For what $c \in \mathbb{Q}$ is $\sqrt{2}+c\sqrt{3}$ a primitive element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$?

Question

I am trying to find the $c\in\mathbb{Q}$ so that $\sqrt{2}+c\sqrt{3}$ is a primitive element of $\mathbb{Q}(\sqrt{2},\sqrt{3})$, i.e. $\mathbb{Q}(\sqrt{2}+c\sqrt{3})=\mathbb{Q}(\sqrt{2},\sqrt{3})$.

In class we have already shown that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$, so I know that c=1 is possible.

If $c=0\in\mathbb{Q}$, then we would have $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2})$, which is not possible since $\sqrt{3}\notin\mathbb{Q}(\sqrt{2})$, I am not sure if I have to prove this further.

So know considering $c\in\mathbb{Q}\setminus\{0\}$, I followed the prove for $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$:

$"\supseteq"$: Since $\sqrt{2},\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$, then $\sqrt{2}+c\sqrt{3}\in\mathbb{Q}(\sqrt{2},\sqrt{3})$, by the closure of $\mathbb{Q}(\sqrt{2},\sqrt{3})$.

$"\subseteq"$: Since $(\sqrt{2}+c\sqrt{3})^{-1}=\frac{1}{\sqrt{2}+c\sqrt{3}}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$ and $\frac{\sqrt{2}-c\sqrt{3}}{\sqrt{2}-c\sqrt{3}}=1\in\mathbb{Q}$, then $\frac{1}{\sqrt{2}+c\sqrt{3}}\frac{\sqrt{2}-c\sqrt{3}}{\sqrt{2}-c\sqrt{3}}=c\sqrt{3}-\sqrt{2}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. Therefore, $\sqrt{2}+c\sqrt{3}+(c\sqrt{3}-\sqrt{2})=2c\sqrt{3}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. And since $\frac{1}{2c}\in\mathbb{Q}\subseteq\mathbb{Q}(\sqrt{2}+c\sqrt{3})$, $\sqrt{3}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$. Using the same procedure we can show that $\sqrt{2}\in\mathbb{Q}(\sqrt{2}+c\sqrt{3})$.

Is this correct?

Answer 1

If you know some galois theory this might be useful, otherwise just ignore it.

Let $K=\mathbb{Q}(\sqrt{2},\sqrt{3})$. By the fundamental theorem of Galois theory there is a bijection

$$\{ \text{subfields } K\supset E\supset \mathbb{Q}\} \to \{\text{subgroups of} \operatorname{Gal}(K/\mathbb{Q})\}$$ given by mapping a subfield $K\supset E\supset \mathbb{Q}$ to $\operatorname{Gal}(K/E)$ and mapping a subgroup $H\subset \operatorname{Gal}(K/\mathbb{Q})$ to its fixed field $K^H=\{ x\in K : \sigma (x)=x \ \forall \sigma\in H\}$.

So the question your asking is: for which $c\in \mathbb{Q}$ do we have: $K_c :=\mathbb{Q}(\sqrt{2}+c\sqrt{3})=K^{\{1\}}$ i.e. $\operatorname{Gal}(K/K_c)=1$.

One can show that $\operatorname{Gal}(K/\mathbb{Q})$ consists of the identity and the field homomorphisms induced by mapping $\sqrt{2}\mapsto -\sqrt{2}, \sqrt{3} \mapsto \sqrt{3}$ or $\sqrt{2}\mapsto \sqrt{2},\sqrt{3}\mapsto -\sqrt{3}$ or $\sqrt{2}\mapsto - \sqrt{2},\sqrt{3}\mapsto -\sqrt{3}$.

Now $K_c = K^{ \{1\}}$ iff no nontrivial element of the galois group fixes $\sqrt{2}+c\sqrt{3}$. Because every galois automorphism fixes $\mathbb{Q}$ one sees that $c\neq 0$ is a sufficient and necessary condition.

Answer 2

Suppose $K(a,b)|K$ is an algebraic extension and $\text{char }K=0$. You want to find $c\in K$, so that $K(c)=K(a,b)$. There's a simple algorithm you can use. Let $f, g$ be the minimal polynomials of $a, b$ over $K$, respectively. Let $Z$ be the splitting field of $f\cdot g$ (so the splitting field of both polynomials). Then you have $$f= \prod_{i=1}^r (x-a_i), \: g=\prod_{i=1}^s (x-b_i), $$ with $r, s$ the degrees of $f, g$. You can assume $a=a_1, b=b_1$. We now want to find $\lambda \in K$, such that $a+\lambda\cdot b\neq a_i+\lambda \cdot b_j$ for all $i=1,\ldots, r$ and $j=2,\ldots, s$. You will always find such a $\lambda$, because you've only excluded finitely many elements of $K$, which is infinite. And then set $c:= a+\lambda\cdot b$. Then you have $K(c)=K(a,b)$.

You're welcome to think about why this works (only in a field with characteristic $0$), or you can comment and I'll help you continue. In your case, you have $$ f=(x-\sqrt{2})(x+\sqrt{2}), \: g=(x-\sqrt{3})(x+\sqrt{3}) $$ over their splitting field, and now you see why $\lambda = 1$ works (among many, many others).

Answer 3

We need to decide whether $\theta=\sqrt{2}+c\sqrt{3}$ has degree $4$ over $\mathbb Q$.

This reduces to deciding whether $1,\theta,\theta^2,\theta^3$ are linearly independent over $\mathbb Q$.

We have $$ \pmatrix{1 \\ \theta \\ \theta^2 \\ \theta^3} = \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & c & 0 \\ 2+3c^2 & 0 & 0 & 2c \\ 0 & 2+9c^2 & 3+6c & 0 \\ } \pmatrix{1 \\ \sqrt2 \\ \sqrt3 \\ \sqrt6} $$ The determinant of the matrix is $2 c (9 c^3 - 4 c - 3)$ and so is zero for rational $c$ iff $c=0$.

This argument relies on knowing that $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $4$ over $\mathbb Q$ and that $1,\sqrt2,\sqrt3,\sqrt6$ is a basis.