minimal polynomials of two elements $\alpha$ and $\beta$ that are algebraically independent.

Question

Consider an extension of fields $K/F$ and $\alpha, \beta \in K$. If $\alpha$, $\beta$ are algebraically independent, is it true that the minimal polynomial of $\alpha$ (resp. $\beta$) on $K(\beta)$ (resp. $K(\alpha)$) are the same as the minimal polynomial of $\alpha$ (resp. $\beta$) on K? How could I prove this?

Answer 1

I'm afraid my answer here becomes a tad on the discursive side at some points, venturing into a few regions which are not, shall we say, exactly on the main line. I'll try to point the places where I stray from the "straight and narrow" path in what follows.

Let

$m_\alpha(x), m_\beta(x) \in F[x] \tag 1$

be the respective minimal polynomials of $\alpha$ and $\beta$ over $F$. The fact that $m_\alpha(x)$ and $m_\beta(x)$ are (tacitly) assumed to exist implies that $\alpha$ and $\beta$ are each algebraic over $F$; thus

$[F(\alpha):F], [F(\beta):F] < \infty; \tag 2$

furthermore, $m_\alpha(x)$ and $m_\beta(x)$ are each irreducible over $F$, as is well-known.

So far, so good. Now the next few lines are not absolutely essential to our main point, but nevertheless:

We observe that

$\alpha \notin F(\beta) \tag 3$

and

$\beta \notin F(\alpha), \tag 4$

since otherwise, for example

$\alpha = \displaystyle \sum_0^{[F(\beta):F]} a_i \beta_i, \; a_i \in F; \tag 5$

that is, $\alpha$ and $\beta$ together satisfy the bivariate polynomial

$p(x, y) = x - \displaystyle \sum_0^{[F(\beta):F]} a_i y^i \in F[x, y], \tag 6$

that is,

$p(\alpha, \beta) = 0 \tag 7$

which is just (5) written out in the form

$\alpha - \displaystyle \sum_0^m a_i \beta^i, \; a_i \in F = 0; \tag 8$

this contradicts the algebraic indendence of $\alpha$, $\beta$; thus (3), and similarly (4), bind. These facts are engaging in their own right, but not central to our major aims.

Back to the main line:

Now since

$F[x] \subset F(\beta)[x], \tag 9$

it follows that

$m_\alpha(x) \in F(\beta)[x]; \tag{10}$

thus if

$t_\alpha(x) \in F(\beta)[x] \tag{11}$

is the minimal polynomial of $\alpha$ over $F(\beta)$, it follows that

$t_\alpha(x) \mid m_\alpha(x) \tag{12}$

in $F(\beta)[x]$; hence we have some

$q_\alpha(x) \in F(\beta)[x] \tag{13}$

with

$m_\alpha(x) = t_\alpha(x) q_\alpha(x), \tag{14}$

which of course implies

$\deg t_\alpha(x) \le \deg m_\alpha(x); \tag{15}$

we obviously also have

$t_\alpha(\alpha) = 0. \tag{16}$

Now suppose

$t_\alpha(x) \ne m_\alpha(x); \tag{17}$

I claim we may thus infer that

$t_\alpha(x) \notin F[x]; \tag{18}$

for if this did not bind, (15)-(16) together contradict the uniqueness of $m_\alpha(x)$ as the minimal polynomial of $\alpha$ over $F$. Thus

$t_\alpha(x) \in F(\beta)[x] \setminus F[x]; \tag{19}$

which implies that some coefficient of

$t_\alpha(x) = \displaystyle \sum_0^{\deg t_\alpha(x)} t_j x^j \tag{20}$

must itself lie in $F(\beta) \setminus F$; but by virtue of (2), such $t_j$ may be expressed in the form

$t_j = \displaystyle \sum_0^{m_j} t_{jk} \beta^k, \; t_{jk} \in F, \; m_j > 0; \tag{21}$

then we may write

$t_\alpha(x) = \displaystyle \sum_0^{\deg t_\alpha(x)} \left (\sum_0^{m_j} t_{jk} \beta^k \right) x^j, \tag{22}$

where at least one of the $m_j > 0$, whence

$0 = t_\alpha(\alpha) = \displaystyle \sum_0^{\deg t_\alpha(x)} \left (\sum_0^{m_j} t_{jk} \beta^k \right ) \alpha^j \tag{23}$

expresses an algebraic dependence 'twixt $\alpha$ and $\beta$; that is, setting

$\phi(x, y) = \displaystyle \sum_0^{\deg t_\alpha(x)} \left (\sum_0^{m_j} t_{jk} y^k \right ) x^j \in F[x, y], \tag{24}$

we have

$\phi(\alpha, \beta) = 0. \tag{25}$

We conclude that

$t_\alpha(x) = m_\alpha(x). \tag{26}$

Interchanging the roles of $\alpha$ and $\beta$ in the above shows that the minimal polynomial $m_\beta(x)$ of $\beta$ over $F$ is the same as $t_\beta(x)$, the minimal polynomial of $\beta$ over $F(\alpha)$. And we are done.