# $End_{\mathbb{C}} ( \mathbb{C}[x])$ and Weyl algebra

How do you see this? $End_{\mathbb{C}} ( \mathbb{C}[x])$

As $M_{n}(\mathbb{C})=End_{\mathbb{C}} ( \mathbb{C}[x])$, so it just a matric with basis of polyonial?

Take the weyl algebra $A_1=\{ {\sum_{i=0}^{n} f_i}(x) \partial^i \ : f_{i}(x) \in \mathbb{C}, n\in \mathbb{N}\}$.

It says this weyl algebra A_1 is a subring of $End_{\mathbb{C}} ( \mathbb{C}[x])$so does that mean weyl algebra just a big matrix with complex numbers in them?

Also, is this the hardest example of a non commutative ring? As I hate to see tricker stuff than this :<

I used to like rings.

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I don't understand the second sentence. What is $n$? – Qiaochu Yuan Oct 6 '11 at 13:40
Fixed it. Was being silly. – weylthis Oct 6 '11 at 13:43

$\text{End}_\mathbb{C}(\mathbb{C}[x])$ is just the space of $\mathbb{C}$-linear maps on the polynomial ring. To verify that a map is $\mathbb{C}$-linear is usually trivial. Yes, you can think of an endomorphism as an infinite dimensional matrix (not $n$-dimensional!), but I don't think that that's the easiest way to think about them.
Now, the Weyl algebra is defined as $A_1=\{ {\sum_{i=0}^{n} f_i}(x) \partial^i \ : f_{i}(x) \in \mathbb{C}[x]\}$ (and not $f_i\in \mathbb{C}$), where $\partial^i$ is the $i$-th derivative operator. Clearly, multiplication by a polynomial is linear, and clearly differentiation is $\mathbb{C}$-linear. Compositions and sums of linear maps are linear, so the elements of the Weyl algebra are indeed $\mathbb{C}$-endomorphisms of $\mathbb{C}[x]$. That's all there is to it.
@weylthis You are mixing several things. The ring $\mathbb{C}[x]$ has a $\mathbb{C}$-basis, $x^i$: $i=0,1,2,\ldots$. Therefore, a $\mathbb{C}$-linear map is determined by its action on this basis and can be written as an infinite matrix. – Alex B. Oct 6 '11 at 14:01