I'm re-reading some material and came to a question, paraphrased below:

Find a first order sentence in $\mathcal{L}=\{0,+\}$ which is satisfied by exactly one of the structures $(\mathbb{Z}\oplus \mathbb{Z}, (0,0), +)$ and $(\mathbb{Z}, 0, +)$.

At first I was thinking about why they're not isomorphic as groups, but the reasons I can think of mostly come down to $\mathbb{Z}$ being generated by one element while $\mathbb{Z}\oplus \mathbb{Z}$ is generated by two, but I can't capture this with such a sentence.

I'm growing pessimistic about finding a sentence satisfied in $\mathbb{Z}\oplus \mathbb{Z}$ but not in the other, since every relation I've thought of between some vectors in the plane seems to just be satisfied by integers, seen by projecting down on an axis.

In any case, this is getting kind of frustrating because my guess is there should be some simple statement like "there exists three nonzero vectors that add to 0 in the plane, but there doesn't exist three nonzero numbers that add to 0 in the integers" (note: this isn't true).

  • 1
    $\begingroup$ Great question, but (I think) a duplicate of this one: math.stackexchange.com/questions/27635/model-theory-problem $\endgroup$
    – Myself
    Commented May 28, 2011 at 16:49
  • $\begingroup$ Ah, yes, it is. I didn't notice that question when I was writing this one out. $\endgroup$
    – matt
    Commented May 28, 2011 at 17:03
  • $\begingroup$ Pardon my necromancy, but the reason that every relation between vectors in the plane is also true in one axis is that infinite vector spaces over a given field are always elementarily equivalent; in particular, ${\bf R}^2\equiv \bf R$. $\endgroup$
    – tomasz
    Commented Dec 9, 2012 at 15:33

1 Answer 1


Here's one: $$ (\forall x)(\forall y)\Bigl[(\exists z)(x=z+z) \lor (\exists z)(y=z+z) \lor (\exists z)(x+y=z+z)\Bigr] $$ This sentence is satisfied in $\mathbb{Z}$, since one of the numbers $x$, $y$, and $x+y$ must be even. It isn't satisfied in $\mathbb{Z}\oplus\mathbb{Z}$, e.g. if $x=(1,0)$, $y=(0,1)$, and $x+y=(1,1)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .