2
$\begingroup$

Setting Let $\mathcal{L} = \{+,0\}$. I want to show that $$\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$$. Note $\equiv$ means elementary equivalence in this question.

Updated Problem

My issue is that given the four ring axioms for addition and identity:

$$ \forall x \forall y \forall z ~ (x + y) + z = x + (y + z)\\ \forall x x + 0 = 0 + x = x\\ \forall x \exists y x + y = y + x = 0\\ \forall x \forall y x + y = y + x$$

it seems like $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$. As someone commented below if I can use multiplication then I may prove $\mathbb{Z} \oplus \mathbb{Z} \not\equiv \mathbb{Z}$, but multiplication is not in $\mathcal{L}$.

$\endgroup$
16
  • 1
    $\begingroup$ That's a mess. $\mathbb Z$ does not "satisfy all field axioms." There are no multiplicative inverses. Second of all, it is possible for $(R,\cdot,+.0,1)$ to be different from $(S,\cdot,+,0,1)$ but $(R,+,0)$ and $(S,+,0)$ to be the same. $\endgroup$
    – Thomas Andrews
    Feb 2, 2015 at 23:43
  • $\begingroup$ If you can use multiplication, then it is easy: consider the formula against divisors of zero $xy=0\implies (x=0\lor y=0)$. [Note that $+$ was not used here.] $\endgroup$
    – Berci
    Feb 2, 2015 at 23:47
  • $\begingroup$ For example, define $S=\mathbb Z$, $m\cdot_S n=0$ for all $m,n\in S$. Then $(S,\cdot_S,+,0,1)$ is different from $(\mathbb Z,\cdot, +,0,1)$, but $(S,+,0)$ is the same as $(\mathbb Z,+,0)$. $\endgroup$
    – Thomas Andrews
    Feb 2, 2015 at 23:47
  • $\begingroup$ @ThomasAndrews sorry for making such an elementary mistake in my proof. The confusion on my side still stands however, since doesn't the statement ℤ⊕ℤ≢ℤ depend on the choice of language? Is that what you were trying to show with S=ℤ, m⋅Sn=0, and (S,+,0) vs (ℤ,+,0)? $\endgroup$
    – chibro2
    Feb 2, 2015 at 23:55
  • 2
    $\begingroup$ Maybe you realized that the answer of Brian M. Scott in your latter question is in fact an answer of this question. $\endgroup$
    – Hanul Jeon
    Feb 3, 2015 at 1:49

2 Answers 2

Reset to default
6
$\begingroup$

Here is Brian Scott's sentence, slightly simplified.

$$\forall x\forall y\exists z ((z+z=x)\lor (z+z=y)\lor (z+z=x+y))$$

This is true in $\mathbb Z$ - it essentially says that given $x,y$, at least one of $x,y,x+y$ is even.

It is not true for $\mathbb Z\oplus\mathbb Z$, for example if $x=(1,0),y=(0,1)$.

$\endgroup$
-1
$\begingroup$

I am not quite what the question is here. But I will answer the question

Show that there does not exist a bijective group homomorphism (isomorphism) between $Z$ and $Z \times Z.$

Note that every element of $Z$ can be written as $n.1$ or $-(n.1)$ where $n.$ means add $n$ copies of together and $-$ means take the additive inverse. i.e. $1$ generates the group.

$Z\times Z$ does not have this property for any element of it. If we take $(a,0)$ with $a \neq 0,$ as our generator we never get non-zero in the second coordinate. If we take $(0,b)$ we never get it the first, If we take $(a,b)$ with both non-zero, we'll never zero in either apart from $(0,0).$

So no element generates.

Since the property of having a group element generate is defined purely in terms of the group actions, it would be preserved by an isomorophism. Hence no isomorphism exists.

(I discuss more of these sorts of arguments in my forthcoming book Proof Patterns. )

$\endgroup$
6
  • $\begingroup$ Your argument does not show that $\mathbb{Z}$ and $\mathbb{Z} \times \mathbb{Z}$ are not elementarily equivalent (the first-order language of groups isn't expressive enough to talk about generating sets). $\endgroup$
    – Rob Arthan
    Feb 3, 2015 at 0:04
  • $\begingroup$ @RobArthan Yeah, unfortunately, the question did not make clear that elementary equivalence was what was being requested. $\endgroup$
    – Thomas Andrews
    Feb 3, 2015 at 0:05
  • $\begingroup$ I think you accidentally the whole question ^^ $\endgroup$
    – Olivier Bégassat
    Feb 3, 2015 at 0:07
  • $\begingroup$ As a general rule, when you are "not quite [sure] what the question is..." ask questions, don't jump into writing an answer. $\endgroup$
    – Thomas Andrews
    Feb 3, 2015 at 0:33
  • $\begingroup$ There is a theorem stating that if some function $j : \mathcal{M} \rightarrow \mathcal{N}$ is an isomorphism, then $\mathcal{M} \equiv \mathcal{N}$. However I am not absolutely sure if proving $j$ is not an isomorphism is sufficient to say $\mathcal{M} \not\equiv \mathcal{N}$ $\endgroup$
    – chibro2
    Feb 3, 2015 at 0:55

You must log in to answer this question.

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