Past exam of the model theory course of the University of Oxford / 2011 / 1 / b / ii

Yes. Zero is the unique absorbing element of the multiplicative first-order structure. It is defined byZero satisfies this universal formula. Conversely, substituting the actual integer zero for $y$ in the semantic verification forces $x=0$. This use of zero in the verification does not insert a constant symbol into the first-order formula.

Yes. One is the unique identity element of multiplication, defined byThe integer one satisfies the first-order formula, and testing at the actual integer one forces any satisfying $x$ to equal one. Thus this is a parameter-free universal formula even though the language has no named identity.

No. Use the prime-sign automorphism of the multiplicative integerswhere $v_2(n)$ is the exponent of two in $|n|$, a P-adic valuation. For nonzero integers, $v_2(mn)=v_2(m)+v_2(n)$, so $s(mn)=s(m)s(n)$; the identity also holds when a factor is zero. The exponent of two is unchanged by changing the sign, hence $s(s(n))=n$. Thus $s$ is a multiplicative structure automorphism.

It fixes one and sends two to minus two. The addition-graph point $(1,1,2)$ is sent to $(1,1,-2)$, outside the graph of addition. By automorphism invariance of parameter-free definability, addition is not definable in the multiplicative reduct.

No. The prime-sign automorphism of the multiplicative integers from the preceding part sends the positive integer two to minus two. It therefore does not preserve the nonnegative integers, contradicting the necessary automorphism invariance of parameter-free definability. This also excludes definability of the strictly positive integers if that convention for the printed notation is used.