## Compute the degree of QQ(i + √2)

Compute the degree of $K = \mathbb{Q}(i + \sqrt{2})$ over $\mathbb{Q}$. Exhibit three distinct subfields of $K$. Deduce that the polynomial found in this previous exercise is irreducible over $\mathbb{Q}$.

Since we know $\theta = i + \sqrt{2}$ is a root of $p(x) = x^4 - 2x^2 + 9$, the degree of $\theta$ over $\mathbb{Q}$ is at most 4. In this previous exercise, we showed that $K$ contains both $i$ and $\sqrt{2}$, so that $\mathbb{Q}(i), \mathbb{Q}(\sqrt{2}) \subseteq K$. Now $i\sqrt{2} \in K$, so we also have $\mathbb{Q}(i\sqrt{2}) \subseteq K$. Note that the minimal polynomial of $i$ over $\mathbb{Q}$ is $a(x) = x^2 + 1$ (irreducible because $a(x+1)$ is Eisenstein at 2), the minimal polynomial of $\sqrt{2}$ is $b(x) = x^2 - 2$ (Eisenstein at 2) and the minimal polynomial of $i\sqrt{2}$ is $c(x) = x^2 + 2$ (Eisenstein at 2). So these subfields all have degree 2 over $\mathbb{Q}$.

Suppose $\alpha,\beta \in \mathbb{Q}$ and $(\alpha + i\beta)^2 = 2$. Comparing coefficients, we see that $\alpha^2 - \beta^2 = 2$ and $\alpha\beta = 0$. If $\alpha = 0$, then $\beta^2 = -2$, a contradiction since $\beta^2 \geq 0$. If $\beta = 0$, then $\alpha^2 = 2$, a contradiction since $\sqrt{2}$ is not rational. In particular, $\sqrt{2} \notin \mathbb{Q}(i)$, so that $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{2})$ are distinct.

Similarly, if $(\alpha + \beta i\sqrt{2})^2 = 2$, then either $\alpha^2 = 2$ or $\beta^2 = -1$. So $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(i\sqrt{2})$ are distinct.

Finally, if $(\alpha + i\beta\sqrt{2})^2 = -1$, then either $\alpha^2 = -1$ or $\beta^2 = 1/2$, again a contradiction. So $\mathbb{Q}(i)$ and $\mathbb{Q}(i\sqrt{2})$ are distinct.

Now we claim that the degree of $\sqrt{2}$ over $\mathbb{Q}(i)$ is 2. To see this, note that (since $\mathbb{Q}(i)$ and $\mathbb{Q}(\sqrt{2})$ are distinct) $x^2-2$ is irreducible over $\mathbb{Q}(i)$, and so is the minimal polynomial of $\sqrt{2}$ over $\mathbb{Q}(i)$. So $\mathbb{Q}(i,\sqrt{2}) = \mathbb{Q}(i+\sqrt{2})$ has degree 2 over $\mathbb{Q}(i)$; hence $K$ has degree 4 over $\mathbb{Q}$.

We can summarize this information using the following labeled lattice diagram.

A lattice of fields.