Homework 7

Problem 1

Let $R$ be a ring (with unity) and suppose we have a morphism of short exact sequence of $R$ -modules:

Prove:

if $τ_{1}$ and $τ_{3}$ are injective, then $τ_{2}$ is injective
if $τ_{1}$ and $τ_{3}$ are surjective, then $τ_{2}$ is surjective

Hints

For the injectivity of $τ_{2}$ , start with an element $y \in Y$ with $τ_{2} (y) = 0$ . Chase that element around the diagram to eventually find an element $x \in X$ with $f (x) = y$ . Then chase $x$ around the diagram to show $x = 0$ and hence $y = 0$ . You will use:

Commutativity of the right square
Injectivity of $τ_{3}$
Exactness of the top row at $Y$
Commutativity of the left square
Exactness of the bottom row at $X^{'}$
Injectivity of $τ_{1}$

For the surjectivity of $τ_{2}$ , start with an element $y^{'} \in Y$ . The goal is to show there is some $y \in Y$ with $τ_{2} (y) = y^{'}$ . First show there is some $y_{1} \in Y$ such that $g^{'} (τ_{2} (y_{1})) = g^{'} (y^{'})$ . Then prove there is an element $x \in X$ such that $f^{'} (τ_{1} (x)) = y^{'} - τ_{2} (y_{1})$ . Finally, show that the element $f (x) + y_{1}$ maps to $y^{'}$ under $τ_{2}$ . You will use:

Surjectivity of $τ_{3}$
Exactness of the top row at $Z$
Commutativity of the right square
Exactness of the bottom row at $Y^{'}$
Surjectivity of $τ_{1}$
Commutativity of the left square

Problem 2

Let $R$ be a commutative ring (with unity) and let $M_{1}$ and $M_{2}$ be two $R$ -modules. Prove that $M_{1} \oplus M_{2}$ is:

projective if and only if both $M_{1}$ and $M_{2}$ are projective
injective if and only if both $M_{1}$ and $M_{2}$ are both injective
flat if and only if both $M_{1}$ and $M_{2}$ are flat

Hints

You'll likely want to exploit the isomorphism $M_{1} \oplus M_{2} ≃ M_{1} \times M_{2}$
Also recall that tensor product commutes with direct sum
You might also want to note that for a pair of morphisms $ϕ : N_{1} \to P_{1}$ and $ψ : N_{2} \to P_{2}$ , there is an isomorphism $\ker (ϕ \oplus ψ) ≃ \ker (ϕ) \oplus \ker (ψ)$

Problem 3

Let $A$ be a nonzero finite abelian group. Prove that:

$A$ is not projective
$A$ is not injective

Hints

If you want to be sneaky, you can use this characterization of projective modules and this characterization of injective modules over a PID.
If you would prefer a direct approach, consider the following strategy:
- Note that a direct summand of a projective/injective module is also projective/injective
- By the Fundamental Theorem of Finite Abelian Groups, $A$ has a direct summand of the form $Z_{p^{k}}$ for some prime $p$ and positive integer $k$
- By considering a certain short exact sequence of the form below, you can show $Z_{p^{k}}$ is neither injective nor surjective:
  
  $0 \to p^{k} Z_{p^{2 k}} \to Z_{p^{2 k}} \to Z_{p^{k}} \to 0$

Problem 4

Suppose $R$ is a commutative ring. Prove that:

the tensor product of two free $R$ -modules is free
the tensor product of two projective $R$ -modules is projective

Hints

Recall that tensor product commutes with direct sums
Use this characterization of projective modules

Problem 5

Suppose $R$ is a commutative ring. Prove that for each cyclic $R$ -module $M$ we have $T (M) ≃ S (M)$ ; i.e., the tensor algebra is already commutative.

Hint

Show the ideal $C (M)$ is trivial.