# 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.

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Since $b^n\in A$, $b$ is integral over $A$. Thus we have an integral extension above. SO the induced map is surjective. How to prove it is injective. I assume the coprime condition should be used to prove that, but i dont have it yet. –  messi Jun 6 '12 at 16:25
Maybe use localisation... Since $n,m$ are coprime, you have that $b=b^{xn+ym}=(b^n)^x(b^m)^y$, for some integers $x,y$, but they will most likely be negative, and so you have to invert. –  M Turgeon Jun 6 '12 at 16:34
@MTurgeon Yes, i did note that but still not sure how to solve the above. –  messi Jun 6 '12 at 16:42
Now i am confused by we need coprime assumption. Suppose we have $p$ in the kernel, then $p\cap A=0$. This would imply that $b\notin p$, so $p=0$. So there is something wrong somewhere. –  messi Jun 6 '12 at 18:12
There is no algebraic structure on the spectra; hence, if you want to check injectivity, you have to assume that $p_1$ and $p_2$ yield the same prime ideal when intersected with $A$. –  M Turgeon Jun 6 '12 at 18:22

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 .

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thanks, that was great. I will read it a few more times carefully and get back if i have any questions regarding the solution. –  messi Jun 7 '12 at 12:19

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).

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thanks Martin, this is nice. Sorry, i cant accept your answer as I already accepted Georges answer. –  messi Jul 1 '12 at 17:05