Exercise 3.6.1 (A concrete free module)

Question

For $ N = R $, we may define a function $ j : A \to R ^{\oplus A} $ by sending $ a \in A $ to the function $ j_a : A \to R $:
$$
j_a(x) = \begin{cases} 
1 & \text{if } x = a \
0 & \text{if } x \ne a 
\end{cases}
$$
That is, for an element $ a \in A $, the mapping into the free group candidate $ R^{ \oplus A} $ for $ F_R(A) $ is given by $ j : A \to R^{ \oplus A} $, $ j(a) = j_a $.

Now we try to prove $ F_R(A) \cong R ^{\oplus A} $.

$$F^R(A) \cong R^{\oplus A}$$

Shinkenjoe · Accepted Answer

By Proposition 1.5.1, $ R^{\oplus A } $ satisfying the free module properties is enough for this. The key point is that every element of $ R^{\oplus A}$ may be written uniquely as a finite sum:
Take an arbitrary $ k \in R{\oplus A} $. Then
$$
k = \sum_{a \in A} k(a) j_a = \sum_{a \in A} r_k(a) j_a
$$
which is often abbreviated as:
$$
k = \sum r_a a
$$
As every function is nonzero only finitely many times, the sum is also finite and thus well-defined.

Let $ f : A \to M $ be any function from $ A $ to an $ R $-module $ M $. Now we define $ \varphi : R^{\oplus A} \to M $ by
$$
\varphi \left( \sum_{a \in A} r_a j_a \right) = \sum_{a \in A} r_a f(a)
$$

Let's check that it is a group homomorphism.
$$
\varphi \left( \sum_{a \in A} r_a j_a \right) + \varphi \left( \sum_{a \in A} s_a j_a \right) = \sum_{a \in A} r_a f(a) + \sum_{a \in A} s_a f(a)
$$
By the $ R $-module properties of $ M $:
$$
\sum_{a \in A} (r_a + s_a) f(a) = \varphi \left( \sum_{a \in A} (r_a + s_a) j_a \right)
$$
By $ R^{\oplus A} $ being distributive too:
$$
\varphi \left( \sum_{a \in A} r_a j_a + \sum_{a \in A} s_a j_a \right)
$$
as needed.

It is also a module homomorphism:
$$
\varphi \left( s \left( \sum_{a \in A} r_a j_a \right) \right) = \varphi \left( \sum_{a \in A} s (r_a j_a) \right)
$$
By scalar associativity in $ R^{\oplus A}$:
$$
\varphi \left( \sum_{a \in A} (sr_a) j_a \right) = \sum_{a \in A} (sr_a) f(a)
$$
By the $ R $-module properties of $ M $:
$$
s \left( \sum_{a \in A} r_a f(a) \right) = s \varphi \left( \sum_{a \in A} r_a j_a \right)
$$

For any element $ a \in A $ we get:
$$
(\varphi \circ j)(a) = \varphi(j(a)) = \varphi(j_a) = f(a)
$$
and $ \varphi \circ j = f $ and the diagram commutes.

Now we need to prove that $ \varphi $ is unique. Let's say there is some other $ \psi $ such that $ \psi $ is a module homomorphism and that $ \psi \circ j = f $. Consider any element $ a \in A $. Then $ \psi \circ j(a) = \psi(j_a) $ has to equal $ f(a) $. That is, we arrive at the definition of $ \varphi $ by force.

Exercise 3.6.1 (A concrete free module)

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Exercise 3.6.1 (A concrete free module)

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