II.1 · Fields & Order — Problem 1 #

← Back to II. Fields & Order, Problem 1

Problem. The set $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt2 \mid a,b\in\mathbb{Q}\}$ is a field. Since it sits inside $\mathbb{R}$, most field axioms come for free. Identify which axioms must actually be checked, and which are inherited from $\mathbb{R}$.

Hint #

A field axiom that is a universally quantified equation (like associativity, commutativity, distributivity) automatically holds in any subset of a field, because it holds for all real numbers in particular. The only axioms you must verify are the ones that assert the existence of an element inside the set — namely that the set is closed under addition and multiplication, and that it contains the identities and the inverses.

Solution #

To check that $\mathbb{Q}[\sqrt{2}]$ is a field, we must verify that all of the defining axioms hold. The important observation is that $\mathbb{Q}[\sqrt{2}]$ is a subset of $\mathbb{R}$, and since the axioms A2–A4, M2–M4, and D hold for all real numbers, they certainly hold for all elements of $\mathbb{Q}[\sqrt{2}]$. Hence we only have to check A1 (closure under $+$), M1 (closure under $\cdot$), and the inverse/identity conditions sitting in A4 and M4 (that the inverses actually land in the set).

To do this, consider $x,y\in\mathbb{Q}[\sqrt{2}]$, so that we can write

$$x = a + b\sqrt{2} \quad\text{and}\quad y = c + d\sqrt{2}$$

where $a,b,c,d$ are all rational. Then

$$x+y = (a+c) + (b+d)\sqrt{2}$$

lies in $\mathbb{Q}[\sqrt{2}]$ since $a+c$ and $b+d$ are rational. This shows A1. Similarly,

$$x\cdot y = (ac+2bd) + (ad+bc)\sqrt{2}$$

lies in $\mathbb{Q}[\sqrt{2}]$ since $ac+2bd$ and $ad+bc$ are rational, showing M1. For the inverses, the additive inverse of $a+b\sqrt{2}$ is $-a-b\sqrt{2}$, and the multiplicative inverse of a nonzero $a+b\sqrt{2}$ is

$$\frac{a}{a^2-2b^2} - \frac{b}{a^2-2b^2}\sqrt{2}.$$

Both have rational coefficients and are therefore in $\mathbb{Q}[\sqrt{2}]$.