Posts Tagged ‘inner derivation’

Let k be a field. We proved here that every derivation of a finite dimensional central simple k-algebra is inner. In this post I will give an example of an infinite dimensional central simple k-algebra all of whose derivations are inner. As usual, we will denote by A_n(k) the n-th Weyl algebra over k. Recall that A_n(k) is the k-algebra generated by x_1, \ldots , x_n, y_1, \ldots , y_n with the relations x_ix_j-x_jx_i=y_iy_j-y_jy_i=0, \ y_ix_j-x_jy_i= \delta_{ij}, for all i,j. When n = 1, we just write x,y instead of x_1,y_1. If \text{char}(k)=0, then A_n(k) is an infinite dimensional central simple k-algebra and we can formally differentiate and integrate an element of A_n(k) with respect to x_i or y_i exactly the way we do in calculus.  Let me clarify “integration” in A_1(k). For every u \in A_1(k) we denote by u_x and u_y the derivations of u with respect to x and y respectively. Let f, g, h \in A_1(k) be such that g_x=h_x=f. Then [y,g-h]=0 and so g-h lies in the centralizer of y which is k[y]. So g-h \in k[y]. For example, if f = y + (2x+1)y^2, then g_x=f if and only if g= xy + (x^2+x)y^2 + h(y) for some h(y) \in k[y]. We will write \int f \ dx = xy+(x^2+x)y^2.

Theorem. If \text{char}(k)=0, then every derivation of A_n(k) is inner.

Proof. I will prove the theorem for n=1, the idea of the proof for the general case is similar. Suppose that \delta is a derivation of A_1(k). Since \delta is k-linear and the k-vector space A_1(k) is generated by the set \{x^iy^j: \ i,j \geq 0 \}, an easy induction over i+j shows that \delta is inner if and only if there exists some g \in A_1(k) such that \delta(x)=gx-xg and \delta(y)=gy-yg. But gx-xg=g_y and gy-yg=-g_x. Thus \delta is inner if and only if there exists some g \in A_1(k) which satisfies the following conditions

g_y=\delta(x), \ \ g_x = -\delta(y). \ \ \ \ \ \ \ (1)

Also, taking \delta of both sides of the relation yx=xy+1 will give us

\delta(x)_x = - \delta(y)_y. \ \ \ \ \ \ \ \ (2)

From (1) we have \delta(x) = - \int \delta(y)_y \ dx + h(y) for some h(y) \in k[y]. It is now easy to see that

g = - \int \delta(y) \ dx + \int h(y) \ dy

will satisfy both conditions in (1). \ \Box

We will assume again that k is a field. We will denote by Z(A) the center of a ring A. The following result is one of many nice applications of the Skolem-Noether theorem (see the lemma in this post!). For the definition of derivations and inner derivations of a k-algebra see Definition 2 in this post.

Theorem. Every derivation of a finite dimensional central simple k-algebra A is inner.

Proof. First note that, since A is simple, M_2(A) is simple. We also have Z(M_2(A)) \cong Z(A) = k. Finally \dim_k M_2(A) = 4 \dim_k A < \infty. Thus M_2(A) is also a finite dimensional central simple k-algebra. Now, let \delta be a derivation of A. Define the map f: A \longrightarrow M_2(A) by

f(a) = \begin{pmatrix} a & \delta(a) \\ 0 & a \end{pmatrix},

for all a \in A. Obviously f is k-linear and for all a,a' \in A we have

f(a)f(a') = \begin{pmatrix} aa' & \delta(a)a'+a \delta(a') \\ 0 & aa' \end{pmatrix} = \begin{pmatrix} aa' & \delta(aa') \\ 0 & aa' \end{pmatrix} = f(aa').

So f is a k-algebra homomorphism and hence, by the lemma in this post, there exists v \in M_2(A) such that v is invertible and f(a)=vav^{-1}, for all a \in A. Let

v = \begin{pmatrix} x & y \\ z & t \end{pmatrix}.

