Since $A$ is finitely generated, we prove this by induction on the number $m$ of generators as a $k$-algebra. === Basis for the induction === If $m = 0$, then $A = k$ and thus there is nothing to prove. === Induction step === Let $A$ be a $k$-algebra generated by the elements $y_1, \dotsc, y_m, y_{m + 1}$.  If $y_1, \dotsc, y_m, y_{m + 1}$ are algebraically independent, then $A$ is isomorphic to $k \sqbrk {x_1, \dotsc, x_m, x_{m + 1} }$. In this case the theorem is obvious, so assume that $y_{m + 1}$ depends algebraically on $y_1, \ldots, y_m$. So there exists a polynomial $P \in k \sqbrk {x_1, \dotsc, x_m, X}$ such that $\map P {y_1, \dotsc, y_n, y_{n + 1} } = 0$ in $A$.  Let $\mu \in \N^n$ and define: :$f_\mu: k \sqbrk {x_1, \dotsc, x_n, X} \to k \sqbrk {x_1, \dotsc, x_n, X}: x_i \mapsto x_i + X^{\mu_i}, X \mapsto X$ This is easily seen to be an isomorphism. {{handwaving|easily seen}} Next, we want to show that there exists $\mu \in \N^n$ such that $\map {f_\mu} P$ is a polynomial in $X$ with leading coefficient in $k$. Let $e$ be the biggest natural number such that  $P$ involves $x_e$ non-trivially.  If there is no such number, define $e$ to be $0$.  We proceed by induction on $e$. The case $e = 0$ is easy since we then have a polynomial $f \in k \sqbrk X$. Thus $\mu = \tuple {0, \dotsc, 0}$ will suffice. Suppose it holds for $e$ and suppose $P$ depends only on $x_1, \dotsc, x_e, x_{e + 1}$. Then we can write $P$ as: :$\displaystyle P = \sum_{i \mathop = 0}^l a_i x_{e + 1}^i$ where $a_i \in k \sqbrk {x_1, \dotsc, x_e, X}$ and $a_l \ne 0$. Hence by the induction hypothesis, there exists a $\mu' \in \N^n $ such that $\map {f_{\mu'} } {a_l}$ has a leading coefficient in $k$. Note that we can choose that $\mu'_l$ for $l > e$ since $\map {f_\mu} {a_i}$ is independent of these components.  Hence we define $\mu$ as $\mu_i = \mu'_i$ for every $i \ne e + 1$. Now observe that: :$\displaystyle \map {f_\mu} P = \sum_{i \mathop = 0}^l \map {f_\mu} {a_i} \, \map {f_\mu} {x_{e + 1} }^i = \sum_{i \mathop = 0}^l \map {f_{\mu'} } {a_i} \paren {x_{e + 1} + X^{\mu_{e + 1} } }^i$ We have that $\map {f_{\mu'} } {a_l}$ has a leading coefficient in $k$. Thus by choosing $\mu_{e + 1}$ big enough, $map {f_\mu} P$ has a leading coefficient in $k$. Consider now thus the elements $z_i = \map {f_\mu^{-1} } {y_i} \in A$.  These elements still generate $A$ since $f_\mu$ is an isomorphism.  On the other hand, we find that: {{begin-eqn}} {{eqn | l = \map {f_\mu} P \tuple {z_1, \dotsc, z_m,  z_{m + 1} }       | r = \map {f_\mu} P \tuple {\map {f_\mu^{-1} } {y_1}, \dotsc, \map {f_\mu^{-1} } {y_m}, \map {f_\mu^{-1} } {y_{m + 1} } } }} {{eqn | r = \map P {y_1, \dotsc, y_m, y_{m + 1} } }} {{eqn | r = 0 }} {{end-eqn}} By the induction hypothesis, we have the existence of: :$\alpha: k \sqbrk {x_1, \dotsc, x_n} \to A'$ where $A'$ is generated by $z_1, \dotsc, z_m$. By extending this $\alpha$ to $A$ by the natural inclusion, we find that $\alpha$ is finite, since $z_i$ is integral by construction and $z_{m + 1}$ is integral over the others since it satisfies $\map {f_\mu} P$.  {{explain|"integral over the others" -- what specifically are the "others"? The above sentence needs to be rewritten as a series of simple statements in standard {{ProofWiki}} style.}} Hence it is the wanted finite injective morphism. {{qed}} {{Namedfor|Emmy Noether|cat = Noether E}}