Question: 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 $1-1$. 

4. 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.

Answer: 
If you have a polynomial

$$ f(t) = \sum\_{n} a\_n t^n $$

then

$$ \psi(f) = \sum\_n a\_n \psi(t)^n. $$

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.