Prove that $\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$.

Question

Let $r,s,t\in\mathbb{Q}$. Prove that $\mathbb{Q}(r+s\sqrt{t})=\mathbb{Q}(\sqrt{t})$.

Ok. So I've fallen a little behind in my algebra class, and I'm a bit confused on how to approach this problem. I know that $\mathbb{Q}(\sqrt{t})=\{a+b\sqrt{t}\mid a,b\in\mathbb{Q}\}$. So I'm assuming that I need to show $\mathbb{Q}(r+s\sqrt{t})\subseteq\mathbb{Q}(\sqrt{t})$ and $\mathbb{Q}(\sqrt{t})\subseteq\mathbb{Q}(r+s\sqrt{t})$. However, I'm unsure of how to go about defining $\mathbb{Q}(r+s\sqrt{t})$ as a set.

Another approach I am considering is to show that $\sqrt{t}$ and $r+s\sqrt{t}$ have the same minimal polynomial in $\mathbb{Q}[x]$. This would prove that $\mathbb{Q}(r+s\sqrt{t})\cong\mathbb{Q}(\sqrt{t})$, which may get me some where closer to where I need to be.

Please steer me in the correct direction. I prefer advice to answers. Thank you in advance for your help.

Additional info: If $K$ is a field and $F$ is a subfield of $K$ and $u\in K$, then $F(u)$ is the simple extension of $F$ - the intersection of all subfields of $K$ that contain both $u$ and $F$. So, in my problem, I'm assuming that $K=\mathbb{C}$ and $F=\mathbb{Q}$. I also know that $u$ is the root of a unique monic irreducible polynomial (called the minimal polynomial of $u$) in $F[x]$- the set of all polynomials with coefficients in $F$.

Answer 1

The question is false when $s = 0$ and $t$ is not a square, for obvious reasons. So suppose $s\neq 0$.

By definition, $\mathbb{Q}(r+s\sqrt{t})$ is some field (actually, the smallest) containing both $\mathbb{Q}$ and $r+s\sqrt{t}$. But if it contains $\mathbb{Q}$, then in particular it contains $r$ and $s$. Now:

• as it contains $r+s\sqrt{t}$ and $r$ and it's a field, we may subtract one from the other to show that it contains $s\sqrt{t}$;
• as it contains $s\sqrt{t}$ and $s\neq 0$, we may divide one by the other to show that it contains $\sqrt{t}$.

To sum up, it's a field, and it contains $\mathbb{Q}$ and $\sqrt{t}$: that is, it must feature somewhere in the big intersection of fields that defines $\mathbb{Q}(\sqrt{t})$, so we have $\mathbb{Q}(\sqrt{t}) \subseteq \mathbb{Q}(r+s\sqrt{t})$.

The converse is basically identical. $\mathbb{Q}(\sqrt{t})$ contains $\sqrt{t}$, which I can multiply by $s$ and then add $r$ to to show that it contains $r+s\sqrt{t}$.

(Sorry for writing up a full answer, but I felt that any hint I could give would give all the details away anyway!)

Answer 2

Taking $s\neq 0$

$r+s\sqrt t \in \mathbb Q(r+s\sqrt t) \implies s\sqrt t\in \mathbb Q(r+s\sqrt t)\implies \sqrt t\in \mathbb Q(r+s\sqrt t)$ since $s\neq 0$

Now $\mathbb Q(\sqrt t)$ is the smallest field containing $\mathbb Q$ and $\sqrt t$

Thus $\mathbb Q(\sqrt t)\subset \mathbb Q(r+s\sqrt t)$

similarly $r,s,\sqrt t\in \mathbb Q(\sqrt t)\implies r+s\sqrt t \in \mathbb Q(\sqrt t)\implies \mathbb Q(r+s\sqrt t)\subset \mathbb Q(\sqrt t)$