MATH 250B: COMMUTATIVE ALGEBRA HOMEWORK #1

JOHN VOIGHT

Some solutions are omitted or sketched.

Problem 4. Let A_{1}, …, A_{r} be row vectors of dimension n over a field k. Let X = (x_{1}, …, x_{n}). Let b_{1}, …, b_{r} ∈ k. By a system of linear equations in k one means a system of type

A_{1} · X = b_{1}, …, A_{r} · X = b_{r}.

If b_{1} = ⋯ = b_{r} = 0, one says the system is homogeneous. We call n the number of variables, and r the number of equations. A solution X of the homogeneous system is called trivial if x_{i} = 0, i = 1, …, n.

(a) Show that a homogeneous system of r linear equations in n unknowns with n > r always has a nontrivial solution.

(b) Let L be a system of homogeneous linear equations over a field k. Let k be a subfield of k′. If L has a non-trivial solution in k′, show that it has a non-trivial solution in k.

Solution. The matrix A with rows A_{i} defines a map L_{A} : k^{n} → k^{r} by X ↦ A · X. By SS2, if W = ker L_{A}, then dim W + rk A = n. Since A has only r rows, it has (row) rank ≤ r, so dim W ≥ n − r > 0, so the system has a nontrivial solution. This proves (a).

For (b), let x = (x_{1}, …, x_{n}) ∈ (k′)^{n} be a solution to Ax = 0. Consider k′ as a vector space over k, and let V be the k-subvector space of k′ generated by the x_{i}. (Note that even if k′ is infinite-dimensional over k, V is necessarily finite dimensional.) Let v_{1}, …, v_{m} be any basis of V and write x_{i} = ∑_{j} c_{ij}v_{j}.

We claim that for each j, c_{j} = (c_{ij})_{i} ∈ k^{n} is a solution; since the x_{i} were not all zero, at least one c_{j} is not zero. The claim follows from the calculation:

Ax = A (∑_{j} c_{1j}v_{j} ; ⋮ ; ∑_{j} c_{nj}v_{j}) = A (c_{11} ; ⋮ ; c_{n1}) v_{1}+⋯+A (c_{1m} ; ⋮ ; c_{nm}) v_{m} = (Ac_{1})v_{1}+⋯+(Ac_{m})v_{m}.

Since the v_{i} are linearly independent, Ac_{j} = 0 for all j.

Problem 5. Let M be an n × n matrix over a field k. Assume that tr(MX) = 0 for all n × n matrices X in k. Show that M = O.

Date: January 30, 2003.

XIII: 4–6, 8–9, 11–18.

Solution. Denote by E_{ij} the matrix whose only nonzero entry is e_{ij} = 1, and let M = (m_{ij})_{i,j}. Then since

tr(MX) = ∑_{i=1}^{n} ∑_{j=1}^{n} m_{ij}x_{ji}

we see that ME_{ij} = m_{ji}. Hence if tr(ME_{ij}) = 0 for all i, j, then M = O.

Problem 6. Let S be a set of n × n matrices over a field k. Show that there exists a column vector X ≠ 0 of dimension n in k such that MX = X for all M ∈ S if and only if there exists such a vector in some extension field k′ of k.

Solution. Since k ⊂ k′, one direction is trivial. For the other, note that if MX′ = X′ for all M ∈ S and X′ ∈ (k′)^{n}, then (M − I)X′ = 0 for all M ∈ S. By the solution to 4(b), choosing a basis for the components for X′ over k, we obtain an element X ∈ k^{n} such that (M − I)X = 0 for all M ∈ S.

Problem 8. Let N be a strictly upper triangular n × n matrix, that is N = (a_{ij}) and a_{ij} = 0 if i ≥ j. Show that N^{n} = 0.

Solution. Treating T as an indeterminant, note that det(N − T I) = (−T)^{n}, since N − T I is (not strictly) upper triangular with −T along the diagonal. By the Cayley-Hamilton theorem, N satisfies its characteristic polynomial T^{n} = 0, so N^{n} = 0.

