II.8 · Fields & Order — Problem 8 #

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Problem. $\mathbb{Q}[\sqrt2]$ inherits the usual order from $\mathbb{R}$, making it an ordered field. Find a different order $\prec$ on $\mathbb{Q}[\sqrt2]$ that still makes it an ordered field.

Hint #

There is a map from $\mathbb{Q}[\sqrt2]$ to itself that swaps $\sqrt2$ and $-\sqrt2$ and preserves addition and multiplication: the conjugation $a+b\sqrt2\mapsto a-b\sqrt2$. Pull back the usual order through this map. Concretely, declare $x\prec y$ iff the conjugate of $x$ is less than the conjugate of $y$ in $\mathbb{R}$.

Solution #

Write $a+b\sqrt2\lt c+d\sqrt2$ when the former is smaller than the latter as a real number. Define a new order on $\mathbb{Q}[\sqrt2]$ by

$$a+b\sqrt2\ \prec\ c+d\sqrt2 \quad\Longleftrightarrow\quad a-b\sqrt2< c-d\sqrt2.$$

In this order $\sqrt2\prec 0$ (since $-\sqrt2<0$), so $\prec$ is genuinely different from the usual ordering.

We must check this is an ordered field.

Order axioms.

Trichotomy is a quick check: $\prec$ satisfies it precisely because $<$ does (for any two elements, exactly one of $<,=,>$ holds for their conjugates).

Transitivity. Suppose $a+b\sqrt2\prec c+d\sqrt2$ and $c+d\sqrt2\prec e+f\sqrt2$. Then $a-b\sqrt2\lt c-d\sqrt2$ and $c-d\sqrt2\lt e-f\sqrt2$. Since $<$ is transitive, $a-b\sqrt2\lt e-f\sqrt2$, i.e. $a+b\sqrt2\prec e+f\sqrt2$.

Ordered field axioms.

Compatibility with $+$. If $a+b\sqrt2\prec c+d\sqrt2$, then adding $e+f\sqrt2$ to both sides yields $(a+e)+(b+f)\sqrt2$ and $(c+e)+(d+f)\sqrt2$. We know $a-b\sqrt2\lt c-d\sqrt2$; adding $e-f\sqrt2$ gives $(a+e)-(b+f)\sqrt2<(c+e)-(d+f)\sqrt2$, which is exactly $(a+e)+(b+f)\sqrt2\prec(c+e)+(d+f)\sqrt2$.

Compatibility with $\cdot$. Consider $a+b\sqrt2\succ 0$ and $c+f\sqrt2\succ 0$. Then $a-b\sqrt2>0$ and $c-f\sqrt2>0$, and their product $ac+2bf-(af+bc)\sqrt2>0$. By definition this means $ac+2bf+(af+bc)\sqrt2\succ 0$. Since that is the product $(a+b\sqrt2)(c+f\sqrt2)$, we are done.

Long solution #

Strategy. The usual order on $\mathbb{Q}[\sqrt2]$ comes from embedding it in $\mathbb{R}$ “as itself.” There is a second map $\mathbb{Q}[\sqrt2]\to\mathbb{R}$ that respects the arithmetic: conjugation $\sigma(a+b\sqrt2)=a-b\sqrt2$, which treats $\sqrt2$ as if it were $-\sqrt2$. The plan: (1) check that $\sigma$ preserves addition and multiplication; (2) define $x\prec y$ to mean $\sigma(x)<\sigma(y)$ and verify the four order axioms; (3) confirm the new order differs from the old, using the sign of $\sqrt2$.

Two number lines showing the two embeddings of Q[sqrt 2] into R: the identity places sqrt 2 at 1.41, the conjugate embedding places it at -1.41

Two ways to superimpose $\mathbb{Q}[\sqrt2]$ on the real line. Top: the identity embedding gives the usual order, with $\sqrt2\approx 1.41$ positive. Bottom: the conjugate embedding $\sigma$ sends $\sqrt2$ to $-\sqrt2$ and $1+\sqrt2$ to $1-\sqrt2$, so under $\prec$ the element $\sqrt2$ is negative. Dashes connect the two placements of the same element.

Step 1: $\sigma$ preserves the field operations. Applying $\sigma$ twice returns the input, so $\sigma$ is its own inverse and hence a bijection. With $x=a+b\sqrt2$ and $y=c+d\sqrt2$,

$$\sigma(x+y)=(a+c)-(b+d)\sqrt2=(a-b\sqrt2)+(c-d\sqrt2)=\sigma(x)+\sigma(y),$$$$\sigma(xy)=(ac+2bd)-(ad+bc)\sqrt2=(a-b\sqrt2)(c-d\sqrt2)=\sigma(x)\,\sigma(y),$$

where both products were expanded using $(\sqrt2)^2=2$. Also $\sigma(0)=0$ and $\sigma(1)=1$.

Step 2: the new order. Declare

$$x\prec y \quad\Longleftrightarrow\quad \sigma(x)<\sigma(y),$$

where $<$ is the usual real order; spelled out, $a+b\sqrt2\prec c+d\sqrt2$ iff $a-b\sqrt2\lt c-d\sqrt2$. Each axiom for $\prec$ now translates into the corresponding axiom for $<$ on $\mathbb{R}$.

(O1) Trichotomy. Exactly one of $\sigma(x)<\sigma(y)$, $\sigma(x)=\sigma(y)$, $\sigma(y)<\sigma(x)$ holds in $\mathbb{R}$. The outer two are $x\prec y$ and $y\prec x$; the middle one is $x=y$, because $\sigma$ is injective.
(O2) Transitivity. $x\prec y$ and $y\prec z$ mean $\sigma(x)<\sigma(y)<\sigma(z)$, so $\sigma(x)<\sigma(z)$, that is, $x\prec z$.
(O3) Compatibility with $+$. If $\sigma(x)<\sigma(y)$, then adding $\sigma(w)$ to both sides gives $\sigma(x)+\sigma(w)<\sigma(y)+\sigma(w)$. Since $\sigma$ preserves addition, this is $\sigma(x+w)<\sigma(y+w)$, that is, $x+w\prec y+w$.
(O4) Compatibility with $\cdot$. If $0\prec x$ and $0\prec y$, then $0<\sigma(x)$ and $0<\sigma(y)$ (using $\sigma(0)=0$), so $0<\sigma(x)\sigma(y)=\sigma(xy)$, that is, $0\prec xy$.

The field axioms are untouched, since $+$ and $\cdot$ are unchanged. So $(\mathbb{Q}[\sqrt2],+,\cdot,\prec)$ is an ordered field.

Step 3: $\prec$ differs from $<$. Since $\sigma(\sqrt2)=-\sqrt2<0=\sigma(0)$, we have $\sqrt2\prec 0$, whereas $\sqrt2>0$ in the usual order. The two orders disagree about the sign of $\sqrt2$. (Likewise $\sigma(1+\sqrt2)=1-\sqrt2\approx-0.41$, so $1+\sqrt2\prec 0\prec 1$, as in the figure.) $\blacksquare$

Remark (are there others?). These are the only two orderings of $\mathbb{Q}[\sqrt2]$. An ordering is determined by its set of positive elements, which must be closed under $+$ and $\cdot$ and must contain exactly one of $x,-x$ for each $x\neq 0$. Squares are forced to be positive, which pins down the order on $\mathbb{Q}$; the only remaining freedom is the sign of $\sqrt2$.