Let $\mathbb{K}$ be a field. Show that the $\mathbb{K}-$isomorphism $\mathbb{K}[t]\to \mathbb{K}[t]$ is given by $t\mapsto t-a.$

I have shown that it is an isomorphism as, Let $\psi : \mathbb{K}[t]\to \mathbb{K}[t] $ such that $\psi(t)=t-a. $

$\mathbf{\psi}$ is an isomorphism.

  1. $\psi(f(t)+g(t))=\psi((f+g)(t))=(f+g)(t-a)=f(t-a)+g(t-a)=\psi(f(t))+\psi(g(t))$

  2. $\psi(f(t)\cdot g(t))=\psi((f\cdot g)(t))=(f\cdot g)(t-a)=f(t-a)\cdot g(t-a)=\psi(f(t))\cdot \psi(g(t))$

  3. $\ker(\psi):=\{f(t)\in \mathbb{K}[t] : \psi(f(t))=0\}=\{f(t)\in \mathbb{K}[t] : f(t-a)=0\}=\{f(t)\in \mathbb{K}[t] : f\equiv 0\} $.

So,$\psi$ is $1-1$.

  1. For surjectivity, if $f(t)\in \mathbb{K}[t]$ then $\psi(f(t+a))=f(t)$.

Hence, this is an isomorphism.

Now for showing that any other isomorphism sends $t$ to a linear polynomial, assume $\phi$ be any isomorphism and $\phi(t)=f(t)\in \mathbb{K}[t].$ Let $f(t)=\sum_{i=0}^na_it^i.$ Now the claim is that $a_i=0, \forall i\ge 2.$

After that I stuck what to do. So please help.


If you have a polynomial

$$ f(t) = \sum_{n} a_n t^n $$


$$ \psi(f) = \sum_n a_n \psi(t)^n. $$

This implies the formula

$$ \deg \psi(f) = \deg \psi(t) \deg f. $$

If $\deg \psi(t) \ne 1$ then this causes issues.


The usual tool for this kind of problem is an euclidean function (in french, a "stathme"), for this particular euclidian domain, this is the degree $d$.

Here, if $\phi(t)=f(t)$, then for all $P\in K(t)$, $\phi(P)=P\circ f(t)$, so $d(\phi(P))=d(P)\times d(f)$.

If $d(f)>1$, then $d(\phi(P))>1$ for all non-constant polynomial $P$, so $t$ is not in the image of $\phi$. And you have your reciprocal.


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