I want to show these two statements:
1- Show that $2 \mathbb Q = \mathbb Q.$
2-Show that $\mathbb Q$ is not a finitely generated $A$ module.
For the first statement I do not know how to show it, so if someone can help me I would appreciate this.
For the second statement I am guessing that the following is the solution:
Assume that $\mathbb Q = (\frac{a}{b})$ i.e., it is finitely generated , then $q = m \frac{a}{b}$ for some integer $m.$ But then I do not know what to do. Could anyone help me in proving this also please?
Since $(1)$ was already handled in the comments...
$(2)$ $\underline{Claim:}$ $\mathbb{Q}$ is an infinitely-generated $\mathbb{Z}$-module.
$\underline{Proof:}$ Suppose $\mathbb{Q}$ has a finite generating set, $\{q_1,...,q_n\}\subseteq \mathbb{Q}$. Note that each $q_i = \frac{r_i}{s_i}$ for $r_i,s_i\in\mathbb{Z}$. We have: $$Span_{\mathbb{Z}}\bigg\{\frac{1}{s_1},...,\frac{1}{s_n}\bigg\} \supseteq Span_{\mathbb{Z}}\bigg\{q_1,...,q_n\bigg\},$$ so without loss of generality, assume the spanning set is of the form on the LHS. Furthermore, suppose this set is linearly independent. Now, consider the element: $$x:= \frac{1}{s_1\cdot ... \cdot s_n}.$$ Clearly $x\in \mathbb{Q}$, but if $x\in Span_{\mathbb{Z}}\bigg\{\frac{1}{s_1},...,\frac{1}{s_n}\bigg\}$ then $\exists \lambda_i\in \mathbb{Z}$ such that: $$\frac{1}{s_1\cdot ... \cdot s_n} = \sum\limits_{i=1}^n\lambda_i\frac{1}{s_i}.$$ Clearing the denominator, subtracting the one over, and noting $1 = \frac{s_j}{s_j}$ we get: $$\implies \sum\limits_{i\neq j = 1}^n\lambda_i\cdot (s_1\cdot ...\cdot s_n)\cdot\frac{1}{s_i} + (\lambda_j\cdot(s_1\cdot ...\cdot s_n)- s_j)\cdot\frac{1}{s_j} = 0$$ Rewriting with new coefficients: $$\sum\limits_{i=1}^n\mu_i\cdot \frac{1}{s_j} = 0$$ $\implies \forall i:\text{ }\mu_i = 0$ (by linear independence of $\frac{1}{s_i}$'s). This in turn yields: $$\forall i\neq j:\text{ }\lambda_i\cdot (s_1\cdot...\cdot s_n) = 0$$ and $$\lambda_j\cdot (s_1\cdot ... \cdot s_n) = s_j.$$ Continuing: $$\implies \forall i\neq j: \lambda_i = 0\text{ }\text{ (integral domain property of integers)}$$ and $$\lambda_j = \frac{1}{(s_1\cdot ...\cdot \widehat{s_j}\cdot... \cdot s_n)}\in \mathbb{Q}\backslash \mathbb{Z}$$ and thus we've reached our first contradiction.
We conclude that $x\in \mathbb{Q}\backslash Span_{\mathbb{Z}}\bigg\{q_1,...,q_n\bigg\}$. So that for the outer contradiction, we have that $\{q_1,...q_n\}$ can't generate all of $\mathbb{Q}$. $\blacksquare$