Minimality in the construction of integers and rationals (Analysis I, Theorem 9.1 & 9.2).

The book Analysis I by Herbert Amann & Joachim Escher can be found here on page 85 & 86.

When constructing $\mathbb{Z}$ from $\mathbb{N}$ the author defines an equivalence relation on $\mathbb{N}^2$, $$(m,n)\sim(m',n'):\Leftrightarrow m+n'=m'+n,$$

and let $\mathbb{Z}=\mathbb{N}^2/\sim$.

Similarly for rationals, an equivalence relation on $\mathbb{Z}\times\mathbb{Z}^{\times},$ $$(a,b)\sim(a',b'):\Leftrightarrow ab'=a'b,$$

and let $\mathbb{Q}=\mathbb{Z}\times\mathbb{Z}^{\times}/\sim$.

In both of the proofs, the author then writes that the sets ($\mathbb{Z}$ and $\mathbb{Q}$ respectively) constructed are "by construction" minimal. Why is that true?

• I think you're going to say way more in order to understand what you mean. First, refer to some specific page(s) in the book ("that manner" is too vague). Second, try to condense the main issue here for those who don't have the book. Commented Oct 2, 2016 at 9:09
• @DonAntonio Made the edit. Please tell me if the question is still unclear. Thanks. Commented Oct 2, 2016 at 9:25
• @Masacroso I understand the intuition behind the notion of minimality, but how to present it rigorously, i.e. how to prove that the numbers added are just the needed? Commented Oct 2, 2016 at 10:33

From the book Analysis I by Herbert Amann & Joachim Escher:

9.1 Theorem There is a smallest domain (commutative ring without zero divisors) with unity, $$\mathbb{Z}$$, such that $$\mathbb{N} \subset \mathbb{Z}$$ and the ring operations on $$\mathbb{Z}$$ restrict to the usual operations on $$\mathbb{N}$$. This ring is unique up to isomorphism and is called the ring of integers.

We argue, not following the book, that $$\mathbb{Z}$$ is the smallest ring containing $$\mathbb{N}$$ by making several observations, leaving the details to the OP:

Assume that $$\mathbb{N}$$ is contained in $$R$$. Suppose in $$R$$ there exist an element $$g$$ such that $$g + 1 = 0$$. Then for $$m \in \mathbb{N}$$, $$mg + m = m (g+1) = m\text{*} 0 = 0$$, so $$mg$$ is the additive inverse of $$m$$.

The set $$Z = \{mg\;|\; m \gt 0 \} \cup \mathbb{N}$$ must be contained in $$R$$.
But this set can be identified with the set $$\mathbb{Z}$$.

There is only one way to define addition and multiplication on $$Z$$ (it is generated by 'adding' $$g$$ to the natural numbers), so it can be algebraically identified with the construction of $$\mathbb{Z}$$.

9.2 Theorem There is, up to isomorphism, a unique smallest field $$\mathbb Q$$, which contains $$\mathbb{Z}$$ as a subring.

In this second part we want to show that $$\mathbb Q$$ is contained in any field that contains (via an injective ring morphism) $$\mathbb Z$$.

In the first part above, we 'threw in' the additive inverses to get $$\mathbb Z$$. In the same way, we now want to 'throw' $$F = \{2^{-1},3^{-1},4^{-1},5^{-1},\dots, n^{-1},\dots\}$$ into $$\mathbb Z$$ and 'generate' $$\mathbb Q$$.

We leave it to the OP to show that each block (element of $$\mathbb Q = \frac{\mathbb Z \times \mathbb Z^{\text{x}}}{\text{~}}$$) has a representative that is '$$\mathbb Z$$ generated' by an element in $$F$$.

The remaining details are also left as an exercise.

Ok, I think I got it.

They do there one of the classical constructions both of integral domain $\;\Bbb Z\;$ and of the field $\;\Bbb Q\;$. The minimality of $\;\Bbb Z\;$ follows from the fact that any domain containing $\;\Bbb N\;$ has to contain the constructed $\;\Bbb Z\;$ (as the operations defined on $\;\Bbb Z\;$ must restrict to the ones already given in $\;\Bbb N\,$ !). This is what the authors do when they write on page $85$, after $\;(9.4)\;$, "Now let $\;R\supset\Bbb N\;$ be some commutative ring...", and then they pass to prove that $\;\Bbb Z\;$ can be embedded within $\;R\;$.

The minimality for $\;\Bbb Q\;$ follows from the one of $\;\Bbb Z\;$, and the end of the proof, after $\;(9.5)\;$ , is very similar to the one above.

• Well, what I don't understand is why any domain containing $\mathbb{N}$ has to contain the constructed $\mathbb{Z}$? Commented Oct 2, 2016 at 9:57
• @YuxiaoXie Because then the set $\;\Bbb N^2/\sim\;$ must be contained in $\;R\;$. Observe that $\;R\;$ must be a ring, first and foremost. Thus, any pair of natural numbers (which are contained in $R$...) must render a unique element which must be natural as well when added and also when multiplied, in such a way that all natural numbers have an additive inverse in $\;pR\;$ (or $\;\Bbb Z\;$) Commented Oct 2, 2016 at 10:19

The minimality it is because the numbers added are just the needed to pass from a ring to a field, i.e. you need from $\Bbb Z$ just the multiplicative inverses to transform it in the field $\Bbb Q$.

By the other side: if you quit just a number of $\Bbb Q$ then it is not a field anymore. In the case of $\Bbb N$ to $\Bbb Z$ is the same: we construct the needed numbers to transform $\Bbb N$ in a ring ($\Bbb Z$), i.e. the additive inverses.

The proof of minimality is that if you quit a number you lose the algebraic structure. By example: taking some $x\in\Bbb Z_{<0}$ then the natural number $-x$ lost it additive inverse. Then $\Bbb Z$ is minimal.

For the case of $\Bbb Q$ if you quit some $p/q$ then you must quit too $-p/q$, $q/p$ and $-q/p$. But then the number $p\cdot q^{-1}$ doesnt exist, so this altered $\Bbb Q$ is not a field anymore. Then you can try to quit $q^{-1}$ too but then $q$ doesnt have anymore inverse.

To prove this rigorously we can do something based in the above to construct a proof by contradiction. You can prove that doesnt exists some $r\in\Bbb Q$ such that $r\neq p/q$ for any pair $p,q\in\Bbb Z$, but by construction such $r$ cannot exists.

And you can prove that you are unable to quit any $p/q$ to maintain a field based in the operations of addition and multiplication over $\Bbb Z$. Notice that if you quit some $p/q$ this implies that you must quit too $q^{-1}$ for some $q\in\Bbb Z$ because $p/q=p\cdot q^{-1}$.