# minimal polynomial given an algebraic number

I am trying to find the minimal polynomial for the algebraic number $1+\sqrt{2}+\sqrt{3}$. My original thought was just let $\alpha=1+\sqrt{2}+\sqrt{3}$. The method I use though seems very complicated. For example, $$\alpha^2=(1+\sqrt{2}+\sqrt{3})^2=1+\sqrt{2}+\sqrt{3}+\sqrt{2}+2+\sqrt{2}\sqrt{3}+\sqrt{3}+\sqrt{2}\sqrt{3}+3$$ $$\alpha^2=1+\sqrt{2}+\sqrt{3}+1+\sqrt{2}+\sqrt{3}+4+2\sqrt{2}\sqrt{3}$$ $$\alpha^2=4+2\alpha+2\sqrt{2}\sqrt{3}$$ If I substitute for $\sqrt{2}=\alpha-1-\sqrt{3}$ and $\sqrt{3}=\alpha-1-\sqrt{2}$, I'm still going to end up with $\sqrt{2}$ and $\sqrt{3}$. So I figured I'd square again. Moving the 4 over makes the calculation easier since binomials are easier than trinomials.... $$(\alpha^2-4)^2=(2\alpha+2\sqrt{2}\sqrt{3})^2$$ $$\alpha^4-8\alpha^2+16=4\alpha^2 +8\alpha\sqrt{2}\sqrt{3}+24$$ $$\alpha^4-12a^2-8=4\alpha(2\sqrt{2}\sqrt{3})$$ From the above calculation I see that $2\sqrt{2}\sqrt{3}=\alpha^2-2\alpha-4$ $$\alpha^4-12a^2-8=4\alpha(\alpha^2-2\alpha-4)$$ $$\alpha^4-4\alpha^3-4\alpha^2+16\alpha-8=0$$ But my question is, how do I KNOW this is the minimal polynomial? Yes, it is true now, since I constructed it so, that if $f(x)=x^4-4x^3-4x^2+16x-8$, then $f(\alpha)=0$ Do I simply attempt to factor out an $(x-\alpha)$. In that case, $f(x)=(x-\alpha)g(x)$. Then I show that $g(\alpha)\neq 0$?

• The degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ is $4$. Commented Jul 25, 2014 at 3:07
• I'm not sure what you mean... Commented Jul 25, 2014 at 3:10
• I'm not sure about the proof of Nicolas's statement but I've factorized your polynomial via a guess: link. An intuitive way of thinking about it is that all corresponding roots with alternating signs must also be taken to make the radicals disappear. Much like complex conjugate roots of a real quadratic equation. Commented Jul 25, 2014 at 3:12
• A comment on another question has a proof that the degree of minimal polynomial is 4. Commented Jul 25, 2014 at 3:24
• @typesanitizer That only shows the degree must be less than or equal to 4. Commented Apr 6, 2023 at 13:01

Consider the polynomial $$(y-\sqrt{2}-\sqrt{3})(y-\sqrt{2}+\sqrt{3})(y+\sqrt{2}-\sqrt{3})(y+\sqrt{2}+\sqrt{3}).$$ The first two terms have product $y^2-2\sqrt{2}y-1$ and the next two have product $y^2+2\sqrt{2}y-1$. Multiply. We get $y^4-10y^2+1$. We can now see that $\alpha$ is a zero of the polynomial $(x-1)^4-10(x-1)^2+1$.

To show this is minimal, some algebra is useful. One can show that the field $\mathbb{Q}(\sqrt{2})$ has degree $2$ over the rationals, since $\sqrt{2}$ is irrational. And $\mathbb{Q}(\sqrt{2},\sqrt{3})$ has degree $2$ over $\mathbb{Q}(\sqrt{2})$, since $\sqrt{3}$ cannot be expressed as $s+t\sqrt{2}$ with $s$ and $t$ rational. Thus $\mathbb{Q}(\alpha)$ has degree $4$ over the rationals, so the minimal polynomial has degree $4$.

• I am currently taking a graduate number theory course. I've not done any abstract in a while, so please give me a bit. If I understand, $\mathbb{Q}$ is a field and we can extend the field if we consider numbers $a+b\sqrt{2}$, where $a,b$ are rational. The notation for this new extended field is $\mathbb{Q}(\sqrt{2})$. What do you mean that $\mathbb{Q}(\sqrt{2})$ has degree 2? Is that simply due to the fact that a minimal polynomial can not have degree less than two because of the extension with $\sqrt{2}$? Commented Jul 25, 2014 at 3:31
• Let $F$ be a field, and let $\beta$ be algebraic over $F$. Then the field $F(\beta)$ obtained by adjoining $\beta$ to $F$ can be viewed as a vector space over $F$. The dimension of that vector space is called the degree of $F(\beta)$ over $F$, and this is the degree of the minimal polynomial of $\beta$ over $F$. Commented Jul 25, 2014 at 3:35
• Yes, except that we have to either show that they are linearly independent over the rationals, or appeal to the general (and not difficult) theorem that the degree of $F(\alpha,\beta)$ over $F$ is the product of the degree of $F(\alpha)$ over $F$ and $F(\alpha,\beta)$ over $F(\alpha)$. Basically just linear algebra here, nothing hard. Commented Jul 25, 2014 at 3:44
• Yes, the degree is $8$. But to do that we need to show that we do get a genuine tower, that for example $\sqrt{5}$ is not in $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Commented Jul 25, 2014 at 3:46
• Another way of showing that the degree $4$ thing you got is the minimal polynomial is to show it is irreducible over the rationals, or equivalently that it cannot be factored over the integers. For this it is more pleasant to work with the simpler polynomial in $y$. Commented Jul 25, 2014 at 3:52

For $~P,Q,R,S,T\in\mathbb Q^\star,~$ irreducible, with $~R\neq T~$ both squarefree, we have

\begin{align}a~=~P~+~Q\sqrt R~+~S\sqrt T\quad&=>\quad(a-P)^2~=~\ldots~=~K+M\sqrt N~\not\in~\mathbb Q\\\\&=>\quad\Big[(a-P)^2-K\Big]^2=~M~^2\cdot N~\in~\mathbb Q\end{align}

where $~K=Q~^2R+S^2T,~$ $~M=2~Q~S,~$ and $~N=R~T$.

If just want the know the answer, without knowing why it is the answer, you can use a verified implementation of algebraic numbers. For instance, if you load the Isabelle library Algebraic_Numbers/Algebraic_Numbers_Test.thy from the Archive of Formal proofs, you can just enter

value (code) "show (1 + sqrt 2 + sqrt 3)"


and will get your polynomial as output: the output will always deliver the minimal polymial. Moreover, if you want to test whether a given polynomial is minimal, i.e., irreducible, you can similarly invoke a verified polynomial factorization algorithm to test irreducibility.

value (code) "show_sf_factorization (factorize_int_poly [: -8, 16, -4, -4, 1 :])"


As a side remark, at the time of writing, the implementation uses resultants and Berlekamp-Zassenhaus factorization to compute the minimal polynomial.