What does this number theory statement mean?

Question

I recently started studying number theory by myself and I am reading a book about number theory. There is one thing that I don't understand, the statement below:

If $a,b \in \mathbb{Z}$, then there is a $d \in \mathbb{Z}$ such that $(a,b)=(d)$.

I understand everything but the $(a,b)=(d)$. I know it has to do something with set theory probably, but what? Specifically why d is in parentheses and a pair is equal to a single variable?

Answer 1

Notation. Let $R$ be a commutative ring and let $a_1,\ldots,a_n$ be elements of $R$, then the ideal generated by $a_1,\ldots,a_n$ is denoted by $(a_1,\ldots,a_n)$.

In your case, $(a,b)=\mathbb{Z}a+\mathbb{Z}b$ and $(d)=\mathbb{Z}d$.

Answer 2

It's ring-theoretic ideal language for $\,a\Bbb Z + b \Bbb Z = d\Bbb Z,\ $ i.e. $\,|d| = \gcd(a,b).\ $

Answer 3

$(a,b)$ means (in simple words) everything you get by multiplying and adding whatever is in $\mathbb{Z}$. More precisely, it is the ideal generated by $a$ and $b$. Similarly, $(d)$ is generated by $d$ single handedly, i.e., its multiples.

Two elements generate an ideal, which is "as fine as" their gcd. Indeed, this is what gcd means: it is the "finer mesh" that covers their union. The gcd is the "common nature" of them, which may also give this "joined mesh". For example, adding and multiplying whatever by 6 and 10, then you get all multiples of 2. Here, gcd(6,10)=2. This is why gcd is also notated as (6,10).

If you want a proof of this, you probably may find it in the section you ar reading, no matter what book you are referring to. Or you may want to show it yourself.