$k[{X_1},\cdots,{X_m}]$ is not isomorphic to $k[{X_1},\cdots,{X_n}]$ if $m\neq{n}.$ 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!
 A: 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


*

*compute that the transcendence degree of the field of fractions over $k$ is $n$.

*compute that the Krull dimension is $n$ (somewhat harder).

*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.
A: 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 $n\le m$.