So f(a)v=va gives us

\begin{cases} ax+\delta(a)z=xa \\ ay+\delta(a)t=ya \\ az=za \\ at=ta. \end{cases} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)

Since (*) holds for all a \in A, we will get from the last two equations that z,t \in Z(A)=k. Since we cannot have z=t=0, because then v wouldn’t be invertible, one of z or t has to be invertible in k because k is a field. We will assume that z is invertible because the argument is similar for t. It now  follows, from the first equation in (*) and the fact that z^{-1} \in Z(A), that

\delta(a)=(xz^{-1})a - a(xz^{-1}). \ \Box

Throughout k is a field, \text{char}(k)=0 and A is a k-algebra.

Example 2. Suppose that \delta is a derivation of A which is not inner. If A is \delta-simple, then B=A[x;\delta] is simple. In partcular, if A is simple, then A[x; \delta] is simple too.

Proof. Suppose, to the contrary, that B is not simple. So B has some non-zero ideal I \neq B. Let n be the minimum degree of non-zero elements of I. Let J be the set of leading coefficients of elements of I. Clearly J is a left ideal of A because I is an ideal of B. To see that J is also a right ideal of A, let a \in J and b \in A. Then there exists

f=ax^n + \text{lower degree terms} \in I.

But, by Remark 5

fb = abx^n + \text{lower degree terms} \in I

and so ab \in J, i.e. J is also a right ideal. Next, we’ll show that J is a \delta-ideal of A:

if a_n \in J, then there exists some f(x)=\sum_{i=0}^n a_ix^i \in I. Clearly xf - fx \in I, because I is an ideal of B. Now

xf - fx=\sum_{i=0}^n xa_i x^i - \sum_{i=0}^n a_ix^{i+1}=\sum_{i=0}^n (a_ix +\delta(a_i))x^i - \sum_{i=0}^n a_i x^{i+1}

= \sum_{i=0}^n \delta(a_i)x^i.

So \delta(a_n) \in J, i.e. J is a non-zero \delta-ideal of A. Therefore J=A, because A is \delta-simple. So 1 \in J, i.e. there exists g(x)=x^n + b_{n-1}x^{n-1} + \cdots + b_0 \in I. Finally let a \in A. Now, ga - ag, which is an element of I, is a polynomial of degree at most n-1 and the coefficient of x^{n-1} is b_{n-1}a - ab_{n-1} + n \delta(a), which has to be zero because of the minimality of n. Thus, since \text{char}(k)=0, we may let c=\frac{b_{n-1}}{n} to get \delta(a)=ca-ac. That means \delta is inner. Contradiction! \Box

Lemma. If A is simple and \delta = \frac{d}{dx}, then A[x] is \delta-simple.

Proof. Let I \neq \{0\} be a \delta-ideal of A[x]. Let f=\sum_{i=0}^n a_ix^i, \ a_n \neq 0, be an element of I of the least degree. Suppose n > 0. Then, since I is a \delta-ideal, we must have \frac{df}{dx}=\sum_{i=0}^{n-1}ia_ix^{i-1} \in I. Hence na_{n-1}=0, by the minimality of n, and thus a_n=0 because \text{char}(k)=0.  This contradiction shows that n=0 and so f=a_0 \in A \cap I. Hence Aa_0A \subseteq I because I is an ideal of A[x] and A \subset A[x]. But A is simple and so Aa_0A = A, i.e. 1 \in Aa_0A \subseteq I and thus I=A[x]. \Box

Definition 5. The algebra A[x][y, \frac{d}{dx}] is called the first Weyl algebra over A and is denoted by \mathcal{A}_1(A). Inductively, for n \geq 2, we define \mathcal{A}_n(A)=\mathcal{A}_1(\mathcal{A}_{n-1}(A)) and we call \mathcal{A}_n(A) the nth Weyl algebra over A.

Example 3. If A is simple, then \mathcal{A}_n(A) is simple for all n. In particular, \mathcal{A}_n(k) is simple.

Proof. By Remark 3 in part (1), \delta = \frac{d}{dx} is a non-inner derivation of A[x]. So, \mathcal{A}_1(A) is simple by the above lemma and Example 2 in part (1). The proof is now completed by induction over n. \Box

Example 4. In this example we do not need to assume that \text{char}(k)=0. Let A be a simple ring and k its center. Let B be a simple k-algebra but the center of B may or may not be k. The first part of the corollary in this post shows that A \otimes_k B is simple. This is another way of constructing new simple rings from old ones.

