KMB, November 15, 2013
M3P11 Galois Theory, Problem Sheet 4, solutions
1. In the language of rings, the argument goes like this: the kernel is an ideal, and if it is non-zero
then it contains an invertible element (as every non-zero element of a eld is in
KMB, November 29, 2013
M3P11 Galois Theory, Solutions to Problem Sheet 5
1. (x )n =
ni xi so D(x )n ) =
ini xi1 ; now
n and the result
2. We need to check that K is algebraically closed (which is given) and that i
KMB, January 6, 2014
M3P11 Galois Theory, Solutions to problem Sheet 6
(i) > 1 so a has a prime divisor p; now use Eisenstein. Or use uniqueness of factorization to
prove a .
Next, if b Q( a) then write b = x + y a; square, and the fact that a is i
KMB, November 11, 2013
M3P11 Galois Theory, Solutions to problem Sheet 3
1. (a) Set K = E (z1 , z2 , . . . , zm ). Our task is to prove that K (zm+1 , . . . , zn ) = E (z1 , . . . , zm ).
Now K (zm+1 , . . . , zn ) is the intersection of all subelds of F
M3P11 Galois Theory, Problem Sheet 2 Solutions
1. (a) If n = p/q in lowest terms (with p, q Z and q = 0) then we deduce that nq 2 = p2 .
In particular q 2 divides p2 but q 2 and p2 are coprime, so q 2 = 1, so p/q Z.
(b) We know Q( 2) = cfw_a
Course: M345P11 Galois Theory, Solutions to progress Test 2, 3/12/2013.
(a) F is a splitting eld for f (x) if f (x) factors as into linear factors (f (x) = c d=1 (x
i ) and if furthermore F = E (1 , . . . , d ). One mark.
(b) Neither of these are t
M3P11 Galois Theory, Solutions to problem Sheet 1
1. K [x] is easily checked to be an abelian group under + (the group laws are all easy
consequences of the fact that (K, +) is an abelian f 1 = 1 f = f . The reason
(f g )h = f (gh) is
M345P11 Galois Theory, Progress Test 1, 4/11/2013, solutions.
(a) This is standard bookwork. Say f = gh with g, h Q[x]. Clear denominators and
get Df = g0 h0 with g0 , h0 Z[x] and D Z>0 . If we can prove that for any prime p|D
we have that either all