k.  Thus I is contained
in I_k.  But by definition of union,  I_k is contained in I, so I = I_k.

(4A)  False.  x^4 + 1 = (x^2 +x -1)(x^2 -x -1)  over  Z_3.
(4B)  True, by the mod 2 test.  The given quintic poly is irred mod 2,
thus it is irred over Z and also over Q.  To see that the quintic poly
is irreducible mod 2, first notice that it has 
no linear factor mod 2, so one only needs to check that it has 
no irreducible quadratic factor mod 2.  
The only irreducible quadratic poly  mod 2 is x^2 + x + 1,
and one can easily show by long division that this quadratic poly
does not divide the quintic poly.

(5A) Let R = Z_5[x].  Note that p(x) is irreducible in R, 
since p(x) is a cubic with no zero in Z_5.
Thus I = (p(x)) is a maximal ideal in the PID  R.
The ring homom f(x) ---> f(u) maps  R onto Z_5[u]
and has kernel I.  Thus  R/I is isomorphic to Z_5[u].
Since I is maximal in R, R/I is a field.  Thus Z_5[u] is a field.

(5B)   By induction, every power of u can be written in the form
a+bu+cu^2,  where a,b,c are in Z_5.  No two such quadratic polynomials
in u can be equal (unless their coeffs are the same), otherwise u would
be a zero of a polynomial of degree less than 3, which is impossible,
since p(x) is the poly of smallest degree having u as a zero.
Thus Z_5[u] consists of all elements of the form a+bu+cu^2, and there
are 125 different elements of this form (5 choices for a, 5 choices for b,
and 5 choices for c).

(6)  The poly factors as x^3 (x^25 - 1) = x^3 * (x - 1)^25 over Z_5.
To see the last equality, first note that (x^5 - 1) = (x-1)^5 by the binomial
theorem, and then raise both sides to the fifth power.

(7A)  True.  Each of the four factors has a norm that is a prime in Z, 
so each is irreducible in Z[s].  

(7B)  False.  The factorizations are not considered different, because
(1+s) and (1+s)u = 17+7s are associates, and
(s+2) and (s+2)v = s-2 are also associates, where u and v are the units
u = 5+2s,  v= 5-2s  (note that uv=1).

(8A)  Let R = Z[x].  If  f(x) and g(x) are in I and h(x) is in R,
then clearly f(x) - g(x) and h(x)f(x) are both in I  (to see this,
plug in x=1).  Thus  I is an ideal of  R.

(8B)  Let J be the set of all polys f(x) for which f(1) is even.
The argument in (8A) shows that J is an ideal of R.
Note that I is contained in J and J is contained in R, and both containments
are strict, because 2 is in J but not in I, and 1 is in R but not in J.
Thus I is not maximal.