Each worksheet prints cleanly on A4 or Letter. Answer keys are on a separate page.

The Lumenary · First-year graduate

Modules

Name: ______________________

Date: ______________________

You will learn to

Work with module homomorphisms, submodules, quotients, kernels, cokernels and exact sequences, and recognise free, projective and injective modules
Prove that free modules are projective, that projective means "direct summand of a free module", and that over a PID submodules of finite free modules are free
Prove the Smith normal form and the structure theorem for finitely generated modules over a PID, including uniqueness of the invariant factors
Deduce the classification of finitely generated abelian groups and the Jordan normal form
Compute tensor products of modules, prove that $M \otimes_R -$ is right exact via the tensor–hom adjunction, and show that it is not exact in general

1.
Which of these $\mathbb{Z}$ -modules are free? (Choose all that apply.)
- A. $\mathbb{Z}^3$
- B. $2\mathbb{Z}$
- C. $\mathbb{Z}[x]$
- D. $\mathbb{Z} \oplus \mathbb{Z}/2$
- E. $\mathbb{Q}$
- F. $\mathbb{Z}/1 = 0$
2.
How many abelian groups of order $72$ are there, up to isomorphism?
3.
Which group is $\mathbb{Z}^2 / \langle (4, 6), (6, 4) \rangle$ (the quotient by the subgroup generated by these two vectors)?
- A. $\mathbb{Z}/2 \oplus \mathbb{Z}/10$
- B. $\mathbb{Z}/20$
- C. $\mathbb{Z} \oplus \mathbb{Z}/2$
- D. $\mathbb{Z}/4 \oplus \mathbb{Z}/5$
4.
A linear map $T$ on $\mathbb{C}^6$ has characteristic polynomial $(x - 2)^6$ , minimal polynomial $(x - 2)^3$ , and $\dim \ker(T - 2) = 3$ . What are the sizes of its Jordan blocks?
- A. $3, 2, 1$
- B. $3, 3$
- C. $2, 2, 2$
- D. $3, 1, 1, 1$
5.
In the proof that a submodule $N \subseteq R^n$ over a PID is free, why does the sequence $0 \to N \cap R^{n-1} \to N \to \pi(N) \to 0$ split?
- A. $\pi(N)$ is an ideal of a PID, hence free, hence projective
- B.Every short exact sequence of modules splits
- C.By the induction hypothesis, $N \cap R^{n-1}$ is free
- D. $R$ is a domain, so there are no zero divisors
6.
Over which of these rings $R$ does some submodule of a free $R$ -module (of finite rank) fail to be free? (Choose all that apply.)
- A. $\mathbb{Z}/4$
- B. $k[x, y]$
- C. $\mathbb{Z}[\sqrt{-5}]$
- D. $\mathbb{Z}$
- E. $k[x]$ for a field $k$
- F. $\mathbb{Q}$
7.
How many elements does $\mathbb{Z}/12 \otimes_{\mathbb{Z}} \mathbb{Z}/18$ have?
8.
Which Lean statement expresses that tensoring with $\mathbb{Z}/2$ does not preserve injectivity (as proved in tensor_not_left_exact)?
- A.Function.Injective (2 • LinearMap.id : ℤ →ₗ[ℤ] ℤ) ∧ ¬ Function.Injective ((2 • LinearMap.id : ℤ →ₗ[ℤ] ℤ).rTensor (ZMod 2))
- B.¬ Function.Injective ((0 : ℤ →ₗ[ℤ] ℤ).rTensor (ZMod 2))
- C.Function.Surjective (g.rTensor Q) whenever Function.Surjective g
- D.¬ Module.Free ℤ (ZMod 2)
9.
Write invariant_factors(orders): given positive integers $n_1, \dots, n_k$ , return the invariant factors of $\mathbb{Z}/n_1 \oplus \cdots \oplus \mathbb{Z}/n_k$ as a list $[d_1, d_2, \dots]$ with $1 < d_1 \mid d_2 \mid \cdots$ (empty for the zero group). For example $\mathbb{Z}/4 \oplus \mathbb{Z}/6 \cong \mathbb{Z}/2 \oplus \mathbb{Z}/12$ , so invariant_factors([4, 6]) == [2, 12].
Computational — Practice

