Degree of field extension $\mathbb{Q}\subseteq\mathbb{Q}(i,i\sqrt2)$

Question

I have a field extension $\mathbb{Q}\subseteq\mathbb{Q}(i,i\sqrt2)$ that I want to find the degree of. Usually I find it easiest to find the minimal polynomial, but I can't start by saying $x=i,i\sqrt2$ now.

How can I find the minimal polynomial?

Answer 1

$\newcommand{\Q}{\mathbb{Q}}\newcommand{\i}{\mathfrak{i}}$ The first thing you want to do is to show that $\i \notin \Q(\i\sqrt{2})$. After you do this, you can break it up into a tower of simple extensions to find the degree:

$$ \left[\Q(\i,\i\sqrt{2}):\Q\right] = \left[\Q(\i,\i\sqrt{2}):\Q(\i\sqrt2)\right] \left[\Q(\i\sqrt{2}):\Q\right] = (2)(2) = 4 $$

The general theorem that lets you do this looks something like

Given a tower of field extensions $K \subset E \subset F$, we have $[F:K]=[F:E][E:K]$.

We need to show that $\i \notin \Q(\i\sqrt{2})$ so we know $\Q(\i,\i\sqrt{2})\neq\Q(\i\sqrt2)$ (i.e. so we know that $\left[\Q(\i,\i\sqrt{2}):\Q(\i\sqrt2)\right]\neq 1$). For the sake of contradiction if $\i \in \Q(\i\sqrt{2})$, we could write $\i = a + b\i\sqrt{2}$ for some $a,b \in \Q$ (this is because we can think of $\Q(\i\sqrt{2})$ as a vector space over $\Q$ with basis $\{1,\i\sqrt{2}\}$). Squaring both sides of this equation, we can reach a contradiction.

Now calculating $\left[\Q(\i,\i\sqrt{2}):\Q(\i\sqrt2)\right]$ comes down to recalling that we can write $\Q(\i,\i\sqrt{2})$ as $(\Q(\i))(\i\sqrt{2})$. Then if we give $(\Q(\i))$ the name $F$, and note that $2\in F$ and $\sqrt{2}\notin F$, we are really just calculating $[F:F(\sqrt{2})]$ (which is much less intimidating).

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If you still want to talk about minimal polynomials, you can talk about the minimal polynomial of each of the intermediate extensions. For the extension $\Q \subset \Q(\i\sqrt{2})$ the minimal polynomial is $x^2+2$. For the extension $\Q(\i\sqrt2) \subset \Q(\i,\i\sqrt{2})$ the minimal polynomial is $x^2+1$ (since we are just adjoinging $\i$). Now if you wanted to characterize the full extension, you could say that $\Q(\i,\i\sqrt{2})$ is the splitting field over $\Q$ of the polynomial $(x^2+2)(x^2+1) = x^4 + 3x^2 +2$.

Answer 2

Note that $\mathbb{Q}(i,i\sqrt2)=\mathbb{Q}(i,\sqrt2)$ and consider the tower $$\mathbb{Q}\subseteq\mathbb{Q}(\sqrt2)\subseteq\mathbb{Q}(\sqrt2)(i)$$ and use the tower law. In particluar $\sqrt2$ has degree $2$ over $\mathbb{Q}$ while $i$ has degree $2$ over $\mathbb{Q}(\sqrt2)$. This is because $i$ has degree at most $2$ over $\mathbb{Q}(\sqrt2)$ and has degree at least $2$ over $\mathbb{Q}(\sqrt2)$ since $i\notin\mathbb{Q}(\sqrt2)$. Hence the degree of the required extension is $2\times 2=4$. Alternatively show that $$\mathbb{Q}(i,\sqrt2)=\mathbb{Q}(i+\sqrt2)$$ and compute the degree of the minimal polynomial of $i+\sqrt2$ over $\mathbb{Q}$.

Answer 3

A less elegant but completely elementar solution is to prove that $$ F=\{a+b\cdot i+c\cdot \sqrt{2}+d\cdot i\sqrt{2}|a,b,c,d\in\mathbb Q\} $$ (clearly a four dimensional vector space over $\mathbb Q$) is actually a field. This amounts essentially to prove that the moltiplicative inverses of elements of $F$ are still in $F$. In that case $F$ cannot be anything else than $\mathbb{Q}(i,i\sqrt2)$