II.4 · Fields & Order — Problem 4 #

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Problem. Can the complex numbers $\mathbb{C}$ be made into an ordered field?

Hint #

In any ordered field two facts hold (from Problem 3): $1>0$, and $x^2>0$ for every $x\neq 0$. Apply the second to $x=i$ and remember that $i^2=-1$.

Solution #

The set of complex numbers cannot be an ordered field. From the previous problem we know two things that hold in any ordered field:

$1>0$, and
$x^2>0$ for any nonzero $x$.

Assume, for contradiction, that $\mathbb{C}$ is an ordered field. Statement (2) applied to $x=i$ means $i^2>0$, and since $i^2=-1$, we get $-1>0$. Adding $1$ to both sides yields

$$-1+1>0+1,\qquad\text{i.e.}\qquad 0>1.$$

But this contradicts statement (1), $1>0$, together with trichotomy (we cannot have both $1>0$ and $0>1$). Hence our assumption was wrong: $\mathbb{C}$ cannot be an ordered field.

Long solution #

Strategy. The answer is no, and the proof is by contradiction. Note what must be ruled out: not just that the obvious candidate orders on $\mathbb{C}$ (by real part, or lexicographically) fail to respect the arithmetic, but that no order whatsoever can work. The way to rule out every candidate at once is to argue from facts that every ordered field must satisfy, the two proved in Problem 3, and exhibit one element of $\mathbb{C}$ that violates them. That element is $i$: it is nonzero, so its square would have to be positive, yet its square is $-1$, which an ordered field forces to be negative.

Imported facts (Problem 3). In any ordered field $(F,+,\cdot,<)$:

(F1) $1>0$;

(F2) $x^2>0$ for every $x\neq 0$.

Both were derived from the ordered field axioms alone, so they would hold in $\mathbb{C}$ too, should $\mathbb{C}$ admit an order.

Suppose, toward a contradiction, that some relation $<$ makes $(\mathbb{C},+,\cdot,<)$ an ordered field. The operations are the usual complex addition and multiplication; only the order is hypothetical. Then F1 and F2 hold for this order.

Step 1: $i\neq 0$. If $i=0$, then $-1=i^2=0$, so $0=1$, contradicting the field axiom $1\neq 0$.

Step 2: $-1>0$. Apply F2 to $x=i$: $i^2>0$. Since $i^2=-1$, this says $-1>0$, that is, $0<-1$.

Step 3: $1<0$. Add $1$ to both sides of $0<-1$ (axiom O3): $0+1<-1+1$, that is, $1<0$.

Step 4: the contradiction. F1 says $0<1$, and Step 3 says $1<0$. Trichotomy (O1) allows exactly one of $1<0$, $1=0$, $0<1$, yet we have derived two of them. Contradiction.

Therefore no order relation makes $\mathbb{C}$ an ordered field. $\blacksquare$

Remark. The obstruction is purely algebraic: any field in which $-1$ is a square, or even a sum of squares, cannot be ordered, because in an ordered field a sum of nonzero squares is positive. Fields that can be ordered are exactly the formally real fields, where $-1$ is never a sum of squares. Note what the theorem does not say: one can put orders on the set $\mathbb{C}$, for instance lexicographic order on $a+bi$. What fails is compatibility with multiplication (O4). Order and arithmetic cannot coexist on $\mathbb{C}$.