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?

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    $\begingroup$ The answer to your question can probably be found at an earlier point in the book. It would be helpful to have the name of the book so that you can be directed to the part of the book that deals with the concept of the ideal (or subgroup) of $Z$ generated by a collection of elements. $\endgroup$ – user49640 Feb 20 '17 at 0:31
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    $\begingroup$ I realized now that on the beginning of the book it says that basic knowledge of ring, group theories and induction+wellordering is required. $\endgroup$ – KKZiomek Feb 20 '17 at 0:55
  • $\begingroup$ It's okay, you may still continue reading your book, if you manage not to be discouraged, and look up things when necessary. $\endgroup$ – Violapterin Feb 20 '17 at 2:53
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    $\begingroup$ $(4,6)$ (also written as $\langle4,6\rangle$, with angle brackets), for example, means everything of the form $4x+6y$ for integers $x$ and $y$. So, for example, $-2=4(1)+6(-1)\in\langle 4,6\rangle$. It can be checked that it's closed under addition and subtraction. Also, $(2)$ means everything of the form $2x$, so all even integers. You can check that $(4,6)=(2)$. (We also have $(2)=(-2)$.) $\endgroup$ – Akiva Weinberger Feb 20 '17 at 5:08
  • $\begingroup$ I've also seen $(a,b)$ used a s a shorthand for $\gcd(a,b)$, but the $(d)$ and the statement implies it is not the case here. (Anyway, the $\gcd$ interpretation would be rather interesting here, cf. the answers...) $\endgroup$ – JiK Feb 20 '17 at 10:39

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 $(d)=\mathbb{Z}d$.

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    $\begingroup$ @ Bill Dubuque, I think ring-theory notation needs to be changed (Not your fault) as $(a,b)$ has too many other meanings. I've even seen it in number-theory texts to denote $\gcd \{a,b\}$. $\endgroup$ – DanielWainfleet Feb 20 '17 at 0:26
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    $\begingroup$ @user254665 That ambiguity is by design, since it allows one to give proofs that work for both gcds and ideals, e.g. see these proofs of Euclid's Lemma. This will become clearer when one studies Divisor Theory (modern form of Kronecker's work). Btw, your comment was misplaced so it did not ping me (I noticed it by chance). $\endgroup$ – Bill Dubuque Feb 20 '17 at 0:28
  • $\begingroup$ @user254665: $\gcd(a,b$ is the positive generator of the ideal $(a,b)$, so the notation is consistent. $\endgroup$ – Bernard Feb 20 '17 at 0:56
  • $\begingroup$ I meant the comment to be for both A's. I appreciate your points. $\endgroup$ – DanielWainfleet Feb 20 '17 at 1:21
  • $\begingroup$ @user254665 I think this is why my algebra professor uses $<a, b>$ to denote ideal or subgroup generated by $a$ and $b$. You still need to see from context whether it is a group or an ideal though. $\endgroup$ – Alex Vong Feb 20 '17 at 14:20

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


$(a,b)$ 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.


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