$\mathbb{Q}\!\left(\sqrt{2}+\sqrt{3}\right)$ definition question I know the definition of $\mathbb{Q}\!\left(\sqrt{2}\right)=\{a+b\sqrt{2}\mid a,b\in\mathbb{Q}\}$. Why can't you similarly say $\mathbb{Q}\!\left(\sqrt{2}+\sqrt{3}\right)=\{a+b\!\left(\sqrt{2}+\sqrt{3}\right)\mid a,b\in\mathbb{Q}\}$? Any intuitive answers without jargon are much appreciated, as I have minimal background.

Recall polynomials with rational coefficients is the set
$$\mathbb{Q}[x] = \{q_0 + q_1x + q_2x^2 + \dots q_nx^2 : q_i \in \mathbb{Q} \}$$
So if we let $x = \sqrt{2}$, we would get
\begin{align}
\mathbb{Q}[\sqrt{2}] &= \{q_0 + q_1\sqrt{2} + q_2\sqrt{2}^2 + \dots + q_n\sqrt{2}^n \}\\
&\approx\{q_0 + q_1\sqrt{2} \}
\end{align}
But if we let $x = \sqrt{2} + \sqrt{3}$
$$\mathbb{Q}[\sqrt{2} + \sqrt{3}] = \{q_0 + q_1(\sqrt{2} + \sqrt{3})+ q_2(\sqrt{2} + \sqrt{3})^2 + \dots + q_n(\sqrt{2} + \sqrt{3})^n : q_i \in \mathbb{Q}\} $$
You can see there is going to be a problem the further we go. But if you work out the expansion, you can see that
\begin{align}
\mathbb{Q}[\sqrt{2} + \sqrt{3}] &= \{q_0 + q_1(\sqrt{2} + \sqrt{3})+ q_2(\sqrt{2} + \sqrt{3})^2 + \dots + q_n(\sqrt{2} + \sqrt{3})^n \} \\
&\approx \{q_0 + q_1\sqrt{2} + q_2\sqrt{3} + q_3\sqrt{2}\sqrt{3} \} \\
&= \mathbb{Q}[\sqrt{2},\sqrt{3}]
\end{align}
