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I am reading the book on number theory and I have problems understanding this definition:

DEFINITION: Let $a_1, a_2, \cdots, a_n\in\mathbb{Z}$; we define $$\left(a_1, a_2, \cdots, a_n\right):=\left\{a_1x_1+\cdots+a_nx_n\,:\,x_1,\cdots, x_n\in\mathbb{Z}\right\}$$

If I got it right, then every set $A=(a,b)$ contains all integer numbers. Let me make it a little bit more clear. $$A=(5,7)\\ 0*5+0*7=0\\ (-4)*5+3*7=1\\ (-1)*5+1*7=2\\ (-5)*5+4*7=3$$ et cetera...

I can do same thing but swap the $+$ and $-$ signs in coefficients, and get all the negative integers too. And I can do it with every other pair of numbers, and triplet and so on.

So my question is, did I get this right?

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  • $\begingroup$ If you start with the set $\{8,12\}$ you only get the multiples of $4$. If $\gcd(a,b)=1$ you do get everything. $\endgroup$
    – André Nicolas
    Jun 6, 2015 at 16:16
  • $\begingroup$ It simply denotes the ideal of $\mathbf Z$ generated by $a_1,a_2,\dots,a_n$. As in all PIDs, it is the ideal generated by the g.c.d. of $a_1, a_2, \dots,a_n$. $\endgroup$
    – Bernard
    Jun 6, 2015 at 16:38

1 Answer 1

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It is correct that $(5,7)$ is the set of all integers. However, this does not make the definition meaningless.

For example, if all $a_i$ are even, clearly each element in $(a_1, \dots, a_n)$ will be even. So the set $(a_1, \dots, a_n)$ is not always the set of all integers.

Presumably, the book will proceed to show that $(a_1, \dots, a_n)= d \mathbb{Z}$ for some (non-negative) integer $d$ and this $d$ is a GCD of $a_1, \dots, a_n$.

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