Throughout is a field, and is a -algebra.
Example 2. Suppose that is a derivation of which is not inner. If is -simple, then is simple. In partcular, if is simple, then is simple too.
Proof. Suppose, to the contrary, that is not simple. So has some non-zero ideal Let be the minimum degree of non-zero elements of Let be the set of leading coefficients of elements of Clearly is a left ideal of because is an ideal of To see that is also a right ideal of , let and Then there exists
But, by Remark 5
and so i.e. is also a right ideal. Next, we’ll show that is a -ideal of :
if then there exists some Clearly because is an ideal of Now
So , i.e. is a non-zero -ideal of Therefore , because is -simple. So i.e. there exists Finally let Now, which is an element of is a polynomial of degree at most and the coefficient of is , which has to be zero because of the minimality of Thus, since we may let to get That means is inner. Contradiction!
Lemma. If is simple and then is -simple.
Proof. Let be a -ideal of Let be an element of of the least degree. Suppose Then, since is a -ideal, we must have Hence by the minimality of and thus because This contradiction shows that and so Hence because is an ideal of and But is simple and so i.e. and thus
Definition 5. The algebra is called the first Weyl algebra over and is denoted by Inductively, for we define and we call the –th Weyl algebra over
Example 3. If is simple, then is simple for all In particular, is simple.
Proof. By Remark 3 in part (1), is a non-inner derivation of So, is simple by the above lemma and Example 2 in part (1). The proof is now completed by induction over
Example 4. In this example we do not need to assume that Let be a simple ring and its center. Let be a simple -algebra but the center of may or may not be The first part of the corollary in this post shows that is simple. This is another way of constructing new simple rings from old ones.