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Let $R$ be a commutative $k$-algebra, where $k$ is a field of characteristic zero. Let $f$ be a $k$-algebra endomorphism of $R$. ($f$ is not assumed to be either injective nor surjective).

Are there additional conditions (on $R$ or on $f$) that guarantee that $f$ has a fixed point $r \in R-k$? Namely, $r \in R-k$ such that $f(r)=r$.

I do not mind to restrict my question to the special case $R=\mathbb{C}[x,y]$ and $f$ having an invertible Jacobian. (Below I am trying to solve the case $R=\mathbb{C}[x]$ and $f$ having an invertible Jacobian).

Remarks: (1) By definition, $f(\lambda)=\lambda$ for every $\lambda \in k$, hence I required that $r \in R-k$.

(2) 'fixed point theorems' tag shows that there are several nice theorems concerning fixed points of functions, but additional structures on $R$ are needed.

(3) Perhaps this question is relevant. (Notice that here the set of fixed points of $f$ is a $k$-subalgebra of $R$, but it may happen that it equals $k$; the point is to find a fixed point outside $k$).

(4) See also this paper, in which the following is proved: If $f:k[x,y] \to k[x,y]$ has an invertible Jacobian and is not an automorphism, then $\cap f^i(k[x,y])=k$. Then it follows that if $f$ has a fixed point $r \in k[x,y]-k$, then $r \in \cap f^i(k[x,y])$, so $\cap f^i(k[x,y]) \neq k$, hence $f$ is an automorphism of $k[x,y]$.

The special case $R=\mathbb{C}[x]$: Consider $R=k[x]$ and $f: x \mapsto a_mx^m+\cdots+a_1x+a_0$, $a_i \in k$. If $f$ has an invertible Jacobian, namely, $ma_mx^{m-1}+\cdots+a_1= (a_mx^m+\cdots+a_1x+a_0)' \in k^{\times}$, then $a_2=\ldots=a_m=0$, $a_1 \in k^{\times}$, $a_0 \in k$, so $f: x \mapsto a_1x+a_0$; for convenience, write $f: x \mapsto ux+v$.

If $r=\sum_{j=0}^{n} b_jx^j$ is a fixed point of $f$, then $\sum_{j=0}^{n} b_j(u^jx^j+\cdots+v^j)= \sum b_j(ux+v)^j= \sum b_jf(x)^j=f(\sum b_jx^j)=f(r)=r=\sum_{j=0}^{n} b_jx^j$

Therefore, $b_nu^n=b_n$, so $b_n(u^n-1)=0$, hence $u^n=1$. (Also, if $v \neq 0$, then $\sum_{j=0}^{n}b_j=b_0$, so $\sum_{j=1}^{n}b_j=0$).

We obtained that $f$, having an invertible Jacobian, has a fixed point $r \in R-k$ if $f: x \mapsto ux+v$ with $u^n=1$ for some $n$ (and other conditions).

For example, $k=\mathbb{C}$, $f: x \mapsto -x$, here $n=2$, so we take $r$ of degree $2$: $r=x^2$: $f(r)=f(x^2)=f(x)^2=(-x)^2=x^2=r$.

Another example: $k=\mathbb{C}$, $f: x \mapsto ix$, here $n=4$, so we take $r$ of degree $4$: $r=x^4$: $f(r)=f(x^4)=f(x)^4=(ix)^4=(i^4)x^4=x^4=r$.

Thank you very much!

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    $\begingroup$ This question is far too broad. Obviously there are conditions that guarantee that $f$ has a fixed point in $R \setminus k$ (assuming $R \neq k$). E.g., you could require $f$ to be the identity function. What kind of conditions are you looking for? $\endgroup$ – Rob Arthan Sep 16 '18 at 22:15
  • $\begingroup$ ok, I agree with your comment. I think that I am looking for conditions such as: $R$ being a UFD/Noetherian/of finite Krull dimension etc. Actually, even for a specific $R$ like $\mathbb{C}[x,y]$, I am not sure what happens for a general $f$. (It seems that there is an answer to the case $\mathbb{C}[x]$). $\endgroup$ – user237522 Sep 16 '18 at 22:22
  • $\begingroup$ I will immediately elaborate on these two special cases in my question. $\endgroup$ – user237522 Sep 16 '18 at 22:23
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    $\begingroup$ It sounds from your comment that you are interested in conditions on $R$ that guarantee that every endomorphism $f$ has a fixed point in $R \setminus k$. If so, then please reword your question to say that. $\endgroup$ – Rob Arthan Sep 16 '18 at 22:33
  • $\begingroup$ Thanks for your comment. I was hoping to get an answer (though I thought that the chances are quite low), for general $R$ and $f$. However, I will be happy to get an answer for $R=\mathbb{C}[x_1,\ldots,x_N]$ and $f$ having an invertible Jacobian. I am trying to solve the case $N=1$ myself, and it would be nice to know the answer for the case $N=2$ ($N \geq 3$ seems more complicated). Actually, the case $N=2$ should be very very difficult. $\endgroup$ – user237522 Sep 16 '18 at 23:31

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