Exercise 1.2.38 (Extend Theorems 1.6, 1.7, 1.8)

Proof.

1.

We show that, for every positive integer $m$, $$\begin{array}{lll}\hfill m{a}_1\wedge m{a}_2\wedge \cdots \wedge m{a}_n=m(a_1\wedge a_2\wedge \cdots \wedge a_n).& & \hfill \text{(1)}\end{array}$$

Theorem 1.6 shows that (1) is true for $n=2$. Suppose now that (1) is true for some integer $n$. Then

$$\begin{array}{llllll}\hfill m{a}_1\wedge m{a}_2\wedge \cdots \wedge & m{a}_n\wedge m{a}_{n+1}& \hfill & & \hfill & \\ \hfill & =(m{a}_1\wedge m{a}_2\wedge \cdots \wedge m{a}_n)\wedge m{a}_{n+1}& \hfill (\text{by Problem 37})& & \hfill & & \hfill \\ \hfill & =m(a_1\wedge a_2\wedge \cdots \wedge a_n)\wedge m{a}_{n+1}& \hfill (\text{by the induction hypothesis})& & \hfill & & \hfill \\ \hfill & =m[(a_1\wedge a_2\wedge \cdots \wedge a_n)\wedge a_{n+1}]& \hfill (\text{by Theorem 1.6)}& & \hfill & & \hfill \\ \hfill & =m(a_1\wedge a_2\wedge \cdots \wedge a_n\wedge a_{n+1})& \hfill (\text{by Problem 37}).& & \hfill & & \hfill \end{array}$$

The induction is done, so (1) is true for every $n\ge 2$.

2.

Suppose that $d\mid a_i$ for $1\le i\le n$, and $d>0$. We show that $$\begin{array}{lll}\hfill \frac{a_1}{d}\wedge \frac{a_2}{d}\wedge \cdots \wedge \frac{a_n}{d}=\frac{1}{d}(a_1\wedge a_2\wedge \cdots \wedge a_n).& & \hfill \text{(2)}\end{array}$$

By (1),

$$\begin{array}{llll}\hfill d\left(\frac{a_1}{d}\wedge \frac{a_2}{d}\wedge \cdots \wedge \frac{a_n}{d}\right)& =d\frac{a_1}{d}\wedge d\frac{a_2}{d}\wedge \cdots \wedge d\frac{a_n}{d}& \hfill & \\ \hfill & =a_1\wedge a_2\wedge \cdots \wedge a_n.& \hfill & \end{array}$$

So (2) is true. In particular, if $g=a_1\wedge a_2\wedge \cdots \wedge a_n$, then

$$\frac{a_1}{g}\wedge \frac{a_2}{g}\wedge \cdots \wedge \frac{a_n}{g}=1.$$

3.

Suppose that $a_i\wedge m=1$ for $1\le i\le n$. We show by induction that $$\begin{array}{lll}\hfill (a_1{a}_2\cdots a_n)\wedge m=1.& & \hfill \text{(3)}\end{array}$$

By Theorem 18, this is true for $n=2$. Under the hypothesis $a_i\wedge m=1$ for $1\le i\le m+1$, suppose that $(a_1{a}_2\cdots a_n)\wedge m=1$. Since $a_{n+1}\wedge m=1$, Theorem 1.8 shows that $[(a_1{a}_2\cdots a_n)a_{n+1}]\wedge m=1$, so $(a_1{a}_2\cdots a_n{a}_{n+1})\wedge m=1$. The induction is done, so (3) is true for every $(a_1,\dots ,a_n)\in {\mathbb{Z}}^n$ such that $a_i\wedge m=1,i=1,2,\dots ,n$.