About ring $( \frac{F[t]}{t^2})$ I was trying to study some properties of the group $GL_2 ( \frac{F[t]}{t^2})$ where $F$ is a field with $p$ elements. $p$ is a prime. 
I found that $ \frac{F[t]}{t^2}$  is a discrete Valuation ring. Is it correct ? 
How can I find the number of elements in $( \frac{F[t]}{t^2})$ ? 
This ring $ \frac{F[t]}{t^2}$  will have the elements of the form $ \sum a_it_i^{2k+1}$ Is this true ? 
 A: Well in $\frac{F[t]}{t^2}$ you have that $t^2=0$ so $t^n$ with $n\geq 2$ will be null. Hence, the most general element of this ring is :
$$a_0+a_1t\text{ where } a_0,a_1\in F $$
Hence there are $p^2$ elements in it.
If I want to be complete, you can show (I think) that :
$$GL_2(\frac{F[t]}{t^2}) \text{ is isomorphic to } M_2(F)\rtimes_{Ad}GL_2(F) $$
Essentially to $A+Bt\in GL_2(\frac{F[t]}{t^2})$ you associate $(B,A)$.
Suppose $C\in GL_2(\frac{F[t]}{t^2})$ because it is included in $M_2(\frac{F[t]}{t^2})$, I can write each coefficient :
$$C_{i,j}=A_{i,j}+B_{i,j}t\text{ where } A_{i,j},B_{i,j}\in F$$
And I can set $A:=(A_{i,j})\in M_2(F)$ and $B:=(B_{i,j})\in M_2(F)$ so that :
$$C=A+tB$$
Now $det(C)$ is inversible in $\frac{F[t]}{t^2}$, i.e. $det(C) $ mod $t$ is non-zero (you can compute explicitely the condition for an element of this ring to be inversible). And then it suffices to see that :
$$det(C)\text{ mod } t=det(A) $$
So we have for all $C$ :
$$C=A+Bt\text{ with } A\in GL_2(F)\text{ and } B\in M_2(F) $$
And clearly, from what we have done all such elements are non-singular matrices.
Now to get the semi direct structure it is easy, it suffices to consider the following group morphism :
$$GL_2(\frac{F[t]}{t^2})\rightarrow  GL_2(F)$$
$$C\mapsto C \text{ mod }t$$
Its kernel is a normal subgroup which is explicitely :
$$N:=\{I_2+Bt|B\in M_2(F)\} $$
One can show that such a group is isomorphic to $(M_2(F),+)$ (in particular it is abelian). And finally if you set :
$$H:=\{A+0t|A\in GL_2(F)\} $$
Then you easily see that $N\cap H=\{I_2\}$, but because $N\triangleleft GL_2(\frac{F[t]}{t^2})$ and $N,H$ generates $GL_2(\frac{F[t]}{t^2})$ we have that :
$$GL_2(\frac{F[t]}{t^2})\text{ is isomorphic to }M_2(F)\rtimes_{\phi}GL_2(F)\text{ for a certain action } \phi$$
The last thing to do is to determine $\phi$ :
$$\phi(A).B:=(A+0t)(I_2+Bt)(A+0t)^{-1}=(A+ABt)(A^{-1}+0t)=I_2+ABA^{-1}t $$
Clearly it is the natural action of $GL_2(F)$ on $M_2(F)$ by conjugation, it is usually denoted by $Ad$. This explains my notation.
Well a very intersting question, what if you change $\frac{F[t]}{t^2}$ by $\frac{F[t]}{t^n}$ with $n\geq 3$? and by $F[[t]]$ (formal power series)?
To go back to the ring $\frac{F[t]}{t^2}$, this is not a discrete valuation ring but it is an artinian ring :http://en.wikipedia.org/wiki/Artinian_ring.
