Homework 7

Problem 1

Let $R$ be a ring (with unity) and suppose we have a morphism of short exact sequences 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

Solution

Let's do a classic diagram chase! First let's suppose $h_{1}$ and $h_{3}$ are injective (and both rows are exact). We wish to show $h_{2} : Y \to Y^{'}$ is injective, so suppose $y \in Y$ is such that $h_{2} (y) = 0$ . We want to prove $y = 0$ and we'll do so by "chasing" $y$ around the diagram.

First, by the commutativity of the right square we have $h_{3} (g (y)) = g^{'} (h_{2} (y)) = g^{'} (0) = 0$ . In other words, if we send $y$ down $h_{2}$ (where it's sent to 0) and then across $g^{'}$ (which must send 0 to 0) then we land at 0; by commutativity it follows that when we send $y$ across $g$ and down $h_{3}$ we must also get 0. This means that $g (y)$ is in the kernel of $h_{3}$ and since $h_{3}$ was assumed to be injective, we must therefore have that $g (y) = 0$ . But this implies $y$ is in the kernel of $g$ , and since the top row is exact (and so in particular exact at $Y$ ) it now follows that $y$ is in the image of $f$ . Thus, there must exist some $x \in X$ with $f (x) = y$ .

Now follow $x$ around the left commutative square. If we send $x$ across (via $f$ ) and then down (via $h_{2}$ ), it maps to $y$ and then to 0. Thus, when we instead send $x$ down (via $h_{1}$ ) and across (via $f^{'})$ it must map to 0. This means $h_{1} (x)$ is in the kernel of $f^{'}$ . But the bottom row is exact (in particular at $X^{'}$ ) and so $f^{'}$ is injective, hence we must have $h_{1} (x) = 0$ . Since $h_{1}$ is injective this means $x = 0$ . But then $y = f (x) = f (0) = 0$ . Thus, $h_{2}$ is injective.

Now let's try the surjectivity argument. Suppoes $h_{1}$ and $h_{3}$ are surjective (and both rows are exact). We wish to show $h_{2} : Y \to Y^{'}$ is surjective, so fix any element $y^{'} \in Y$ . We must prove there is an element $y \in Y$ with $h_{2} (y) = y^{'}$ .

Let $z^{'} = g^{'} (y^{'})$ . Since $h_{3}$ is surjective there is some $z \in Z$ with $h_{3} (z) = z^{'}$ . Since the top row is exact (in particular at $Z$ ), the map $g$ is surjective and so there is some $y_{1} \in Y$ with $g (y_{1}) = z$ . Now note that $y$ must map down and over to $z^{'}$ , but this is not quite enough to conclude $y_{1}$ maps down to $y^{'}$ . In other words, $h_{2} (y_{1})$ and $y^{'}$ are both elements that map (via $g^{'}$ ) to the same element $z^{'}$ , but that's not enough to conclude $h_{2} (y_{1}) = y^{'}$ .

So, let $\tilde{y} = y^{'} - h_{2} (y_{1})$ . Then $g^{'} (\tilde{y}) = g (y^{'} - h_{2} (y_{1})) = g (y^{'}) - g (h_{2} (y_{1})) = z^{'} - z^{'} = 0$ , so $\tilde{y}$ is in the kernel of $g^{'} .$ Since the bottom row is exact at $Y^{'}$ , this implies $\tilde{y}$ is in the image of $f^{'}$ and hence there is some $x^{'} \in X^{'}$ with $f^{'} (x^{'}) = \tilde{y}$ . Now $h_{1}$ is surjective so there is some $x \in X$ with $h_{1} (x) = x^{'}$ . Finally, we let $y = f (x) + y_{1}$ . Then observe that

$h_{2} (y) = h_{2} (f (x)) + h_{2} (y_{1}) = \tilde{y} + h_{2} (y_{1}) = y^{'} .$

Thus, we've proven $h_{2}$ is surjective.

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 $\ker (ϕ \oplus ψ) ≃ \ker (ϕ) \oplus \ker (ψ)$

Solution

You could use this characterization of projective modules, in which case this property is basically immediate. However, it is not hard to give a direct argument, so let's see that.

