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Annihilating an algebraic number with a polynomial with rational coefficients

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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} .

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