Solutions to Math 103B Test #2, February 25, 2009

1)  x^51 = (x+2)*q(x) + r
where the remainder polynomial r has to be a constant.
Plug in x=3 to get 3^51 = 0 + r  mod 5.
Since 3^4 = 1 mod 5,  3^52 is also 1 mod 5, so 3^51 is the inverse
of 3 mod 5, namely 2.  Thus r=2.

2)  The exponent of the leading term of x^2 f^3  has the form 3k+2.
The exponent of the leading term of g^3  has the form 3m.
Since a number that is 2 mod 3 cannot equal a number that is 0 mod 3,
these two leading terms cannot cancel each other out, so x^2 f^3 - g^3
cannot have all of its coefficients equal to zero.

3)  By the division algorithm,  r(x) = f(x) - g(x) q(x) for some 
q(x) in Q[x].  Since g is in the ideal I, so is the product gq.
Since f is also in I, so is the difference f - gq.  Thus r(x) is in I.
If r(x) were nonzero, it would be a polynomial in I with degree 
smaller than the degree of g(x), a contradiction 
(as g(x) is the monic poly in I of smallest degree).
Thus r(x) is the zero polynomial.

4i)  This poly is x^3 + x + 1 mod 2, which is irreducible mod 2 because
it is a cubic poly with no linear factor 
(since neither 0 nor 1 is a zero of this poly).
Thus this poly is irreducible over Z, by the mod 2 test.
For primitive polys, irreducibility over Z is equivalent to irreducibility
over Q.

4ii)  It suffices to show this poly is irreducible over Z.  Replace x
by x+1 and show the new polynomial (x+1)^8 + 1 is irreducible.
Since all the binomial coefficients  8!/k!(8-k)!  are even for 
k=1,2,3,4, and since the constant term of the new poly is  2 
(which is not divisible by 4),
the new poly is irreducible by Eisenstein's criterion with p=2.

5) X^999 + p is irreducible for any prime p of your choice,
by Eisenstein's criterion.

6i)  True.  It suffices to show that for all n > 2, u^n has the form
a + bu + cu^2 for some rational coeffs a,b,c.  This is true for n=3,
since u^3 = -u -1  by definition of u.  Assume as induction hypothesis
that it is true for n, i.e., u^n = a + bu + cu^2.  Multiply by u to get
u^(n+1) = au + bu^2 + c(-1 - u) = -c + (a-c)u + bu^2, and these coeffs
are rational.  Thus the answer to (6i) is true, by induction on n.

6ii)  Following the hint, consider the ring homomorphism 
that maps f(x) to f(u).
This map is ONTO D, so by the first isomorphism theorem for rings,  D
is isomorphic to Q[x]/I, where I is the kernel.  Since Q[x] is a PID,
I = <g(x)> for some monic g(x) in Q[x].  Since by definition of kernel,
x^3 + x + 1 is in I, it follows that x^3 + x + 1  must be divisible by g(x).  
But x^3 + x + 1 is irreducible, since
it is a cubic with no integer zeros.  Thus g(x)= x^3 + x + 1.
Since g(x) is irreducible, I is maximal, so Q[x]/I is a field.