I have two unrelated questions regarding two claims made in Ireland and Rosen, A Classical Introduction to Modern Number Theory (2nd edition). Altough common for both is that there seems to be a logical step I'm missing.
Q1.
Let $\Omega$ denote the ring of algebraic integers. In Chapter 6 we have the following statements.
Proposition 6.1.8. If $\alpha \in \Omega$ then $\mathbb{Q}(\alpha) = \mathbb{Q}[\alpha]$.
Corollary. If $\alpha$ is an algebraic number of degree $n$ then $[\mathbb{Q}(\alpha) : \mathbb{Q}] = n$.
PROOF: By the proposition it is enough to show $[\mathbb{Q}[\alpha] : \mathbb{Q}] = n$. …
The $\alpha$ in the proposition is an algebraic integer, but the $\alpha$ in the corollary is an algebraic number (thus not necessarily an algebraic integer). So how can we apply the Proposition so immediately? Doesn't this require some more justification?
Q2.
Let $F$ be an algebraic number field, and let $D = \Omega \cap F$ be its ring of integers. In Chapter 12 we have the following.
Proposition 12.2.1. Every ideal $A$ of $D$ contains a basis for $F$ over $\mathbb{Q}$.
PROOF. Let $\beta_1, \dots, \beta_n$ be a basis for $F$ over $\mathbb{Q}$. By the preceding lemma there is a $b \in \mathbb{Z}, b \neq 0$, such that $b\beta_1, \dots, b\beta_n \in D$. Choose $\alpha \in A, \alpha \neq 0$. Then the elements $b\beta_1\alpha, \dots, b\beta_n\alpha$ are in $A$ and are a basis for $F$ over $\mathbb{Q}$.
It's the last part that I'm not seeing (immediately). Let $\beta \in F$. Since $\beta_1, \dots, \beta_n$ is a basis for $F$ over $\mathbb{Q}$, we have
$$ \beta = \sum_{i = 1}^{n} q_i \beta_i, $$
for unique $q_i \in \mathbb{Q}$. Now, for $b\beta_1\alpha, \dots, b\beta_n\alpha$ to be a basis for $F$ over $\mathbb{Q}$, we need to find $q_1', \dots, q'_n \in \mathbb{Q}$ such that
$$ \beta = \sum_{i = 1}^{n} q'_i b\beta_i\alpha = \sum_{i = 1}^{n} q_i \beta_i. $$
In other words, we need to have $q'_i = \frac{q_i}{b\alpha}$. The $1 / b$ part is OK, since the $q'_i$ are rational numbers. However, it is the $1 / \alpha$ part I don't understand, because $\alpha$ is not guaranteed to be invertible. In fact, if it were, we would have $A = D$.
Best Answer
I don't know if they've proven this already (I don't immediately see it in Chapter 6), but:
Proposition. If $\alpha$ is an algebraic number, then there exists an integer $d$ such that $d\alpha$ is an algebraic integer.
Proof. Let $$p(x) = a_nx^n + \cdots + a_1x + a_0$$ be a polynomial with integer coefficients such that $a_n\neq 0$ and $p(\alpha)=0$ (just clear denominators from a monic polynomial with rational coefficients that has $\alpha$ as a root). Multiplying $p(\alpha)$ through by $a_n^{n-1}$ we get $$a_n^{n-1}p(\alpha) = (a_n\alpha)^n + a_{n-1}a_n(a_n\alpha)^{n-1} + \cdots + a_1a_n^{n-1}(a_n\alpha) + a_0a_n^n = 0.$$ But this is $q(a_n\alpha)$, with $$q(x) = x^n + a_{n-1}a_nx^{n-1} + \cdots + a_1a_n^{n-1}x + a_0a_n^n.$$ Thus, $a_n\alpha$ is an algebraic integer. $\Box$
Thus, you can replace $\alpha$ with $d\alpha$, since $\mathbb{Q}[d\alpha] = \mathbb{Q}[\alpha]$, and apply the Theorem.
The second question is more easily understood by linear independence. The new elements are still linearly independent (you just multiplied each of them by the same nonzero number), so you know you still have a basis: it's the correct number of elements for a basis, and it's linearly independent.