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Let $\mathbb K$ be a finite field and let $x, y$ be two indeterminates. My intuition tells me that $$\mathbb K[x, y]\cong\mathbb K[x, y+1],$$ but $\mathbb K[x, y]\ncong\mathbb K[x, xy],$ or $\mathbb K[x, y]\ncong\mathbb K[x, x+y]$.

So, my question is:

Let $\mathbb K$ be a finite field and let $x_1, x_2, ..., x_n$ indeterminates. If $f_1, f_2, ..., f_n$ are polynomials in $n$ variables, what conditions must satisfy the polynomials so that we have an isomorphism $$\mathbb K[x_1, ..., x_n]\cong \mathbb K[f_1(x_1, ..., x_n),\dots, f_n(x_1,..., x_n)]?$$

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    $\begingroup$ It's not true that $K[x,y]\not\simeq K[x,x+y]$. $\endgroup$ – Matt Samuel Apr 9 '17 at 13:54
  • $\begingroup$ I wonder if you assume that the isomorphism is identity on $K$. (Btw, I don't think the finiteness of $K$ plays a role here.) $\endgroup$ – user26857 Apr 9 '17 at 14:38
  • $\begingroup$ Dear user26857, yes I assume that the isomorphism is identity on $\mathbb K$ $\endgroup$ – carreira Apr 9 '17 at 15:24
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Your intuition is in error. For any ring $R$, the homomorphism $$ R[x,y] \to R[x,xy] : f \mapsto f(x, xy) $$ is an isomorphism. It's clearly a homomorphism since the domain is a polynomial ring and the map is given by evaluation. One way to see that it's an isomorphism is to observe

  • $R[x,y]$ is a free $R$-module with basis $\{ x^i y^j \mid i,j \in \mathbb{N} \}$
  • $R[x,xy]$ is a free $R$-module with basis $\{ x^i y^{i+j} \mid i,j \in \mathbb{N} \} = $

and the map $f(x,y) \mapsto f(x, xy)$ is a bijection between the two bases.

For the other example you can even say something stronger: you actually have equality $R[x,y] = R[x,x+y]$, because $y = (x+y) - x$, and so $y \in R[x,x+y]$.


Regarding your actual question, $\mathbb{K}$ being finite is irrelevant. Allowing $\mathbb{K}$ now to be any field, the condition you are looking for is

There is an isomorphism fixing $\mathbb{K}$ if and only if the $f_i$ are algebraically independent — that is, there does not exist a nonzero polynomial $g$ of $n$-variables over $\mathbb{K}$ with the property that $$ g(f_1, \ldots, f_n) = 0 $$

You always have a surjective homomorphism given by

$$\mathbb{K}[x_1, \ldots, x_n] \to \mathbb{K}[f_1, \ldots, f_n] : h \mapsto h(f_1, \ldots, f_n)$$

So the above condition is clearly sufficient, because it is equivalent to saying that the kernel of this homomorphism is the zero ideal.

As for being necessary, if such a $g$ existed, then the field extension $\mathbb{K}(f_1, \ldots, f_n) / \mathbb{K}$ would have transcendence degree strictly less than $n$. However, $\mathbb{K}(x_1, \ldots, x_n) / \mathbb{K}$ has transcendence degree exactly $n$.

Any isomorphism $\mathbb{K}[x_1, \ldots, x_n] \cong \mathbb{K}[f_1, \ldots, f_n]$ that fixes $\mathbb{K}$ would extend to an isomorphism of the above field extensions, which would contradict the fact they have different transcendence degree.

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