Let $\phi:k[x_1,\dots,x_n]\mapsto k[x_1,\dots,x_n]$ be a $k$-algebra homomorphism with $\phi(x_i)=f_i$, where $k$ is algebraically closed and has characteristic zero. I have the following questions:

(1) If $\phi$ is surjective, i.e. $k[f_1,\dots,f_n]=k[x_1,\dots,x_n]$, is $(f_1,\dots,f_n)$ a maximal ideal in $k[x_1,\dots,x_n]$? Conversely, if $(f_1,\dots,f_n)$ is a maximal ideal in $k[x_1,\dots,x_n]$, is $\phi$ surjective?

(2) If $\phi$ is surjective, is $\phi$ injective? If this is not true, is there any counterexample?

(3) Now, let $X,Y$ be two affine variety, $\mu :X\mapsto Y$ is a morphism (polynomial map), $\mu$ induce a $k$-algebra homomorphism between the coordinate ring of $X,Y$ $$\mu^*:A(Y)\mapsto A(X):f\mapsto f\circ \mu.$$ If $\mu$ is injective, is $\mu^*$ always surjective? at least in the case where $X=Y=\mathbb{A}^n$?

  • $\begingroup$ Nice questions:+1 $\endgroup$ – Georges Elencwajg Aug 15 '14 at 22:54

(2) Yes, if $\phi$ is surjective, it is injective.
Indeed, the dual morphism $f=\phi^*:\mathbb{A}^n_k\to \mathbb{A}^n_k$ is a closed embedding and since $f(\mathbb{A}^n_k)\subset \mathbb{A}^n_k$ are irreducible of dimension $n$ they are equal and $f$ is an isomorphism (since $\mathbb{A}^n_k$ is reduced ).
Thus $\phi$ is an isomorphism too and, in particular, $\phi$ is injective.

(1) If $\phi$ is surjective, it is an isomorphism by answer (2) so that, yes, $(f_1,\dots,f_n)$ is a maximal ideal in $k[x_1,\dots,x_n]$ .
The converse is false: the ideal $(x_1,(x_1+1)x_2)\subset k[x_1,x_2]$ is maximal but the corresponding morphism $\phi: k[x_1,x_2]\to k[x_1,x_2]$ determined by $\phi(x_1)=f_1=x_1,\phi(x_2)=f_2=(x_1+1)x_2$ is not surjective, since $x_2$ is not in the image of $\phi$ .

(3) is false. A counterexample is obtained by considering the cusp $C\subset \mathbb A^2_k$ of equation $y^2=x^3$ and the bijective morphism given by $$\mu:\mathbb A^1\to C:t\mapsto (t^2,t^3)$$ This bijective morphism is not an isomorphism and the dual algebra morphism $\mu^*:k[x, y]/(y^2-x^3)\to k[t]$ is injective but not surjective: its image is $k[t^2,t^3]$.

Notice that all of the above is valid for an arbitary field $k$ of any characteristic, algebraically closed or not.

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  • $\begingroup$ @user26857: I believe Georges' argument is that an isomorphism preserves all ideal-theoretic properties, such as being maximal. Georges: do you have a counterexample for (3) when $X = Y = \mathbb{A}^n$? $\endgroup$ – zcn Aug 16 '14 at 18:19
  • $\begingroup$ Dear @zcn: first, many thanks for answering user26857 exactly the way I intended to! Second, I think that if $k$ is algebraically closed of characteristic zero then any injective $\mu:\mathbb{A}^n _k\to \mathbb{A}^n_k$ will be an isomorphism (so that $\mu^*$ will certainly be surjective) . This requires difficult results, like Ax-Grothendieck's theorem. The result is however false in char. $p$, as demonstrated by the Frobenius morphism. $\endgroup$ – Georges Elencwajg Aug 16 '14 at 22:49
  • $\begingroup$ Yes, Ax-Grothendieck seemed to be the most immediately relevant fact, although I must say I couldn't quite see the full line of reasoning (specifically, why bijectivity would be sufficient). I believe it suffices to show: must a universally injective map $\mu : \mathbb{A}^n_k \to \mathbb{A}^n_k$ be a monomorphism? $\endgroup$ – zcn Aug 16 '14 at 23:59
  • $\begingroup$ @zcn: I would use the fact that the degree of $\mu$ is the generic number of points in the fiber (in char. zero!) to get birationality and then Zariski's main theorem. $\endgroup$ – Georges Elencwajg Aug 17 '14 at 9:45

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