# Find the minimal polynomial of the number $\sqrt{2} + \sqrt{3}$ on $\mathbb{Q}$ and on $\mathbb{Q}\sqrt{2}$

Find the minimal polynomial of the number $\sqrt{2} + \sqrt{3}$ on $\mathbb{Q}$ and on $\mathbb{Q}\sqrt{2}$.

I found for $x= \sqrt{2} + \sqrt{3}$ the polynomial $(\frac{x^2-5}{2})^2-6$.

I know my question may seem silly, but how can I ensure that this polynomial is minimal? How to find the minimal polynomial for $\mathbb{Q}\sqrt{2}$?

• Slightly nit-picky, but technically a minimal polynomial ought to be monic. Re: showing minimality, you're best off showing that it's irreducible, which you can do by showing it has no factors of degree 1 or 2 (consider why). For the second question, the minimal polynomial for $\sqrt2+\sqrt3$ and $\sqrt3$ ought to be of the same degree over this field (again, consider why), which hints at the answer somewhat. – πr8 Jan 29 '16 at 14:57
• – MJD Jan 29 '16 at 17:43

Noting that if $x=\sqrt2+\sqrt3$, $x^2=5+2\sqrt6\implies(x^2-5)^2=24\implies x^4-10x^2+1=0$, giving us $p(x)=x^4-10x^2+1$ as a good candidate for minimal polynomial over $\mathbb{Q}$. We now show that it is irreducible.

Noting that $p(x)$ is degree $4$, we see that any nontrivial factorisation will contain a factor of degree $1$ or $2$.

• A factor of degree $1$ corresponds to a linear factor of $p(x)$, and hence to a rational root of $p$. But the rational root theorem tells us that the only possible such roots are $\pm1$, which we can verify are not roots. So, we have no linear factors.
• A factor of degree $2$ would necessarily lead to a factorisation of the form $p(x)=(x^2-ax+1)(x^2+ax+1)$ or $p(x)=(x^2-ax-1)(x^2+ax-1)$ (constant terms must be $\pm1$ to match with $p(x)$, the $\pm ax$ factors must match up to guarantee $p(x)$ is an even function.

• $p(x)=(x^2-ax+1)(x^2+ax+1)=(x^2+1)^2-a^2x^2\implies2-a^2=-10$ by comparing the coefficient of $x^2$, which isn't soluble in $\mathbb{Q}$, so this isn't a valid reduction.

• $p(x)=(x^2-ax-1)(x^2+ax-1)=(x^2-1)^2-a^2x^2\implies-2-a^2=-10$ by comparing the coefficient of $x^2$, which also isn't soluble in $\mathbb{Q}$, so this also isn't a valid reduction.

So, $p(x)$ is irreducible, and thus is the minimal polynomial of $\sqrt2+\sqrt3$ over $\mathbb{Q}$.

Re: the same question over $\mathbb{Q}[\sqrt2]$, note that $\sqrt2+\sqrt3$ isn't in this field, so can't have a minimal polynomial of degree 1. It does vanish on a polynomial of degree 2 (namely $(x-\sqrt2)^2-3$), though, so it's immediately minimal, and we're done.

• I think it's enough to remark that $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q]=4$ and that $p(x)=x^4-10x^2+1$ is s.t. $p(\sqrt 2+\sqrt 3)=0$ to prove that $p$ is irreducible over $\mathbb Q$. – Surb Jan 29 '16 at 15:22
• @Surb Yep, that's true - I suppose the quickest route through that path is to show $[\mathbb Q(\sqrt 2,\sqrt 3):\mathbb Q(\sqrt2)]=2$ and $[\mathbb Q(\sqrt 2):\mathbb Q]=2$, then stitching those two together? – πr8 Jan 29 '16 at 15:30

Write your polynomial in monic for gives $p(x)=(x^2-5)^2-24 = x^4-10x^2+1$.

Show that there are no polynomial factors. If there is a linear factor, then it must be of the form $x-1$ or $x+1$ by the rational root test, which means that either $-1$ or $1$ must be a root.

Otherwise, there must be a factorization as two quadratics, and a factorization over the integers.

So, you want: $a,b\in\mathbb Z$ so that:

$$(x^2+ax\pm 1)(x^2+bx\pm 1)= x^4-10x^2+1$$

We quickly see that $a+b=0$ and the coefficint of $x^2$ is $ab\pm 2=-10$. Substituting $b=-a$ we need $a^2=8,12$.

There are no integer solutions for $a$, but interestingly, $\sqrt{8}\in\mathbb Z[\sqrt{2}]$ and $\sqrt{12}\in \mathbb Z[\sqrt{3}]$. That might show you how to find a minimal polynomial in those rings.

First of all, you can observe (or proof easily) that $[\mathbb Q(\sqrt 3,\sqrt{2}):\mathbb Q(\sqrt 2)]=2$ since $\sqrt 3\notin \mathbb Q(\sqrt 2)$ (why) and thus that $X^2-3$ is the minimal polynomials of $\sqrt 3$ over $\mathbb Q(\sqrt 2)$. Moreover, $\sqrt 3+\sqrt 2\in\mathbb Q(\sqrt 3,\sqrt 2)$ therefore its minimal polynomial over $\mathbb Q(\sqrt 2)$ has degree at most $2$ (and in fact exactly $2$ since $\sqrt 2+\sqrt 3\notin\mathbb Q(\sqrt 2)$ since $\sqrt 3\notin \mathbb Q(\sqrt 2)$) So it's necessarily of the form $$X^2+aX+b$$ where $a,b\in\mathbb Q(\sqrt 2)$. So now, just compute $a,b\in\mathbb Q(\sqrt 2)$ using the fact that $$(\sqrt 2+\sqrt 3)^2+a(\sqrt 2+\sqrt 3)+b=0$$ and you will have the result.