Using dimension arguments to prove the existence of polynomials

Question

Let $V=<1,\sqrt{2},\sqrt{3},\sqrt{6}> \subset \mathbb{R}$ be a $\mathbb{Q}$-vector space. Using a dimension argument prove that there exists an $f \in \mathbb{Q}[X]$ such that $\deg(f)\leq4$ and $f(\sqrt{2}+\sqrt{3})=0$. Find $f$. Find $\dim_{\mathbb{Q}}V.$

Any hints as to how to approach this exercise? I know that the polynomial is $x^4-10x^2+1$.

What I don't know is how to derive an existence proof from the dimension facts. The dimension of the space of polynomials of degree $4$ or less is $5$ and $\dim V \leq 4$... But I'm unsure how to relate these facts.

Should it be $\geq 4$? Otherwise, couldn’t you just use $f=0$? — Joe, Sep 22 '19 at 18:34
Oh, I see. So a proof by contradiction, if there is no such polynomial, the $1, x, x^2, x^3, x^4$ evaluated at $x= (\sqrt{2} + \sqrt{3})$ are all linearly independent vectors in your 4-dimensional space $V$. — Joe, Sep 22 '19 at 18:45

score 1 · Answer 1 · edited Sep 22 '19 at 18:57

We know that $\Bbb Q(\sqrt{2}+\sqrt{3})$ is a $4$-dimensional vector space over $\Bbb Q$, see for example

Is $\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})$?

How to show that $\mathbb{Q}(\sqrt{p},\sqrt{q}) \subseteq \mathbb{Q}(\sqrt{p}+\sqrt{q})$

Basis for $\mathbb Q (\sqrt2 , \sqrt3 )$ over $\mathbb Q$

Hence there is a monic polynomial $f$ of degree $4$ such that $f(\sqrt{2}+\sqrt{3})=0$, namely the minimal polynomial of $\sqrt{2}+\sqrt{3}$. As you said, this polynomial is known to be $x^4-10x^2+1$:

Find the minimal polynomial of $\sqrt2 + \sqrt3 $ over $\mathbb Q$

Using dimension arguments to prove the existence of polynomials

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