First suppose $M_{1}$ and $M_{2}$ are projective. To prove $M_{1} \oplus M_{2}$ is projective, let $g : K \to L$ be any surjection and $ϕ : M_{1} \oplus M_{2} \to L$ be any morphism. Then consider the morphism $ϕ \circ i_{1} : M_{1} \to L$ . Since $M_{1}$ is projective, this morphism factors through $g$ , i.e., there is some morphism $ψ_{1} : M_{1} \to K$ such that $ϕ \circ i_{1} = g \circ ψ_{1}$ . This is pictured below:

We can repeat this same argument for the projective module $M_{2}$ , obtaining a morphism $ψ_{2} : M_{2} \to K$ such that $ϕ \circ i_{2} = g \circ ψ_{2}$ . By a universal property for the direct sum $M_{1} \oplus M_{2}$ , we then get a unique module morphism $ψ : M_{1} \oplus M_{2} \to K$ with $ψ \circ i_{1} = ψ_{1}$ and $ψ \circ i_{2} = ψ_{2}$ :

By construction, we have

$(g \circ ψ) \circ i_{1} = g \circ (ψ \circ i_{1}) = g \circ ψ_{1} = ϕ \circ i_{1},$

and similarly

$(g \circ ψ) \circ i_{2} = g \circ (ψ \circ i_{2}) = g \circ ψ_{2} = ϕ \circ i_{2} .$

By our universal property for $M_{1} \oplus M_{2}$ , it follows that $g \circ ψ = ϕ$ . Thus, we've proven $M_{1} \oplus M_{2}$ is projective.

Conversely, suppose $M_{1} \oplus M_{2}$ is projective. To prove $M_{1}$ is projective, let $g : K \to L$ be any surjection and $ϕ : M_{1} \to L$ be any morphism. Let $π_{1} : M_{1} \oplus M_{2} \to M_{1}$ be the projection (available via the isomorphism $M_{1} \oplus M_{2} ≃ M_{1} \times M_{2}$ and the canonical projection $M_{1} \times M_{2} \to M_{1}$ ). Since $M_{1} \oplus M_{2}$ is projective, the morphism $ϕ \circ π_{1} : M_{1} \oplus M_{2} \to L$ must factor through $g$ , i.e., there is some $ψ : M_{1} \oplus M_{2} \to K$ such that $g \circ ψ = ϕ \circ π_{1}$ .

Now let $ψ_{1} = ψ \circ i_{1}$ , where $i_{1} : M_{1} \to M_{1} \oplus M_{2}$ is the canonical injection. Then observe that

$\begin{aligned} g \circ ψ_{1} & = g \circ (ψ \circ i_{1}) \\ = (g \circ ψ) \circ i_{1} \\ = (ϕ \circ π) \circ i_{1} \\ = ϕ \circ (π_{1} \circ i_{1}) \\ = ϕ \circ 1_{M_{1}} \\ = ϕ . \end{aligned}$

Thus, $M_{1}$ is indeed projective. The same argument, mutatis mutandis, proves $M_{2}$ is also projective.
This argument is basically the dual of the projective argument. First suppose $M_{1}$ and $M_{2}$ are both injective. To prove $M_{1} \oplus M_{2}$ is injective, let $g : L \to K$ be any injection and $ϕ : L \to M_{1} \oplus M_{2}$ be any morphism. Then consider the morphism $π_{1} \circ ϕ : L \to M_{1}$ . Since $M_{1}$ is injective, this morphism factors through $g$ , i.e., there is some morphism $ψ_{1} : K \to M_{1}$ such that $π_{1} \circ ϕ = ψ_{1} \circ g$ . This is pictured below:

We can repeat this same argument for the projective module $M_{2}$ , obtaining a morphism $ψ_{2} : K \to M_{2}$ such that $π_{2} \circ ϕ = ψ_{2} \circ g$ . By a universal property for the direct product $M_{1} \times M_{2} ≃ M_{1} \oplus M_{2}$ , we then get a unique module morphism $ψ : K \to M_{1} \oplus M_{2}$ with $π_{1} \circ ψ = ψ_{1}$ and $π_{2} \circ ψ = ψ_{2}$ :

By construction, we have

$π_{1} \circ (ψ \circ g) = (π_{1} \circ ψ) \circ g = ψ_{1} \circ g = π_{1} \circ ϕ,$

and similarly

$π_{2} \circ (ψ \circ g) = (π_{2} \circ ψ) \circ g = ψ_{2} \circ g = π_{2} \circ ϕ .$

By our universal property for $M_{1} \times M_{2} ≃ M_{1} \oplus M_{2}$ , it follows that $ψ \circ g = ϕ$ . Thus, we've proven $M_{1} \oplus M_{2}$ is injective.

Conversely, suppose $M_{1} \oplus M_{2}$ is injective. To prove $M_{1}$ is injective, let $g : L \to K$ be any injection and $ϕ : L \to M_{1}$ be any morphism. Since $M_{1} \oplus M_{2}$ is injective, the morphism $i_{1} \circ ϕ : L \to M_{1} \oplus M_{2}$ must factor through $g$ , i.e., there is some $ψ : K \to M_{1} \oplus M_{2}$ such that $ψ \circ g = i_{1} \circ ϕ$ .

