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-l} \}$ 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 not even is 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.

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.