= Solution
If $f$ is an <irreducible polynomial>, any two roots $\alpha,\beta$ have the same <minimal polynomial>. Thus the map $\mathbb Q(\alpha)\to\mathbb Q(\beta)$ carrying $\alpha$ to $\beta$ is a field embedding. The <extension theorem for field embeddings> extends it to an embedding of the <splitting field> $L$ into an algebraic closure. Since $L/\mathbb Q$ is normal, its image is $L$, so it is an element of the <Galois group>.
Conversely, factor $f$ into irreducible factors over $\mathbb Q$. An automorphism preserves the vanishing of each factor, so it cannot send a root of one factor to a root of another. Since there are no repeated roots, different factors have disjoint root sets. Transitivity on all roots therefore forces a single irreducible factor. This proves \b[the equivalence].
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