Can the complex field structure define a non-trivial partial order?

Question

Consider the structure $(\mathbb{C};+,-,*,0,1)$. Certainly, it can't be ordered to make an ordered field. However, I am asking whether that structure can define any partial order other than equality. It does not have to be an order that makes it into an ordered field(which is impossible), it can be any partial order. Also, if the answer is no, then it will answer in the negative my follow-up question as to whether the structure can define any linear order.

Answer 1

Yes: every structure admits a definable partial order by $x\leq y$ iff $x = y$.

The complex field admits many other rather trivial definable partial orders. For example, we can define $x\leq y$ iff $(x = 0) \lor (x = y)$. This order has a least element $0$, and every other pair of elements is incomparable.

But the complex field admits no definable partial order (in fact no definable preorder) with infinite chains. This is exactly the content of the statement that the theory $\mathrm{ACF}_0$ does not have the strict order property, and it follows from the fact that the theory $\mathrm{ACF}_0$ is stable (equivalently, does not have the order property), which in turn follows from the fact that $\mathrm{ACF}_0$ is $\omega$-stable, which in turn follows from the fact that $\mathrm{ACF}_0$ is strongly minimal (every definable set is finite or cofinite), which has the cute consequence that Noah Schweber points out in the comments below.

Answer 2

Let R be the real field with a non-Archimedean total order. Define a complex number positive if the real part is positive and the imaginary part is infinitesimally smaller than the real part. It is a partial order on complex field. The same idea may be used to construct partial orders on real quaternions H to make it into a partially ordered division algebra.