einsetzungshomomorphismus zwischen Polynomringen über Integritätsbereichen als Isomorphismus

Since \(\varphi\) is \(R\)-linear, we get \(\varphi(r) = r\) for \(r \in R\).
Thus, since \(R\) is an Integral domain, it follows, that

\begin{align*} \mathrm{deg}(\alpha g(x) \cdot h(x)) = \mathrm{deg}(g(x)) + \mathrm{deg}(h(x)) \end{align*}

(cf. Grad von Polynomen unter Multiplikation in einem Integritätsbereich)
By injectivity, we may assume, that \(\mathrm{deg}(\varphi_g(x)) > 0\), because otherwise \(\varphi_g(X + \varphi_g(X)) = 0\)

Since \(\varphi_x\) is surjective, there exists a polynomial \(f = \sum_{i=0}^{n} \beta_i X^i\) with

\begin{align*} \varphi_g(f) =& X \\ \sum_{i=0}^{n} \beta_i \varphi_g(X)^i =& X \end{align*}

here, since \(R\) is an integral domain, we get homogenous parts \(\beta_i \varphi(X)^i\) with different degre.
Hence by degree of homogeneous polynomials under addition (and again integrity of \(R\)) we get

\begin{align*} \mathrm{deg}(X) =& \mathrm{deg}(\sum_{i=0}^{n} \beta_i \varphi_g(X)^i) \\ =& \mathrm{max}(\mathrm{deg}(\beta_i \varphi_g(X)^i)) \\ =& \mathrm{deg}(\varphi_g(X)^i) \end{align*}

Here, \(X^1\) with \(\mathrm{deg}(\varphi_g(X)) = 1\) is the only possible choice, hence

\begin{align*} \varphi_g(X) = \alpha X + b \end{align*}

By assumption, there exists a polynomial \(f = \beta_1 X + \beta_0\) (w.l.o.g. \(\mathrm{deg}(f) = 1\), see above) with

\begin{align*} \varphi_g(f) = X \end{align*}

evulation results in

\begin{align*} \beta_1 (\alpha X + \beta ) + \beta_0 =& \beta_1 \cdot \alpha X + \beta_1\beta + \beta_0 \end{align*}

where \(\beta_1 \cdot \alpha = 1\)
hence \(\beta_1 = \alpha^{-1}\)

Let \(\beta X^n \in R[X]\) be arbitrary.
Then we get

\begin{align*} \varphi_g(\beta \cdot \alpha^{-n} X^n) =& \beta \cdot \alpha^{-n} \cdot \alpha^n X^n + r \\ =& \beta \cdot X^n + r \end{align*}

with \(\mathrm{deg}(r) < n\), hence we can inductively construct for any polynomial \(g = \sum_{i=0}^{n} \beta_i X^i\) a polynomial \(h \in R[X]\) with \(\varphi_g(h) = g\)

Let \(f = \sum_{i=0}^{n} \beta_i X^i \in R[X]\) with \(\varphi_g(f) = 0\).
Suppose \(f \neq 0\), then since a unit is never a zero divisor (cf. unit is not a zero divisor), we get for

\begin{align*} \varphi_g(f) = \beta_i \alpha^n X^n + ... \mathrm{deg}(\varphi_g(f)) > 0 \end{align*}

contradiction