Definition 1. A ring R with 1 is called simple if (0) and R are the only two-sided ideals of R.

Remark 1. The center of a simple ring R is a field.

Proof. Let a be any non-zero element of the center of R. Then Ra is a non-zero two-sided ideal of R and hence, since R is simple, Ra=R. Thus there exists some r \in R such that ra=1, i.e. a is invertible. Since a is in the center, r is in the center too and we’re done. \Box

Obviously commutative simple rings are  just fields. So only non-commutative simple rings are interseting.

Note 1. If k is the center of a simple ring R, then R is both a vector space over k and a ring. So it is a k-algebra. The term simple algebras is commonly used instead of simple rings.

We now give the first example of simple rings.

Example 1. Let R be a ring with 1 and let M_n(R) be the ring of n \times n matrices with entries from R. It is a well-known fact, and easy to prove, that J is a two-sided ideal of M_n(R) if and only if J=M_n(I) for some two-sided ideal I of R. In particular, M_n(R) is simple if and only if R is simple. So, for example, since every division ring D is simple, M_n(D) is simple too. In fact, by the Wedderburn-Artin theorem, a ring R is simple and (left or right) Artinian if and only if R = M_n(D) for some division ring D and some integer n \geq 1.

Example 1 gives all simple rings which are Artinian. But what about simple rings which are not Artinian? But first some definitions and remarks. From now on we will assume that k is a field and A is a k-algebra.

Definition 2. A k-linear map \delta : A \longrightarrow A is called a derivation of A if \delta(ab)=\delta(a)b + a \delta(b), for all a,b \in A. If there exists some c \in A such that \delta(a)=ca-ac, for all a \in A, then \delta is called an inner derivation.

Note that if c \in A, then the map \delta : A \longrightarrow A defined by \delta(a)=ca-ac, for all a \in A, is a derivation.

Remark 2. If \delta is a derivation of A, then \delta(1)=\delta(1 \cdot 1)=\delta(1)+\delta(1) and thus \delta(1)=0. Therefore if \alpha \in k, then \delta(\alpha)=\alpha \delta(1)=0.

Remark 3. Consider the polynomial algebra R:=A[x] and define the map \delta : R \longrightarrow R by \delta(f(x))=\frac{df}{dx}. Then \delta is a derivation which is not inner. The reason is that if there was g \in R such that \delta(f)=gf - fg, for all f \in R, then, since x is central, we’d get 1 = \delta(x) = gx-xg=0, which is non-sense.

Definition 3. Let \delta be a derivation of A. An ideal I of A is called a \deltaideal if \delta(I) \subseteq I. If the only \delta-ideals of A are \{0\} and A, we’ll say that A is \deltasimple.

Obviously if A is simple, then A is \delta-simple for every derivation \delta.

Remark 4. If \delta is an inner derivation of A, then every ideal of A is also a \delta-ideal. An obvious example of  a \delta-simple algebra is A=k[x] with \delta = \frac{d}{dx}.

Defintion 4. If \delta is a derivation of A, then we define the  k-algebra A[x; \delta] to be the set of all polynomials in the indeterminate x and left coefficients in A, with the usual addition and multiplication and the rule xa=ax + \delta(a). The algebra A[x; \delta] is also called a differntial polynomial algebra.

Note 2. So a typical element of A[x; \delta] is in the form a_nx^n + a_{n-1}x^{n-1} + \cdots + a_0, where a_i \in A. When we multiply two of these polynomials we will have to use the rule given in Definition 4. For example

x(ax+b)=(xa)x + xb = (ax + \delta(a))x + bx +\delta(b)=ax^2 + (\delta(a)+b)x + \delta(b).

Of course, we need to prove that A[x;\delta] is a k-algebra but we won’t do it here.

Remark 5. Let a \in A and m \geq 1 be an inetger. Then an easy induction shows that in A[x; \delta] we have

x^ma=\sum_{i=0}^m \binom{m}{i} \delta^{m-i}(a)x^i,

where \delta^0(a)=a. So x^m a is a (left) polynomial of degree m with the leading coefficient a.

In part (2) we will give three important examples of simple rings.