Math 593 Assignment # 7 Solutions

Bryden R. Cais

November 8, 2002

1.

2. Put R = ℚ[x, y] and let M = (x, y) ⊂ R (considered as an R module). We claim that we have an exact

sequence

0 ⟶ R ⟶^{A} R^{2} ⟶^{B} M ⟶ 0

where

Ar = r (\begin{matrix}−y \\ x \end{matrix})

and

B (\begin{matrix}f \\ g \end{matrix}) = fx + gy.

To verify this assertion, it suffices to check that Im A = ker B since it is obvious that B is surjective. Suppose that fx + gy = 0. Then fx = −gy and y|f since R is a UFD and y is irreducible. Similarly, x|g and it follows that f = −ry, g = rx for some r ∈ R, that is, ker B = Im A. It follows that the presentation matrix for M is the matrix of multiplication by A as a map R → R^{2}. It is clear that this matrix is

(\begin{matrix}−y \\ x \end{matrix}) .

3. The presentation matrix is

(\begin{matrix}3 & 2 \\ 6 & −4 \\ 12 & 10 \end{matrix}) .

We use row and column operations to simplify this matrix. We have

(\begin{matrix}3 & 2 \\ 6 & −4 \\ 12 & 10 \end{matrix}) ⟶^{col 1−col 2} (\begin{matrix}1 & 2 \\ 10 & −4 \\ 2 & 10 \end{matrix}) ⟶^{row 2−10·row 1} (\begin{matrix}1 & 2 \\ 0 & −24 \\ 2 & 10 \end{matrix}) ⟶^{row 3−2·row 1}

(\begin{matrix}1 & 2 \\ 0 & −24 \\ 0 & 6 \end{matrix}) ⟶^{col 2−2·col 1} (\begin{matrix}1 & 0 \\ 0 & −24 \\ 0 & 6 \end{matrix}) ⟶^{row 2+4·row 3} (\begin{matrix}1 & 0 \\ 0 & 0 \\ 0 & 6 \end{matrix}) ,

from which it is evident that

M ≃ ℤ ⊕ ℤ/(6).

4. Let A be an integral domain and M an A module.

(a) Let T (M) ⊂ M denote the set of all torsion elements and let m_{1}, m_{2} ∈ T (M). Then there exist a_{1}, a_{2} ∈ A \ {0} such that a_{i}m_{i} = 0 for i = 1, 2. Then for any a ∈ A we have

a_{1}a_{2}(m_{1} + m_{2}) = a_{2}(a_{1}m_{1}) + a_{1}(a_{2}m_{2}) = 0

and

a_{1}(am_{1}) = a(a_{1}m_{1}) = 0.

Since A is an integral domain, a_{1}a_{2} ≠ 0 and therefore m_{1} + m_{2}, am_{1} ∈ T (M) which shows that T (M) is a submodule of M.

(b) Suppose that m ∈ T (M/T (M)). Let m̃ be any lift of m to M. Then there exists some a ∈ A \ {0} such that am = 0, or equivalently, such that am̃ ∈ T (M). Then there exists b ∈ A \ {0} such that bam̃ = 0 in M. Since A is an integral domain, ba ≠ 0 so that m̃ ∈ T (M). Therefore, m = 0 and we are done.

(c) Let A = ℂ[x, y]. Considering M = (x, y) as an A module, it is not hard to see that A is torsion free: this follows directly from the fact that A is an integral domain and M ⊂ A. However M is not free since we have a nontrivial relation between the generators x, y; namely x(y) − y(x) = 0 and since M is not generated by any single element of ℂ[x, y]. Now consider the A module

M = A ⊕ A/(x).

It is clear that N = A/(x) considered as an A submodule of M by the obvious inclusion is the torsion submodule of M and is a proper submodule.

5. Let A be a Euclidean ring and let M be a finitely generated torsion free A module. Let S = {m_{i}}_{i=1}^{s} generate M and without loss of generality suppose that m_{1}, …, m_{n} is a maximal linearly independent subset of S. Let F be the free A module generated by m_{1}, …, m_{n}.

(a) By our choice of m_{j} for 1 ≤ j ≤ n, for any n + 1 ≤ i ≤ s the set m_{1}, …, m_{n}, m_{i} is linearly dependent. We therefore have a nontrivial relation

∑_{j=1}^{n} a_{j}m_{j} + a_{i}m_{i} = 0,

where not all a_{j} are zero. In particular, a_{i} ≠ 0 since m_{1}, …, m_{n} are linearly independent. It follows that

a_{i}m_{i} ∈ F.

(b) Set a = a_{n+1} ⋯ a_{s} ∈ A. Since each a_{i} ≠ 0 and A is an integral domain, we have a ≠ 0. Define ϕ_{a} : M → M by ϕ_{a}(m) = am. Since a ≠ 0 and M is torsion free, ϕ_{a} is injective. Moreover, any m ∈ M has the form

m = ∑_{j=1}^{n} b_{j}m_{j} + ∑_{i=n+1}^{s} b_{i}m_{i}

for some b_{l} ∈ A. We have

am = a ∑_{j=1}^{n} b_{j}m_{j} + ∑_{i=n+1}^{s}(b_{i}a_{n+1} ⋯ â_{i} ⋯ a_{s})(a_{i}m_{i}),

where â_{i} denotes the omission of a_{i} from the product. Since a_{i}m_{i} ∈ F by construction, it follows that am ∈ F and hence that Im ϕ_{a} ⊂ F.

(c) Since ϕ_{a} is injective is gives an isomorphism

ϕ_{a} : M → Im ϕ_{a} ⊂ F.

Since F is free and finitely generated, it follows that Im ϕ_{a} and hence M is also free.