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We have a structure: $\mathbb Q=\langle \mathbb Q,\,+,\,-,\,*,\,0,\,1 \rangle$

Question: Does $\mathbb Q$ contain a substructure which is not a field?

I have in my notes that there is such a substructure and it is $\mathbb Z$ because $\mathbb Z \subseteq \mathbb Q$

But I don't think that I fully understand it.

We wrote the axioms of field and that

$x \neq 0 \rightarrow (\exists y)(x*y=y*x=1)$ does not satisfy rest of the theory.

but I don't know exactly why.

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    Let $x = 2$ then what is the solution to $2*y = y*2 = 1; y \in \mathbb Z$? – 2017-01-08
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    What definition of "substructure" are you using? – 2017-01-08
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    Definition: A is substructure of B, if they have same language and $A \subseteq B$ and for every symbol from language is $S^A = S^B $ – 2017-01-08
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    @BillDubuque: what choice does the OP have about the definition of substructure? See my comments on the answers below. – 2017-01-08
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    @Rob I have encountered some uncommon definitions of (sub)structure, so my point was merely to nudge the OP to be a bit more precise. – 2017-01-08
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    @BillDubuque: where have you encountered these alternative definitions? The notions as used in universal algebra and model theory are uncontentious. – 2017-01-08
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    @Rob Over a few decades in forums analogous to this (e.g. sci.math). The matter has to do with convention, not contention. – 2017-01-08
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    @Bill: I'm intrigued to know what alternatives you have in mind that could possibly be relevant to this question. – 2017-01-08
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    @Rob I don't recall specific examples, but probably they had to do with whether or not the notion of substructure is determined completely by the signature or by some other context/convention. – 2017-01-08

3 Answers 3

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A possible "substructure" here is "subring" in this context. The integers are a subring of the field $\mathbb{Q}$. But they are not a subfield, because not all nonzero elements are invertible, e.g. the equation $2x=1$ has no solution in the integers.

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    Well,, I didn't downvote but considering the nature of the question, this vocabulary is probably over the OP's head. – 2017-01-08
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    @fleablood OK, thanks for your help. I tried to explain it without technical terms. – 2017-01-08
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    I honestly wouldn't have downvoted and disapprove of the one who did. – 2017-01-08
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    Well, I can't find a reason why this answer has negative vote. Clearly fleablood's answer is better and very nice, but I keep this answer here for others to see how some users vote. – 2017-01-09
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The answer to your question depends on what kind of substructure you're looking for. (That might have been explained in class but didn't get to your notes.)

There is no subset of $\mathbb{Q}$ that is itself a field with the usual addition and multiplication.

$\mathbb{Z}$ is a subring of $\mathbb{Q}$ when you think of the latter as a ring.

Any subset of $\mathbb{Q}$ is a "substructure" if "being a set" is the only structure you care about.

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    See my comment on Fleablood's answer. The fact that the OP has written out the signature of this structure makes it clear that "structure" is being used in the usual sense it has in universal algebra and model theory. The OP doesn't have any choice about what "substructure" means as it is part of the subject matter of the branch of mathematics he or she is studying. – 2017-01-08
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I don't know what "structure" is supposed to mean. But I can easily explain to you why $\mathbb Z$ is not a field.

It is, as you state, because the multiplicative inverse axiom fails. For $\mathbb Z$ to be a field, it must be true that for all $x \ne 0$ in $\mathbb Z$, then exists a $y \in \mathbb Z$ so that $x*y = y*x = 1$. Such a number we would call $y = \frac 1x$. It is called the "multiplicative inverse of $x$" because $x*\frac 1x = \frac 1x *x = 1$.

That simply is not true for $\mathbb Z$. Take any $x \ne \pm 1$. Say for instance $x = 2$. Then there does not exist any integer $y$ so that $x*y = y*x = 1$. (For example: if $2*y = y*2 =1$ then $y = \frac 12$ and $y = \frac 12 \not \in \mathbb Z$).

$\mathbb Z$ is not a field, because it is not true that ever element other than $0$ has a multiplicative inverse. In fact, other than $1$ and $-1$, no integer has a multiplicative inverse.

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    Actually from a strictly axiomatic and definition point of view it might not be as trivial as it sound to prove $0 \ne x \in \mathbb Z$ and $x \ne \pm 1$ and $1/x \in \mathbb Q$ that it is so that $1/x \not \in \mathbb Z$. – 2017-01-08
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    .... which to prove would involve formally defining $\mathbb Z$ and I'm not entirely sure how that was done in your class. Which I guess goes back to the issue: what does you class define as a "structure". – 2017-01-08
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    [Structure](https://en.wikipedia.org/wiki/Structure_(mathematical_logic)) as used in this question is a fundamental notion in model theory and universal algebra. In this case a substructure of the structure $\mathbb{Q}$ is a subset that contains the constants $0$ and $1$ and is closed under the operations $+$, $-$ and $\times$ (not $*$, please!). Any subring (with $1$) of $\Bbb{Q}$ other than $\Bbb{Q}$ itself is a substructure but not a field. – 2017-01-08
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    Right. So basically a ring with unity. To be closed under multiplication means all $n\times m \in Z$ but it need not be the case for any $m \in \mathbb Z$ that there are elements $j,k; j\times k = m$ (other than $j$ or $k = 1$). I don't know whether this class formally defines Z and Q or just assumes we can presume everything we know is true. Proving that Z as a structure does *not* have the multiplicative inverse property may not be trivial but if we define Q to have order we know $0 < 1/n < 1$ and there are no integers between 0 and 1. But it does depend on our definitions. – 2017-01-08
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    The definitions relevant to this question are well-known and uncontentious. A subset $X$ of a ring $R$ is closed under multiplication iff $ab \in X$ whenever $a\in X$ and $b\in X$. – 2017-01-08