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Let $Q = R[x,y]$ be the ring of polynomials in $x$ and $y$ over the ring $R$. Then isn't the homomorphism defined by $x\mapsto p_1(x,y), \,\,y\mapsto p_2(x,y)$ an automorphism of $Q$, where $p_1,p_2$ are any distinct polynomials in the ring?

Proof: Let $Q' = R[p_1, p_2]$, which can clearly be identified with a subset of $Q$. Define $\phi : Q' \rightarrow Q$ as sending constant polynomials to themselves, and $p_1 \mapsto x, \,\, p_2 \mapsto y$. Then it must be that $\phi (p(p_1, p_2)) = \phi(\sum_{i\in I} a_i p_1^{i_1} p_2^{i_2}) = \sum_{i\in I} a_i x^{i_1} y^{i_2}$. Is this a homomorphism, let's check: let coefficients of $p$ be $a_i$ and coefficients of $q$ be $b_i$. Let $i,j$ be multi-indexes, e.g. $i = (i_1, i_2)$, and let $i,j\in I$ mean they range over all pairs of non-negative integers. $$ \phi(p + q) = \phi(\sum_{i\in I} (a_i + b_i)p_1^{i_1}p_2^{i_2}) = \sum_{i\in I}(a_i+b_i)x^{i_1}y^{i_2} = \phi(p) + \phi(q) \\ \phi(pq) = \phi(\sum_{i,j \in I} (a_ib_j)p_1^{i_1+j_1}p_2^{i_2 + j_2}) = \sum_{i,j\in I}(a_ib_j)x^{i_1+j_1}y^{i_2 + j_2} = \phi(p) \phi(q) $$ So it appears to be a homomorphism. It's clearly surjective. And if $\phi(p) = \phi(q)$ then $p,q$ have the same degrees and coefficients so it's injective. Since this is an isomorphism of $Q'$ onto $Q$ and $Q' \subset Q$ we actually have an automorphism of $Q$, right? Then the inverse of the automorphism $\phi$ is $\psi$ that was first talked about is also an automorphism.

But take this simple example: $Q = \mathbb{Z}[x,y]$, and $p(x,y) = x^2 - y^3$ and let $\psi$ be given by $x\mapsto x^4, \,\, y\mapsto y^5$. Then there isn't a pre-image of $p$ under $\psi$ contradicting its surjectivity.

So I'm wrong somewhere. Please give a hint. Thanks.

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This can't be true for arbitrary $p_1$ and $p_2$. It's false for example if they are elements of $R$, or if $p_1=-p_2$. There is a unique $R$-algebra endomorphism of $R[x,y]$ sending $x$ to $p_1$ and $y$ to $p_2$, and the image of this endomorphism if $R[p_1,p_2]$, the $R$-subalgebra of $R[x,y]$ generated by $p_1$ and $p_2$ (you aren't identifying this with a subring of $R[x,y]$, it is one; the notation $R[p_1,p_2]$ means the image of this homomorphism). Anyway, when you "define" $\phi$, you are assuming that $p_1$ and $p_2$ behave in the same way that $x$ and $y$ do, i.e., that they are – Keenan Kidwell Aug 23 '13 at 23:26
I think that $p_1(x,y)=x^2$, $p_2(x,y)=y^2$ is not surjective. Worse, $p_1(x,y)=x$, $p_2(x,y)=2x$ is an even egregiouser counterexmaple. – Lubin Aug 23 '13 at 23:26
algebraically independent, which means precisely that the map $x\mapsto p_1$, $y\mapsto p_2$, is injective. If they satisfy an algebraic relation over $R$, you can't define $\phi$ this way. This is a continuation of my comment above Lubin's (in case that's not clear). – Keenan Kidwell Aug 23 '13 at 23:27
What about the example given at the bottom. They don't seem to share an algebraic relation. – Enjoys Math Aug 23 '13 at 23:30
@EnjoysMath: Take it down a notch: the map $t \mapsto t^2$ on $\mathbb{C}[t]$ much more clearly has this property. – Pete L. Clark Aug 23 '13 at 23:50
up vote 3 down vote accepted

You have defined an endomorphism of the polynomial ring $R[x,y]$. It certainly need not be an automorphism. All you say about surjectivity is that it is "clear", but it isn't clear and indeed may certainly be false. (Added: Reading your question more closely, you don't actually say this. Rather you define a putative inverse map, which if it were well-defined would indeed clearly be surjective. But defining the inverse map as you have essentially assumes what you want to prove. Because there are counterexamples to the claimed bijectivity, the inverse map is thus not always definable.)

It is not always easy to tell whether you have surjectivity, but sometimes it is obviously false: e.g. if $p_1$ and $p_2$ are both polynomials in $x$ only, then the image of the map clearly lands in $R[x] \subsetneq R[x,y]$. Your argument for the injectivity isn't much of an argument and also can certainly be false, as many of the commenters have pointed out: at an extreme, you haven't even said that $p_1$ and $p_2$ must be nonzero!

Let me take $R = \mathbb{C}$; this is already an interesting case. Then the Inverse Function Theorem shows that a necessary condition for your map to be invertible is for the Jacobian determinant

$J(p_1,p_2) = \left[ \begin{array}{ccc} \frac{\partial p_1}{\partial x} & \frac{\partial p_1}{\partial y} \\ \frac{\partial p_2}{\partial x} & \frac{\partial p_2}{\partial y} \end{array} \right]$

to be nonzero at every $(x,y) \in \mathbb{C}^2$ (when this fails, at some points your function is not even locally invertible). Since $\mathbb{C}^2$ is algebraically closed and $J(p_1,p_2)$ is a polynomial, the only way it can have this property is if it is a nonzero constant. So this gives a necessary condition for invertibility. (I leave it to you as an easy but amusing exercise to show that none of the counterexamples exhibited here or in the comments satisfy this necessary Jacobian condition.) In fact it is a notorious open problem that this necessary condition is also sufficient, the Jacobian Conjecture.

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