$Im(\phi)$ is closed subset of $\mathbb{A}^2$ let $\alpha(t)$ and $\beta(t)$  $\in$ $K[t]$ ,  $\phi(t)=(\alpha(t),\beta(t))$ is a morphism from $\mathbb{A}^1$ to $\mathbb{A}^2$ show that $Im(\phi)$ is closed subset of $\mathbb{A}^2$.
it seems easy question but i can't see it
İn this question should i find polynom or i need to show it from topology i just want clue .
 A: This is obvious if both $\alpha(t)$ and $\beta(t)$ are constants, since the image is then a point.     
If at least one of those polynomials, say $\alpha(t)$, is not a constant the dual morphism of rings  $f:K[x,y]\to K[t]$ sending $x$ to $\alpha(t)$ and $y$ to $\beta(t)$ makes of $K[t]$ a finite algebra over $K[x,y]$ (since $K[t]$ is already a finite $K[x]$-algebra) and thus a fortiori an integral algebra.   
The conclusion follows by remembering that a morphism of varieties is closed as soon as its dual morphism is  integral  (Atiyah-Macdonald, Chapter 5, Exercise 1).
A: If you know some basics about projective varieties, you may also argue as follows.
Embed $\Bbb A^1$ into $\Bbb P^1$ and $\Bbb A^2$ into $\Bbb P^2$ (for example by $t \mapsto [t:1]$ and $(u,v) \mapsto [u:v:1]$).
Your morphism
\begin{split}
\phi: \Bbb A^1 &\rightarrow \Bbb A^2\\
t &\mapsto (\alpha(t),\beta(t))
\end{split}
can then be "compactified" to a morphism $\bar{\phi}: \Bbb P^1 \rightarrow \Bbb P^2$ (cf. Hartshorne Prop.I.6.8.).
Now every morphism from a projective variety is closed, and thus $Im(\bar{\phi}) \subset \Bbb P^2$ is closed. But $Im(\phi) \cong Im(\bar{\phi}) \cap U$, where $U=\{[u:v:1]\} \subset \Bbb P^2$ is the image of the embedding $\Bbb A^2 \subset \Bbb P^2$ chosen above. Hence $Im(\phi)$ is closed in $\Bbb A^2$ as well.
