If \(F\) is a field and \(h\) is a polynomial over \(F\) of degree \(\geq 1\), show that the mapping \(f \to f(h)\) is a one-one linear transformation of \(F[x]\) into \(F[x]\). Show that this transformation is an isomorphism of \(F[x]\) onto \(F[x]\) if and only if \(\deg h = 1\).
Solution
1. The map \(\phi: F[x] \to F[x]\), \(\phi(f) = f(h)\), is linear.
\((f + g)(h) = f(h) + g(h)\) and \((cf)(h) = c f(h)\) for \(c \in F\).
2. \(\phi\) is injective.
2.1. If \(f \neq 0\) has degree \(d\), then \(f(h)\) has degree \(d \cdot \deg h \ge d \ge 0\), so \(f(h) \neq 0\).
the leading term of \(f(h)\) is \(a_d h^d\), whose degree is \(d \cdot \deg h\) (since \(\deg h \ge 1\)), so \(f(h)\) is nonzero.
2.2. Hence \(\ker \phi = 0\), so \(\phi\) is injective.
1.
3. \(\phi\) is surjective iff \(\deg h = 1\).
3.1. If \(\deg h = 1\), then \(h = ax + b\) with \(a \neq 0\), and \(\phi\) is surjective.
the inverse map is \(g \mapsto g((x - b)/a)\), i.e. \(\phi\) has inverse \(\psi(g) = g(h^{-1})\) where \(h^{-1} = (x-b)/a\); indeed \(\phi(\psi(g)) = g(h(h^{-1})) = g(x) = g\).
3.2. If \(\deg h > 1\), then \(\phi\) is not surjective.
the image of \(\phi\) consists of polynomials \(f(h)\) whose degree is a multiple of \(\deg h\) (or \(0\)); in particular, the polynomial \(x\) is not in the image (since \(x = f(h)\) would require \(\deg f \cdot \deg h = 1\), impossible when \(\deg h > 1\)).
4. Q.E.D.
1, 2, and 3.