Smallest subring containing $\sqrt{5}$

Question

I want to find the smallest subring of $\mathbb R$ which contains $\mathbb Q$ and $\sqrt 5$.

I am sure that$\{a+b\sqrt{5}:a,b \in \mathbb Q \}$ is the right candidate.

I already showed that this is a subring (i guess even a field?)

But I dont know how to show that there is no other "smaller" subring.

Any tips?

Answer 1

Assuming you mean $\{a+b\sqrt{5}:a,b\in\Bbb Q\}$ as I asked, then it's not hard. Let $R\subseteq\Bbb R$ be a subring such that

$$\begin{cases}\Bbb Q\subseteq R\\\sqrt 5\in R\end{cases}.$$

Then because $R$ is a ring, $b\sqrt 5\in R$ for every $b\in\Bbb Q$ because rings are closed under multiplication. And since $b\sqrt 5\in R$, we have $a+b\sqrt 5\in R$ for each $a\in\Bbb Q$ since rings are also closed under sums. But then $\{a+b\sqrt{5}:a,b\in\Bbb Q\}\subseteq R$, so the set you showed is contained in every other ring with the properties you describe, hence it is the smallest.