Context is a course covering field theory (ch. 13) and Galois theory (ch. 14) of Abstract Algebra by Dummit & Foote.
Let $f, g ∈ \mathbb{Q} [x]$ be non-constant, irreducible polynomials. Let $α$ be a root of $f$, and put $K := \mathbb{Q} (α)$. Let $β$ be a root of $g$, and put $L := \mathbb{Q} (β)$. Show that $f$ is irreducible in $L[x]$ if and only if $g$ is irreducible in $K[x]$.
I could use a hint to get started with this. I suppose I only have to show one way due to symmetry.
Let $\deg(f)=m, \deg(g)=n$. Then $[\mathbb{Q}(\alpha):\mathbb{Q}]=m$ and $[\mathbb{Q}(\beta):\mathbb{Q}]=n$. Now suppose $f$ is irreducible over $\mathbb{Q}(\beta)$. Then the minimal polynomial of $\alpha$ over $\mathbb{Q}(\beta)$ has degree $m$, so $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\beta)]=m$. By the tower rule we conclude that $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}]=mn$, and hence $[\mathbb{Q}(\alpha,\beta):\mathbb{Q}(\alpha)]=n$. This means that the minimal polynomial of $\beta$ over $\mathbb{Q}(\alpha)$ has degree $n$, just like the polynomial $g$. It follows that $g$ must be irreducible over $\mathbb{Q}(\alpha)$, because otherwise the minimal polynomial would have smaller degree.