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

Yes. In the additive first-order structure, the quantifier-free formulaholds exactly at zero, because cancellation in the additive group gives $x=0$. This first-order formula uses only the permitted function symbol and equality; it does not require a constant symbol for zero.

No. The map $s(x)=-x$ is an automorphism of a first-order structure for the additive integers, but it sends one to minus one. By automorphism invariance of parameter-free definability, every parameter-free definable singleton is fixed by every structure automorphism. The singleton containing one is not fixed, so it is not definable.

No. Definability of a function means definability of its graph. The same additive structure automorphism $s(x)=-x$ would have to preserve the graph of multiplication. It sends the graph point $(1,1,1)$ to $(-1,-1,-1)$, which is not a multiplication-graph point because $(-1)(-1)=1\ne -1$. This contradicts automorphism invariance of parameter-free definability.

No. Interpreting the printed ${\mathbb{Z}}^{⩾}$ as the nonnegative integers, negation is an additive structure automorphism carrying the positive integer one outside that set. The set is therefore not invariant under structure automorphisms and cannot have parameter-free definability. The same argument would apply if the notation were taken to mean the strictly positive integers.

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.

For the additive zero singleton, $x+x=x$ is already a quantifier-free formula, so it is also an existential formula and a universal formula with an empty quantifier block. If a nonempty block is required, add a dummy $\exists y$ or $\forall y$; first-order domains are nonempty.

For the two definable multiplicative singletons, existential formulas areIn the integers, the equation $x^2=x$ restricts $x$ to zero or one. In the first line, one never satisfies the inequality, whereas zero does with $y=1$. In the second, zero never satisfies it, whereas one does with $y=0$. The respective universal formulas are $\forall y(xy=x)$ and $\forall y(xy=y)$, as already verified.

Neither multiplicative singleton has a quantifier-free formula. In this constant-free, multiplication-only language, every term in the sole free variable $x$ is $x^k$ for a positive integer $k$. Every atomic formula $x^r=x^s$ is true at both zero and one. Every Boolean combination consequently has the same truth value at those two points, and cannot separate their singletons. This is the fact that zero and one have the same quantifier-free type in the multiplicative integers.

Thus additive zero admits all three forms; multiplicative zero and one admit existential and universal forms, but no quantifier-free form. All remaining cases were excluded from definability itself by structure automorphisms.