$x^{2} - 2$ has no root in $\mathbb{Q}$, so we enlarge $\mathbb{Q}$ to a field that has one. There are two ways to do it, and they give the same field: adjoin $\sqrt{2}$ inside $\mathbb{R}$, or build the root out of nothing as a quotient ring.
Definitions. A field $K$ containing a field $F$ is an extension of $F$. For $\alpha \in K$, $F(\alpha)$ is the smallest subfield of $K$ containing $F$ and $\alpha$. Since $K$ is a vector space over $F$, the degree $[K : F]$ is its dimension.
Why it works
$\mathbb{Q}(\sqrt{d})$ for $d$ not a perfect square is $\{a + b\sqrt{d} : a, b \in \mathbb{Q}\}$, which is $\mathbb{Q}[x]/(x^{2} - d)$. The inverse of $a + b\sqrt{d} \ne 0$ is $$\frac{a - b\sqrt{d}}{a^{2} - db^{2}},$$ where $a^{2} - db^{2} \ne 0$ because $\sqrt{d}$ is irrational.
Worked example: Multiplying
- $3 - \sqrt{3} + 6\sqrt{3} - 2 \cdot 3 = -3 + 5\sqrt{3}$.
Answer: $-3 + 5\sqrt{3}$.
Worked example: Inverting
- Multiply top and bottom by the conjugate $1 - 2\sqrt{3}$: the bottom becomes $1 - 4 \cdot 3 = -11$.
- $\dfrac{1 - 2\sqrt{3}}{-11} = -\dfrac{1}{11} + \dfrac{2}{11}\sqrt{3}$.
- Check: $(1 + 2\sqrt{3})(-1 + 2\sqrt{3}) = -1 + 2\sqrt{3} - 2\sqrt{3} + 12 = 11$.
Answer: $-\frac{1}{11} + \frac{2}{11}\sqrt{3}$.
Worked example: A cubic extension
- Expand: $\theta + \theta^{2} + \theta^{2} + \theta^{3} = \theta + 2\theta^{2} + \theta^{3}$.
- $\theta^{3} = 2$: the product is $2 + \theta + 2\theta^{2}$.
- Every element of $\mathbb{Q}(\theta)$ is $a + b\theta + c\theta^{2}$, and $[\mathbb{Q}(\theta) : \mathbb{Q}] = 3$.
Answer: $2 + \theta + 2\theta^{2}$.
The same recipe gives $\mathbb{C} = \mathbb{R}[x]/(x^{2} + 1)$, with $i$ the class of $x$, and it gives the finite fields of the last two lessons.
Common mistake
Key idea