Annihilating an algebraic number with a polynomial with rational coefficients

1,016pages on
this wiki

We will examine algebraic numbers ($\mathbb{A}$) and a way to determine a polynomial with rational coefficients that will annihilate a given algebraic number. In particular, we will focus on algebraic numbers that can be expressed as a finite combination of additions, subtractions, multiplications, divisions, and n-th roots of integers.

One such method of annihilating an algebraic number is the use of linear algebra: viewing the set of algebraic numbers as being a vector space over the rational numbers.

In review of algebraic numbers, we define $\alpha$ to be an algebraic number if there exists a non-trivial (that is, with at least one non-zero coefficient) polynomial function $P\in\mathbb{Q}[x]$ with rational coefficients $p_0,p_1,\ldots,p_n$ with $p_n\ne0$ such that $\alpha$ is a root. That is:

$P(\alpha)=p_0+p_1\alpha+p_2\alpha^2+\cdots+p_n\alpha^n=0$

We then say that $P$ is an annihilator of $\alpha$ , and $n$ is the smallest degree possible for such a polynomial, we will say that $\alpha$ is an $n$-th degree algebraic number.

To that end, consider $\alpha=\sqrt2+\sqrt[3]3$ . To find an annihilator for this particular $\alpha$, we can simply use some algebra:

\begin{align}\alpha&=\sqrt2+\sqrt[3]3\\ \implies\quad\alpha-\sqrt2&=\sqrt[3]3\\ \implies\quad\left(\alpha-\sqrt2\right)^3&=\left(\sqrt[3]3\right)^3\\ \implies\quad\alpha^3-3\sqrt2\alpha^2+6\alpha-2\sqrt2&=3\\ \implies\quad\alpha^3+6\alpha-3&=\sqrt2(3\alpha^2+2)\\ \implies\quad\left(\alpha^3+6\alpha-3\right)^2&=\left(\sqrt2\left(3\alpha^2+2\right)\right)^2\\ \implies\quad\alpha^6+12\alpha^4-6\alpha^3+36\alpha^2-36\alpha+9&=18\alpha^4+24\alpha^3+8\\ \implies\quad\alpha^6-6\alpha^4-6\alpha^3+12\alpha^2-36\alpha+1&=0 \end{align}

And so we have a polynomial of degree 6. However, simple algebraic manipulations do not always work for any arbitrary algebraic number. For instance, if $\gamma=\sqrt[3]2+\sqrt[3]3$ , this method would not work.

Instead, we must explore another approach, using linear algebra. We will now view $\mathbb{A}$ as being a vector space over $\Q$ . That is, we will be concerned with vectors as being algebraic numbers, while the scalars are the rational numbers.

Now for an arbitrary $\alpha\in\mathbb{A}$ with degree $n$ , consider the subspace $V_{\alpha}=\text{span}\left(\big\{\alpha^m\mid m\in\Z\big\}\right)$ .

As $\alpha$ is an $n$-th degree algebraic number, we have $P(\alpha)=p_0+p_1\alpha+p_2\alpha^2+\cdots+p_n\alpha^n=0$ for some polynomial function $P\in\Q_n[x]$ with coefficients $p_0,p_1,p_2,\cdots,p_n$ . This implies that $\left\{1,\alpha,\alpha^2,\ldots,\alpha^n\right\}$ is a linearly dependent set, and so we take $B_\alpha'=\left\{1,\alpha,\alpha^2,\ldots,\alpha^{n-1}\right\}$ as a basis for $V_\alpha$ . However, we cannot solve for $\alpha^n$ in terms of basis $B_\alpha'$ , nor may we necessarily know $n$ .

What we are able to do, however, is perform a change of basis. This can be achieved by expanding powers of $\alpha$ and keeping track of possible basis vectors. Once we expand up to a particular power of $\alpha$ , say $\alpha^n$ , and found $n$ possible basis vectors, we can infer a new basis $B_\alpha=\left\{ \beta_1,\ldots,\beta_n\right\}$ , including 1 as a basis vector.

Back to the case of $\gamma=\sqrt[3]2+\sqrt[3]3$ , experience will inform us that in expanding powers of $\gamma$ , we should expect to see powers of $\sqrt[3]2,\sqrt[3]3$ , and products thereof. Thus, we can expect the following basis: $B_\beta=\left\{\beta_1,\ldots,\beta_n\right\}=\left\{\big(\sqrt2\big)^r\big(\sqrt[3]3\big)^s\Big|r,s\in\Z,0\le r\le1,0\le s\le2\right\}$

Thus, we will have identified 9 new basis vectors after expanded up to $\gamma^9$ in this example. Clearly, these are computations that should be left to a computer algebra system.

Once we have expanded $\alpha^0$ to $\alpha^n$ , we will have identified $B_\alpha$ , as well as determined each of $\alpha^0$ to $\alpha^n$ with respect to $B_\alpha$ . To that end, we can then also express $P(\alpha)$ in terms of basis $B_{\alpha_i}$ .