## Derivations of Weyl algebras are inner

Posted: September 23, 2011 in Noncommutative Ring Theory Notes, Weyl Algebras
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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 $(1). \ \Box$

## Derivations of central simple algebras

Posted: February 2, 2011 in Noncommutative Ring Theory Notes, Simple Rings
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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 $n$th 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.

## Examples of simple rings (1)

Posted: June 12, 2010 in Noncommutative Ring Theory Notes, Simple Rings
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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 obviously 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?

Notation. 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 $\displaystyle \delta(f(x))=\frac{df}{dx}.$ 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 $\delta$ideal if $\delta(I) \subseteq I.$ If the only $\delta$-ideals of $A$ are $\{0\}$ and $A,$ we’ll say that $A$ is $\delta$simple.

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 $\displaystyle \delta = \frac{d}{dx}.$

Definition 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 differential polynomial algebra. Note that if $\delta=0,$ i.e. $\delta(a)=0$ for all $a \in A,$ then $A[x; \delta]=A[x],$ the ordinary 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.