Now let $ψ_{1} = π_{1} \circ ψ$ . Then observe that

$\begin{aligned} ψ \circ g & = (π_{1} \circ ψ) \circ g \\ = π_{1} \circ (ψ \circ g) \\ = π_{1} \circ (i_{1} \circ ϕ) \\ = (π_{1} \circ i_{1}) \circ ϕ \\ = 1_{M_{1}} \circ ϕ \\ = ϕ . \end{aligned}$

Thus, $M_{1}$ is indeed injective. The same argument, mutatis mutandis, proves $M_{2}$ is also injective.
To make our lives easier, let's first make an observation. Let $f : J \to K$ be any morphism and consider the result of tensoring with $(M_{1} \oplus M_{2})$ over $R$ :

$(M_{1} \oplus M_{2}) \otimes_{R} J \overset{1_{M_{1} \oplus M_{2}} \otimes f}{\to} (M_{1} \oplus M_{2}) \otimes K$

Recall that tensor product commutes with direct sum, and so the above morphism is naturally isomorphic to the morphism

$(M_{1} \otimes_{R} J) \oplus (M_{2} \otimes_{R} J) \overset{(1_{M_{1}} \otimes f) \oplus (1_{M_{2}} \otimes f)}{\to} (M_{1} \otimes_{R} K) \oplus (M_{2} \otimes_{R} K)$

An immediate consequence of a universal property of the direct sum is that the kernel of the above morphism is (isomorphic to ) the direct sum of the kernels of the component morphisms:

$\ker ((1_{M_{1}} \otimes f) \oplus (1_{M_{2}} \otimes f)) ≃ \ker (1_{M_{1}} \otimes f) \oplus \ker (1_{M_{2}} \otimes f) .$

As a consequence, the morphism $(1_{M_{1}} \otimes f) \oplus (1_{M_{2}} \otimes f)$ is injective if and only if the morphisms $1_{M_{1}} \otimes f$ and $1_{M_{2}} \otimes f$ are both injective.

From this observation, it is now immediate that $M_{1} \oplus M_{2}$ is flat if and only if $M_{1}$ and $M_{2}$ are both flat.

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:
  
  $0 \to p^{k} Z_{p^{2 k}} \to Z_{p^{2 k}} \to Z_{p^{k}} \to 0$

Solution

Let $A$ be a nonzero finite abelian group of order $n$ , say. Then $n a = 0_{A}$ for every $a \in A$ , and so every element in $A$ is torsion and $n A = 0 \neq A$ .

Now suppose $A$ is projective. Then by this characterization of projective modules it follows that $A$ is (isomorphic to) a direct summand of a free $Z$ -module. In other words, there is an isomorphism of the form $Z^{k} ≃ A \oplus B$ . Since every element in $A$ is torsion, but there are no nonzero torsion elements in $Z^{k}$ , this is a contradiction.

Now suppose $A$ is injective. Then by this characterizations of injective modules over PIDs (which includes $Z$ ), we must have $m A = A$ for every nonzero $m \in Z$ . But we've already noted that $n A = 0 \neq A$ , so we've once again arrived at a contradiction.

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

Solution

Suppose $M$ and $N$ are free $R$ -modules, say $M ≃ F (X)$ and $N ≃ F (Y)$ for some sets $X$ and $Y$ . Recalling that $F (A) ≃ ⨁_{a \in A} R$ for any set $A$ , and that tensor product commutes with direct sum, observe that

$\begin{aligned} M \otimes_{R} N & ≃ (⨁_{x \in X} R) \otimes_{R} (⨁_{y \in Y} R) \\ ≃ ⨁_{x \in X} (R \otimes_{R} (⨁_{y \in Y} R)) \\ ≃ ⨁_{x \in X} (⨁_{y \in Y} (R \otimes_{R} R)) \\ ≃ ⨁_{(x, y) \in X \times Y} R \\ ≃ ⨁_{z \in X \times Y} R \\ ≃ F (X \times Y) . \end{aligned}$
Suppose $M$ and $N$ are projective $R$ -modules. By this characterization of projective modules, there exist isomorphisms of the form $F (X) ≃ M \oplus M^{'}$ and $F (Y) ≃ N \oplus N^{'}$ for some sets $X, Y$ and modules $M^{'}, N^{'}$ . Then observe that

$\begin{aligned} F (X ⊔ Y) & ≃ F (X) \otimes_{R} F (Y) \\ ≃ (M \oplus M^{'}) \otimes_{R} (N \oplus N^{'}) \\ ≃ (M \otimes_{R} N) \oplus (M \otimes_{R} N^{'}) \oplus (M^{'} \otimes_{R} N) \oplus (M^{'} \otimes_{R} N^{'}) . \end{aligned}$

So, $M \otimes_{R} N$ is isomorphic to a direct summand of a free module and hence is projective.

Problem 5

Moved to next assignment!

Since we haven't covered symmetric algebras yet, this problem is being moved to the next homework assignment.

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.