$\mathbb{Q}\!\left(\sqrt{2}+\sqrt{3}\right)$ definition question

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.

Answer 1

That's true. There are two ways to define $\mathbb{Q}(\sqrt{2})$.

Option 1: The way you did, $\mathbb{Q}(\sqrt{2}) = \{a+b\sqrt{2} : a,b\in\mathbb{Q}\}$.

Option 2: $\mathbb{Q}(\sqrt{2})$ is the minimal field which contains $\sqrt{2}$ and $\mathbb{Q}$.

Those two definitions are equivalent in this case. However if you replace $\sqrt{2}$ with $\sqrt{2}+\sqrt{3}$ they're no longer the same!

If $\mathbb{Q}(\sqrt{2}+\sqrt{3})$ is a field which contain $\sqrt{2}+\sqrt{3}$ it must also contain the product, that is $(\sqrt{2}+\sqrt{3})^2 = 5+\sqrt{6}$ and so must contain $\sqrt{6}$. However it is not hard to show that $\sqrt{6}\not\in \{a+b(\sqrt{2}+\sqrt{3}) : a,b\in\mathbb{Q}\}$.

You should try to multiply two elements in $\mathbb{Q}(\sqrt{2})$ and see how you get another element in $\mathbb{Q}(\sqrt{2})$ this is a really nice property $\sqrt{2}$ has but $\sqrt{2}+\sqrt{3}$ does not.

Answer 2

The issue is that $\mathbb Q(\sqrt2+\sqrt3)$ should be a field that contains both $\mathbb Q$ and $\sqrt2+\sqrt3$, so in particular it should be closed under multiplication, and therefore should contain $(\sqrt2+\sqrt3)^2=5+2\sqrt6$; but you see this element cannot be written in the form $a+b(\sqrt2+\sqrt3)$ for elements $a,b\in\mathbb Q$.

Answer 3

Because that's not the definition. The definition of $\mathbb Q(\sqrt 2)$ is the smallest field that extends the rationals and contains $\sqrt 2.$ And $\mathbb Q(\sqrt 2+\sqrt 3)$ is the smallest field that extends the rationals and contains $\sqrt 2+\sqrt 3.$

Intuitively, if you play around with $\{a+b\sqrt 2: a,b\in\mathbb Q\},$ you will see that it is a field, in particular that it is closed under addition, subtraction, multiplication and division. However, that is not true for $\{a+b(\sqrt 2+\sqrt 3):a,b\in \mathbb Q\}.$ For instance, $(\sqrt 2+\sqrt 3)^2$ cannot be written in the form $a+b(\sqrt 2+\sqrt 3)$ where $a$ and $b$ are rational, so it is not closed.

Answer 4

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}