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Hypothesis: Let $E$ be a two-dimensional separable extension of $F$. Say $E = F[\alpha, \beta]$. Then $\exists \gamma \in E$ s.t. $E = F[\gamma]$. Note that the case for beyond two dimensions follows from this result via induction.

Proof:

  1. Let $m_\alpha$ and $m_\beta$ be the minimal polynomials of $\alpha$ and $\beta$ respectively over $F$.

  2. Let $\alpha_1, \ldots , \alpha_m$ be the (distinct) roots of $m_\alpha$ and $\beta_1, \ldots , \beta_n$ be the (distinct) roots of $m_\beta$.

  3. Let $\gamma = \alpha + c \beta$ s.t. $\alpha + c\beta \ne \alpha_i + c\beta_j$ for all $(i,j) \ne (1,1)$. We can choose such a $c$ since the $a_i$ and the $b_j$ are distinct and $E$ is infinite (separable fields are always infinite).

    Question1: Why do these two facts yield that such a $c$ exists?

  4. To prove that $E = F[\gamma]$, it suffices to prove that $\beta \in F[\gamma]$, for then $\gamma - c\beta = \alpha \in F[\gamma]$ so that it would be clear $E = F[\gamma]$.

  5. Let $K$ be a splitting field for both $m_\alpha$ and $m_\beta$.

  6. Define $h(x) = m_\alpha(\gamma - cx) \in K[x]$.

  7. We have that $\beta$ is a root of both $h,m_\beta$ in $K[x]$.

  8. By choice of $c$, no other $\beta_j$ can be a root of $h$.

  9. Then $\gcd(h, m_\beta) = (x - \beta)$ in $K[x]$.

  10. Yet $h$ has coefficients in $F[\gamma]$ by construction.

  11. Then $h$ and $m_\beta$ have $(x - \beta)$ as a divisor in in $F[\gamma][x]$.

    Question 2: Why is this so?

  12. Then $\beta \in F[\gamma]$ as desired.

user1770201
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  • Here, dimension (DimensioN from now on) is used in a non-standard way to refer to a (non unique) number $n$ satisfying $E=F[x_1,\dots, x_n]$. There is no relation between that meaning and any other usual use of the word dimension. For example, the dimension of $E$ as an $F$-vector space is quite independent of DimensioN – eduard Jan 31 '19 at 09:50

1 Answers1

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This proof of Primitive element theorem is based on B. L. van der Waerden 's classical book Algebra: Volume I, pp 139-140, $\S 6.10$.

For question 1, I think you have a typo: $c$ is chosen from $F$ instead of $E$. So the assumption is that $F$ is infinite.

Such a $c$ exists, because on $E$, $α+cβ=αi+cβj$ for all $(i,j)≠(1,1)$ has 1 root for each equation, there are totally $mn-1$ equations. So just pick any $c$ on $F$, avoiding these $mn-1$ roots.

For question 2, to get $\texttt{gcd}(h(x), m_\beta(x))$, one only needs to follow the Euclidean division, a form of $x-\beta$ will always be driven out. Since $h(x), m_\beta(x) \in F[\gamma][x]$, the coefficient in the calculation will all be in $F[\gamma]$, hence $\beta \in F[\gamma]$.

athos
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