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In the book we can read

Lemma 5.14. Let $K(\alpha) : K$ be a simple algebraic extension, let the minimal polynomial of a over $K$ be $m$, and let $\partial m = n$. Then $ \{ 1 , \alpha ,\dots \alpha^{n-1} \}$ is a basis for $K(\alpha)$ over $K$.

Proof The theorem is a restatement of Lemma 5.9.

Of course I reread such lemma, that states,

Lemma 5.9. Every polynomial $\alpha\in K[t]$ is congruent modulo $m$ to a unique polynomial of degree $< \partial m$.

This lemma is not even about dimension. I think there is some error, if not, I am unable to restate lemma 5.9 to arrive to lemma 5.14. In any case I'd like to understand some proof. I can figure out in the examples that a basis for the extension is that the lemma states (we can get $\alpha^2$, $\alpha^3$ etc from the product, but the e.v. operations only allow sums), but a more formal proof will be welcome.

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  • $\begingroup$ What does $\partial m$ mean? $\endgroup$ Commented May 27, 2023 at 17:49
  • $\begingroup$ The degree of the polynomial $m$ $\endgroup$ Commented May 27, 2023 at 20:17

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I'm not sure either what they meant by a restatement of lemma $5.9$. Let's look at lemma $5.14$. Consider the minimal polynomial of $\alpha$ over $K$, denoted by $f^{\alpha}_K := f = \sum^n_{i=0} b_i x^i \in K[x]$, where $b_n =1$.

The intuitive idea behind lemma $5.14$ is that you can express any power of $\alpha$ larger than (or equal to) $n$ by the elements $\{1,\alpha, \alpha^2,...\alpha^{n-1}\}$, because of the relation given by $f$ (you can rewrite the expression $f(\alpha) = 0$ to $\alpha^n = \sum^{n-1}_{i=0}b_i\alpha^i$). Therefore they already generate $K(\alpha)$ over $K$. The linear independence for the elements $\{1,\alpha, \alpha^2,...\alpha^{n-1}\}$ follows from the fact that the degree of the extension $K \subset K(\alpha)$ is exactly equal to the degree of $f$ (which is equal to $n$).

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  • $\begingroup$ For the l.i. how to prove that about the equality of degrees? $\endgroup$ Commented May 28, 2023 at 9:07
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    $\begingroup$ @RafaBudría Ah you can prove that $K(\alpha) \cong K[x] /(f)$. Where $K[x]/(f)$ is a field and it is generated over $K$ by the elements $\{ \overline{1}, \overline{x},..., \overline{x^{n-1}} \}$. $\endgroup$ Commented May 28, 2023 at 15:45
  • $\begingroup$ Oh, my. In fact it is truly a rewording of that lemma. What a joke. $\endgroup$ Commented May 28, 2023 at 15:57

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