Problem 9. Let E be a vector space over k of dimension n. Let T : E → E be a linear map such that T is nilpotent, that is T^{m} = 0 for some positive integer m. Show that there exists a basis of E over k such that the matrix of T with respect to this basis is strictly upper triangular.

Solution. We construct such a basis inductively. Since T is nilpotent, there exists a v_{1} such that T v_{1} = 0 (e.g. since (det T)^{m} = det T^{m} = 0).

Now for i = 1, …, n − 1, given linearly independent vectors v_{1}, …, v_{i}, we claim there exists a vector v_{i+1} independent of the v_{j} such that T v_{i+1} ∈ span{v_{1}, …, v_{i}}. Otherwise, T v ∉ span{v_{j}} for all v ∉ span{v_{j}}, and this contradicts T^{m}v = 0.

This gives a basis v_{1}, …, v_{n} of E. In this basis, the ith column of T has t_{ij} = 0 for i ≥ j since T v_{i+1} ∈ span{v_{1}, …, v_{i}}, so T is strictly upper triangular.

Problem 11. Let R be the set of all upper triangular n × n matrices (a_{ij}) with a_{ij} in some field k, so a_{ij} = 0 if i > j. Let J be the set of all strictly upper triangular matrices. Show that J is a two-sided ideal in R. How would you describe the factor ring R/J?

Solution. Clearly, J is an additive subgroup of R. If M = (m_{ij}) ∈ J and A = (a_{ij}) ∈ R, writing AM = B we see that

b_{ij} = ∑_{k} a_{ik}m_{kj}.

For i ≥ j, a_{i1} = ⋯ = a_{i(i−1)} = 0 while m_{ij} = m_{(i+1)j} = ⋯ = m_{nj} = 0, so b_{ij} = 0. A similar result holds for MA, so J is a two-sided ideal.

Now R/J ≅ k^{n} as rings. Clearly, any matrix A ∈ R can be written A = D + M where D is a diagonal matrix and M ∈ J; the ring of diagonal matrices is isomorphic to k^{n} because addition and multiplication are componentwise.

Problem 14. Let F be any field. Let D be the subgroup of diagonal matrices in GL_{n}(F). Let N be the normalizer of D in GL_{n}(F). Show that N/D is isomorphic to the symmetric group on n elements.

Solution (sketch). The group N is the group of matrices with exactly one nonzero element in each row and column. Every such matrix can be written as a permutation matrix (a matrix of zeros and ones contained in N) and a diagonal matrix, so N/D ≅ S_{n} by the action of the permutation matrix on the standard basis vectors.

Problem 16. Let F be a finite field with q elements. Show that the order of SL_{n}(F) is

q^{n(n−1)/2} ∏_{i=2} n(q^{i} − 1)

and that the order of P SL_{n}(F) is

\frac{1}{d}q^{n(n−1)/2} ∏_{i=2} n(q^{i} − 1).

Solution. The determinant map det : GL_{n}(F) → F^{∗} is clearly surjective, and has kernel SL_{n}(F), so SL_{n}(F) ≅ GL_{n}(F)/F^{∗}. Hence #SL_{n}(F) = #GL_{n}(F)/(q − 1), which is the result.

We know that P SL_{n}(F) = SL_{n}(F)/Z, where Z is the group of scalar matrices which are nth roots of unity (see after Theorem 9.2). Since F^{∗} is cyclic, #μ_{n}(F^{∗}) = gcd(n, q − 1) = d. This gives the second statement.

Problem 17. Let F be a finite field with q elements. Show that the group of all upper triangular matrices with 1 along the diagonal is a Sylow subgroup of GL_{n}(F) and SL_{n}(F).

Solution. We know from Exercise 12 that this is a subgroup. It has order q^{n(n−1)/2}, since we can choose the strictly upper triangular elements arbitrarily from F. If q = p^{m}, then we see it is a p-subgroup; by Exercise 15, this is the largest power of p that divides #GL_{n}(F) (and #SL_{n}(F)), so it is a p-Sylow subgroup as well.