Let $f \in \mathbb{Q}[X]$ have distinct roots $\alpha_1, \ldots, \alpha_d$. Let $L = Q(\alpha_1, \ldots, \alpha_d)$.
Then we know there is an natural injection $$\operatorname{Gal}(L / \mathbb{Q}) \rightarrow S_d: \sigma \mapsto \pi_{\sigma}$$ with $\pi_{\sigma}(i) = j\;$ if and only if $\sigma(\alpha_i) = \alpha_j$.
Define the discriminant of $f$ as $$\Delta(f) := \prod_{1 \leq i < j \leq d} (\alpha_i - \alpha_j)^2. $$
I've proven the following lemma:
Lemma: $\operatorname{Gal}(L / \mathbb{Q}) \subset A_d$ if and only if $\Delta(f)$ is the square of an element in $\mathbb{Q}$.
I now wish to show the following:
Problem: Suppose $d = 3$ and $f$ is irreducible. Prove that $\operatorname{Gal}(L/ \mathbb{Q})$ equals $S_3$ or $A_3$, where the latter holds if and only if $\Delta(f)$ is a square in $\mathbb{Q}$. Show with an example that irreducibility of $f$ is a necessary assumption.
I wasn't really sure why irreducibility of $f$ is needed. Can't I just conclude from the lemma that $\operatorname{Gal}(L/ \mathbb{Q}) = A_3$ when $\Delta(f)$ is a square in $\mathbb{Q}$ (since both sets have the same number of elements)? Do I need irreducibility of $f$ here? And then, if this is not the case, how do I use the natural injection above to show that $\operatorname{Gal}(L/ \mathbb{Q}) = S_3$?
Also, another question. Can I show from these results that there are an infinite amount of Galois extensions $\mathbb{Q} \subset K$ of degree $3$ which are not isomorphic to each other?