An extension with the induced map on Spec being bijective. Let $A$ be a commutative ring with unit. Let $A\subset A[b]$ be an extension of rings such that $b^n, b^m\in A$, where $m,n$ are positive integers that are coprime with each other. Show that $SpecA[b]\to SpecA$ is bijective.
 A: Preliminaries:setting-up
Let $b^n=c , b^m=d\in A$ . Since $b$ is integral over $A$ (it is a zero of $T^n=c$), so is the extension $A\subset A[b]$ and $SpecA[b]\to Spec A$ is thus surjective.There remains to see that this map is injective. 
First reduction: to the general case
It is sufficient to consider the general case $A[x]=A[X]/(X^n-c,X^m-d)$ because the quotient morphism $q:A[x]\to A[b]:x\mapsto b$ gives rise to the maps $Spec(A[b])\to Spec(A[x])\to Spec(A)$ of which the first is injective since $q$ is is surjective. Thus the result for $A[x]$ (that the second map is bijective) will imply the result  for $A[b]$.  
Second reduction: to a field
We have reduced to proving that for every $p\in Spec(A)$ the fiber over $\kappa(p)$ of 
$Spec(A[x])\to Spec(A)$ is a singleton set.
That fiber is the spectrum of $A[x]\otimes_A{\kappa(p)}$, and the algebra is isomorphic to the $\kappa(p)$-algebra $\kappa(p)[x]=\kappa(p)[X]/(X^n-c,X^m-d)$.
So we may suppose $A=k=$ a field and study the $k$-algebra $k[x]=k[X]/(X^n-c,X^m-d)$.
Resolution of the problem
If $k[x]=k[X]/(X^n-c,X^m-d)$, the spectrum of $k[x]$ consists of the (automatically maximal) kernels of the $k$-morphisms $k[x]\to \bar k$, where $\bar k$ is an algebraic closure of $k$.
It is thus sufficient to show that there exists at most  one such morphism $f:k[x]\to\bar k$.
But these morphisms correspond to elements $\xi\in \bar k$ satisfying $\xi^n=c$ and $\xi^m=d$.
So finally we have reduced to showing unicity of $\xi$.
$\bullet$ Unicity of $\xi$ is obvious if $c=0$ or $d=0$.
$\bullet \bullet$ Else $\xi\neq 0$ so $\xi$ is invertible in $\bar k$.
Write  then $\nu n+\mu m=1$ (here is where we  use the hypothesis of coprimeness !) to conclude that necessarily $\xi=\xi^{\nu n+\mu m}=c^\nu d^\mu$: only one $\xi$ is possible, hence we have our unicity .
A: The following proof is quite similar to the one by Georges.
A morphism of schemes $X \to Y$ is called radicial if for every field $K$ the induced map $X(K) \to Y(K)$ is injective (EGA I, 3.5.4). I claim that $\mathrm{Spec}(A[b]) \to \mathrm{Spec}(A)$ is radicial, i.e. that every homomorphism $\phi : A \to K$ has at most one extension to $\psi : A[b] \to K$, i.e. that  $x:=\psi(b)$ is unique. Choose $p,q \in \mathbb{Z}$ such that $pn+qm=1$. Then $x=(x^n)^p (x^m)^q = \phi(b^n)^p \phi(b^m)^q$ is unique. This works when $x \neq 0$, equivalently, $\phi(b^n) \neq 0$, but otherwise $x=0$ is also unique.
Now, we just have to cite EGA IV, 2.4.5, which says that an integral surjective radicial morphism is a universal homeomorphism (actually the converse is also true: EGA IV, 18.12.11).