Answer key — Modules

1.
Free: $\mathbb{Z}^3$ , $2\mathbb{Z} \cong \mathbb{Z}$ , $\mathbb{Z}[x]$ (basis of monomials) and the zero module (empty basis). Not free: $\mathbb{Z} \oplus \mathbb{Z}/2$ (torsion) and $\mathbb{Q}$ (torsion-free but not free: a basis would have one element $x$ , and $x/2 \notin \mathbb{Z}x$ ).
2.
$72 = 2^3 \cdot 3^2$ . The $2$ -part is $\mathbb{Z}/8$ , $\mathbb{Z}/4 \oplus \mathbb{Z}/2$ or $(\mathbb{Z}/2)^3$ ; the $3$ -part is $\mathbb{Z}/9$ or $(\mathbb{Z}/3)^2$ . So there are $3 \cdot 2 = 6$ groups, with invariant factors $(72)$ , $(2, 36)$ , $(2, 2, 18)$ , $(3, 24)$ , $(6, 12)$ , $(2, 6, 6)$ .
3.
The relation matrix $\begin{pmatrix} 4 & 6 \\ 6 & 4 \end{pmatrix}$ has $d_1 = \gcd$ of the entries $= 2$ and $d_1 d_2 = |\det| = 20$ , so its Smith form is $\operatorname{diag}(2, 10)$ and the group is $\mathbb{Z}/2 \oplus \mathbb{Z}/10 \cong (\mathbb{Z}/2)^2 \oplus \mathbb{Z}/5$ . (The first experiment prints this example.)
4.
The elementary divisors are $(x - 2)^{e_j}$ with $\sum e_j = 6$ (characteristic polynomial), $\max e_j = 3$ (minimal polynomial) and the number of blocks $= \dim \ker(T - 2) = 3$ . The only possibility is $3 + 2 + 1$ . The invariant factors are $(x-2), (x-2)^2, (x-2)^3$ .
5.
$\pi(N)$ is an ideal $(a)$ of the PID $R$ , so it is $0$ or isomorphic to $R$ ; either way it is free, hence projective, and the sequence splits: $N \cong (N \cap R^{n-1}) \oplus \pi(N)$ .
6.
In a domain, two elements $f, g$ satisfy the relation $g \cdot f - f \cdot g = 0$ , so a free ideal has rank $\le 1$ , i.e. is principal. So non-principal ideals — $(x, y) \subseteq k[x, y]$ and $(2, 1 + \sqrt{-5}) \subseteq \mathbb{Z}[\sqrt{-5}]$ — are non-free submodules of $R$ . Over $\mathbb{Z}/4$ , the ideal $2\mathbb{Z}/4$ is too small to be free. Over the PIDs $\mathbb{Z}$ , $k[x]$ and the field $\mathbb{Q}$ , submodules of finite free modules are free.
7.
$\mathbb{Z}/12 \otimes \mathbb{Z}/18 \cong (\mathbb{Z}/18) / 12(\mathbb{Z}/18) = \mathbb{Z}/(12, 18) = \mathbb{Z}/6$ , which has $6$ elements, generated by $\bar 1 \otimes \bar 1$ .
8.
The statement must exhibit an injective map ( $2 \cdot$ on $\mathbb{Z}$ ) and show that its tensor with $\mathbb{Z}/2$ is not injective. The Lean proof computes $(2 \cdot)(1) \otimes 1 = 1 \otimes 2 = 0$ while $1 \otimes 1 \neq 0$ , using $\mathbb{Z} \otimes \mathbb{Z}/2 \cong \mathbb{Z}/2$ (TensorProduct.lid).

def invariant_factors(orders):
    powers = {}                      # prime -> list of prime powers
    for n in orders:
        p = 2
        while n > 1:
            q = 1
            while n % p == 0:
                n //= p
                q *= p
            if q > 1:
                powers.setdefault(p, []).append(q)
            p += 1
    k = max((len(v) for v in powers.values()), default=0)
    factors = [1] * k
    for v in powers.values():
        v.sort(reverse=True)
        for i, q in enumerate(v):
            factors[k - 1 - i] *= q
    return factors

The largest invariant factor collects the largest power of each prime, the next one the second largest, and so on; this is exactly how the uniqueness proof of the structure theorem recovers invariant factors from elementary divisors. For $[12, 18, 8]$ the $2$ -powers are $8, 4, 2$ and the $3$ -powers $9, 3$ , giving $d_3 = 72$ , $d_2 = 12$ , $d_1 = 2$ .