Show that there are models of $\mathbf{Q}$ where the following sentence is false
\begin{align*} \forall x \forall y (x \cdot y = y \cdot x) \\ \end{align*}
For reference, the ten finite axioms of $\mathbf{Q}$ are as follows. Presume $\forall x \forall y$ for all the following; the universal quantifiers are omitted for simplicity.
\begin{align*} (Q1) & & \mathbf{0} \neq x' \\ (Q2) & & x' = y' \implies x = y \\ (Q3) & & x + \mathbf{0} = x \\ (Q4) & & x + y' = (x + y)' \\ (Q5) & & x \cdot \mathbf{0} = \mathbf{0} \\ (Q6) & & x \cdot y' = (x \cdot y) + x \\ (Q7) & & \lnot x < 0 \\ (Q8) & & x < y' \leftrightarrow (x < y \lor x = y) \\ (Q9) & & \mathbf{0} < y \leftrightarrow y \neq \mathbf{0} \\ (Q10) & & x' < y \leftrightarrow (x < y \land y \neq x') \\ \end{align*}
My first intuition is to use the compactness theorem. But that doesn't seem to work. Any hints on what I can try?