# Let $A=k[x]$ and $V=k[x]/((x-λ)^n)$. Find a filtration $V=V_0 ⊃ V_1 ⊃ \dots⊃ V_n=0$ such that the subsequent quotients $V_{i-1}/V_i$ are irreducible.

Let $A$ be the algebra $A=k[x]$ and let $V$ be the representation $V=k[x]/((x-\lambda)^n)$ for some $\lambda \in k$ and $n\in\Bbb N$.

Find a filtration $V=V_0 \supset V_1 \supset \dots \supset V_n=0$ such that the subsequent quotients $V_{i-1}/V_i$ (where $i=1,...,n$) are irreducible. Describe the $A$-module structure of these irreducible "subquotients" of $V$.

I know that the irreducible representations of $k[x]$ are 1 dimensional. So therefore $V_{i-1}/V_i$ must be $1$-dimensional. My intuition says, that probably $V_i= \overline{(x-\lambda)^i}$. Can someone put some light for me in this matter ?

• Did you check to see if your guess is correct? – whacka Oct 1 '14 at 20:35

If you mean by your notation that $V_i$ should be the subrepresentation of $V$ generated by $\overline{(x-\lambda)^i}$, then you are correct.
To shed some more light to prove this there are several things you must show. That the $V_i$ are subrepresentations is cLear with this definition. It is easy to see that they are contained in each other with $V_0=V$ and $V_n=0$.
E.g. using the Euclidean algorithm one can show that a basis of $V_i$ is given by $\overline{(x-\lambda)^i},\dots, \overline{(x-\lambda)^{n-1}}$. With this fact, it is obvious that $V_{i-1}/V_i$ is one-dimensional, since there is only one basis vector not contained in $V_i$.
To describe the $A$-module structure, observe that $x$ acts on these irreducibles as multiplication by $\lambda$