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I am trying to prove this. Assume that $k$ is a field and that $k[{X_1},\cdots,{X_r}]$ denotes the ring of polynomials in $r$ variables with coefficients in $k$. If ${\phi}:k[{X_1},\cdots,{X_m}]\rightarrow{k[{X_1},\cdots,{X_n}]}$ is a ring isomorphism such that ${{\phi}|_k}={({\rm{}id})_k}$, (the restriction of $\phi$ to $k$ is the identity mapping), then it must be the case that $m=n$. Not sure how to approach this because I don't know if anything can be said about the set $\{{\phi{(X_1)}},\cdots,{\phi{(X_m)}}\}.$ Does the image of this set be contained in $\{{X_1},\cdots,{X_n}\}$? If so, then $\phi$ being a bijection will force $m=n$. Thanks!

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    $\begingroup$ Do you know about dimension of commutative Noetherian rings? $\endgroup$ Commented Sep 18, 2019 at 18:13
  • $\begingroup$ $\phi(X_1), \dots \phi(X_m)$ can be any polynomials whatsoever, so you'll have to work harder than that. $\endgroup$ Commented Sep 18, 2019 at 18:20
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    $\begingroup$ I don’t know the dimension of a commutative ring! $\endgroup$
    – student
    Commented Sep 18, 2019 at 18:31

2 Answers 2

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There are lots of ways to do this by formalizing the intuitive idea that the spectrum $\text{Spec } k[x_1, \dots x_n]$, also known as affine $n$-space $\mathbb{A}^n$, has dimension $n$. For example, you can

  1. compute that the transcendence degree of the field of fractions over $k$ is $n$.
  2. compute that the Krull dimension is $n$ (somewhat harder).
  3. compute that the dimension of the Zariski tangent space at any $k$-point is $n$ (easier).

These are all isomorphism invariants.

You can also work very directly on the level of polynomials by assuming that $m > n$ and showing that $\phi : k[x_1 \dots x_m] \to k[x_1, \dots x_n]$ cannot be injective by finding a nonzero polynomial in the kernel. The argument is a generalization of this math.SE answer.

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    $\begingroup$ Similarly to 3, the global $\Omega^1_{k[x_1,\ldots,x_n] / k}$ is a free $k[x_1,\ldots,x_n]$-module of rank $n$. $\endgroup$ Commented Sep 18, 2019 at 18:49
  • $\begingroup$ @Daniel: yes, that's also very nice! And it implies 3). $\endgroup$ Commented Sep 18, 2019 at 18:51
  • $\begingroup$ Awed by your insights! I am struggling to learn the basics of algebraic curves, so I will revisit these useful comments when I have learned more. The last (direct) approach is very elegant. Trying to generalize your argument "this math.SE answer" above. in the case of two variables, am wondering whether the "volume" of the $m$-dimensional simplex formed can be evaluated by a multiple integral. (The calculation would have to give your answer for the two variable case obviously!) $\endgroup$
    – student
    Commented Sep 19, 2019 at 5:24
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    $\begingroup$ @student: yes. The volume for $n$ variables is $\frac{D^n}{n! d_1 d_ 2 \dots d_n}$. $\endgroup$ Commented Sep 19, 2019 at 5:27
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HINT:

Consider a $k$-morphism from $k[x_1, \ldots,x_m]$ to $k[x_1, \ldots, x_n]$, $x_i\mapsto a_i(x_1, \ldots, x_n)$, $i=1,\ldots, m$.

Assume that the map is surjective. Then there exist $P_1$, $\ldots$, $P_n$ polynomials in $m$ variables $a_1$, $\ldots$, $a_m$ so that $$P_j(a_1(x_1,\ldots, x_n), \ldots, a_m(x_1, \ldots, x_n)) = x_j$$ Using the chain rule we obtain for the jacobians $$\frac{\partial P}{\partial a}(a(x)) \cdot \frac{ \partial a}{\partial x}(x)=(\delta_{ij})\ (=I_n) $$

and so the rank of both matrix factors must be $n\le m$.

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  • $\begingroup$ One can regard the Jacobian matrices as linear transformations ${K^{m}}\rightarrow{K^{n}}$ where $K=k({X_1},\cdots,{X_n})$ the field of rational functions in $m$ indeterminates. The rank then has the usual definition. I suppose we work in the ring directly, and think in terms of modules over the ring $k[{X_1},\cdots,{X_n}]$, but in this case I am not sure about the notion of dimension. $\endgroup$
    – student
    Commented Sep 19, 2019 at 5:13
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    $\begingroup$ @student: if you specialize the Jacobians to a point they become linear maps on Zariski tangent spaces, and this becomes an elaboration of the third proof I suggest in my answer. $\endgroup$ Commented Sep 19, 2019 at 5:26
  • $\begingroup$ I will struggle to understand your three proofs. I have a feeling I will learn a lot!! $\endgroup$
    – student
    Commented Sep 19, 2019 at 5:28
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    $\begingroup$ @student: Assume that $m<n$. On the left hand side the determinant is $0$ (something that is not hard to check), on the right hand side is $1$. No need to look at ranks. $\endgroup$
    – orangeskid
    Commented Sep 19, 2019 at 5:30
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    $\begingroup$ @student: Check out the second answer here math.stackexchange.com/questions/2787362/… $\endgroup$
    – orangeskid
    Commented Sep 19, 2019 at 5:53

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