Q:

Isomorphism between polynomial with indeterminate and integral element

If $x$ is an indeterminate, $A$ is a ring and $\alpha$ is in some ring containing $A$, is $A[x]$ is isomorphic to $A[\alpha]$ if $\alpha$ is integral over $A$? What about if $\alpha$ is transcendental over $A$? Explain.
My interpretation of this is that it is asking if there is an isomorphism $I:A[x] \rightarrow A[\alpha]$ in the two cases for $\alpha$. However I am not quite sure how to tackle it. 
Usually finding a map between the domains and proving it is bijective is enough. Since $x$ is an indeterminate, I assume I can simply replace it, so would the evaluation map be appropriate here? And if so, I'm not sure how I would go about proving its bijectivity. It also would not use whether $\alpha$ is integral over $A$ or not.
So far, I know that $\alpha$ integral over $A \iff A[\alpha]$  is a finitely generated A-module. Would finding its finite generating set be useful?
I also know
$\alpha$ integral $\implies$ satisfies a monic polynomial in $\mathbb Z[x]$, and 
$\alpha$ transcendental $\implies$ not algebraic over A, ie. can only express it as the null polynomial.
But I am not sure how they would come into play here since we know nothing in terms of a relation of $A$ to $\mathbb Z$.

A:

$A[x]\cong A[\alpha]$ if $\alpha$ is transendental.  If $\alpha$ is integral over $A$,  then
$A[\alpha]$ will be isomorphic to some quotient ring of $A[x]$.
Suppose $\alpha$ is transcendental, and $\phi:A[x]\to A[\alpha]$ be the unique homomorphism $x\mapsto\alpha$. It's not hard to see that this will be bijective.
If $\alpha$ is integral, with minimal polynomial say $p(x)$, then $\phi$ will have a non-trivial kernel since $\phi(p(x))=p(\alpha)=0$.  In fact, $\ker\phi=(p)$, the ideal generated by $p$.  So, $A[\alpha]\cong A[x]/(p)$.

