Fall 1993 Algebra Prelim Solutions

  1. It is clear that Re is a left R-submodule of R, so we need to prove it is projective. It will be sufficient to show that Re is a direct summand of a free R-module. Since R(1 - e) is also a left R-submodule of R, the result will be proven if we can show that R = Re $ \oplus$ R(1 - e). If re $ \in$ R(1 - e), then re = s(1 - e) for some s $ \in$ R and so re = ree = s(1 - e)e = 0. This shows that Re $ \cap$ R(1 - e) = 0. Finally if r $ \in$ R, then r = r(e + 1 - e) = re + r(1 - e), so R = Re + R(1 - e).

    1. Since we are in characteristic zero, everything is separable so we may use the theorem of the primitive element. This tells us that there exists a $ \in$ K such that K = F(a). If p is the minimal polynomial of a over F, then K $ \cong$ F[X]/(p).

    2. Since p is the minimal polynomial, it is monic and irreducible in F[X]. Also we are in characteristic zero, so p is separable. This means that when we write p = p1...pn in K[X] where the pi are monic irreducible polynomials, the pi are distinct. Now write c = f[X] + (p) where f $ \in$ K[X]. Since c2 = 0, we see that f2 $ \in$ (p). Thus f2 = p1...pnq for some q $ \in$ K[X]. By uniqueness of factorization, we have pi divides f for all i and we deduce that f $ \in$ (p). Therefore c = 0 as required.

    1. Suppose the action of M 2($ \mathbb {Q}$) has a finite orbit with at least two elements. Using the formula that the number of elements in an orbit is the index of the stablizer of any element in that orbit, we see that M 2($ \mathbb {Q}$) has a nontrivial subgroup H of finite index. Then qM 2($ \mathbb {Q}$) $ \subseteq$ H for some positive integer q. However if a $ \in$M 2($ \mathbb {Q}$), then a = q(a/q), which shows that qM 2($ \mathbb {Q}$) = M 2($ \mathbb {Q}$). We conclude that M 2($ \mathbb {Q}$) = H and the result follows.

    2. To check that we have an action, we must show that (gh)*l = g*(h*l) for all g, h $ \in$GL 2($ \mathbb {Q}$). This is true because
      (g*h)(l) = det(gh)l/| det(gh)| = det(g)det(h)l/(| det(g)|| det(h)|)  
        = det(g)(h*l)/| det(g)| = g*(h*l).  

      Finally we see that all orbits have the form l}, and so in particular there are finite orbits which are not singletons.

  2. By Sylow's theorem a group of order 34 has a normal subgroup of order 17, hence G has a normal subgroup H of order 17. Again by Sylow's theorem, this subgroup is the unique subgroup of G which has order 17. Since we are in characteristic zero, everything is separable and so L is a Galois extension of LG. Therefore there is a one-one correspondence between the subfields of L containing LG and the subgroups of G. This correspondence has the property that if A is a subgroup of G, then the dimension of L over LA is | A|. The result follows by setting K = LH.

  3. Suppose S is not a field. Then it has a nonzero prime ideal P. Note that S/P is an integral domain. Since S[X]/P[X] $ \cong$ (S/P)[X], we see that S[X]/P[X] is an integral domain which is not a field. We deduce that P[X] is a nonzero nonmaximal prime ideal of S[X]. But nonzero prime ideals in a PID are maximal and since we are given that S[X] is a PID, we now have a contradiction and the result follows.

    1. First H has an identity, namely the zero homomorphism defined by 0(a) = 0 for all a $ \in$ A. If f, g $ \in$ H, then
      (f + g)(a + b) = f (a + b) + g(a + b) = f (a) + f (b) + g(a) + g(b)  
        = f (a) + g(a) + f (b) + g(b) = (f + g)(a) + (f + g)(b)  

      which shows that f + g is a homomorphism, and so f + g $ \in$ H. Also if f, g, h $ \in$ H, then
      ((f + g) + h)(a) = (f + g)(a) + h(a) = f (a) + g(a) + h(a)  
        = f (a) + (g + h)(a) = (f + (g + h))(a),  

      so (f + g) + h = f + (g + h) which establishes the associative law. Finally for f $ \in$ H, the inverse of f is - f, where (- f )(a) = - f (a). Since (- f )(a + b) = - f (a + b) = - f (a) - f (b) = (- f )(a) + (- f )(b), we see that - f $ \in$ H, and we have established that H is a group.

    2. We first show that H is torsion free. If f $ \in$ H has order n$ \ne$1, then f (a)$ \ne$ 0 for some a $ \in$ A. But then (nf )(a) = n(fa)$ \ne$ 0, a contradiction. Therefore the subgroup generated by f1,..., fm is a finitely generated torsion free abelian group, so by the fundamental structure theorem for finitely generated abelian groups it is free.

    1. A 2 by 2 matrix with entries in $ \mathbb {Z}$/p$ \mathbb {Z}$ will be invertible if and only if its columns are linearly independent over $ \mathbb {Z}$/p$ \mathbb {Z}$. So there are p2 - 1 choices for the first column (we cannot choose (0, 0) for the first column) and p2 - p choices for the second column (we cannot choose the vector in the first column). It follows that | G| = (p2 - 1)(p2 - p).

    2. Note that H is a Sylow p-subgroup of G, so the number of conjugates of H is congruent to 1 modulo p. Therefore 1 is congruent to 8 modulo p, which can only happen if p = 7. In the case p = 7, we have by Sylow's theorem that the number of conjugates of H in G is congruent to 1 modulo 7, which is of course congruent to 8 modulo 7. Thus the answer is p = 7.

Peter Linnell