# Derivations of Polynomial Algebra

Let $A=\mathbb{F}_p[x,y]$, the commutative polynomial algebra on two variables over the finite field $\mathbb{F}_p$. Define a derivation on an algebra as a map which satisfies the Leibniz rule, ie if $d$ is a derivation then

$$d(ab)=ad(b)+bd(a)$$

Let $Der(A)$ be the $\mathbb{F}_p$-module (algebra?) of derivations of $A$. I'm interested in knowing what $Der(A)$ is. First of all, I'm not sure exactly how complicated the algebraic structure it carries is. I know that it's at least a $\mathbb{F}_p$-module. But is it an algebra? I suspect no, but I can't nail down a reason why. Is it a module over a bigger ring? $\mathbb{F}_p[x,y]$ seems like it might be a good choice- but again, I'm not very sure of this. Finally, what's an explicit description of elements of $Der(A)$? Is it just linear combinations of $\frac{d}{dx}$ and $\frac{d}{dy}$?

I apologize if this is a bit overly broad, but I'm encountering derivations over finite fields for the first time, and I'd like to really understand what's going on.

## 2 Answers

Let $\Lambda$ be an associative algebra over the field $k$. Then $\operatorname{Der}(\Lambda)$ is a $k$-Lie algebra under the operation $[f,g]:=fg-gf$ (on the right $fg$ means the composition of the two maps $f$ and $g$). $\operatorname{Der}(\Lambda)$ has an ideal $\operatorname{IDer}(\Lambda)$, the inner derivations, consisting of those derivations of the form $d_\lambda(x):= \lambda x - x \lambda$ (this is trivial in your case).

The quotient $\operatorname{Der}(\Lambda)/\operatorname{IDer}(\Lambda)$ is isomorphic to the first Hochschild cohomology group $\operatorname{HH}^1(\Lambda)$. The Hochschild cohomology has the structure of a Gerstenhaber algebra, in particular $\operatorname{HH}^1(\Lambda)$ is a Lie algebra. The Lie algebra structure defined above induces a Lie algebra structure on $\operatorname{Der(\Lambda)}/\operatorname{IDer}(\Lambda)$ that agrees with the Lie algebra structure coming from the Gerstenhaber algebra structure.

Each Hochschild cohomology group is a module over $\operatorname{HH^0}(\Lambda)$. Zeroth Hochschild cohomology is the centre of the algebra, so in your case the whole algebra. The action of the centre is $(z\cdot \partial) (x) := z\partial(x)$.

In your case ($A = \mathbb{F}_p[x,y]$) the derivation algebra is much bigger than just linear combinations of the two derivative operators. In fact given any $a,b \in A$ you can define a derivation $\partial$ on $A$ by setting $\partial(1)=0, \partial(x)=a, \partial(y)=b$ and extending according to the derivation rules. It's clear that a derivation is completely determined by its action on $x$ and $y$, thus $\partial = a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}$ where the action of $a,b \in A$ is the action of the centre defined in the last paragraph. So these two partial derivatives do generate the derivations, but as an $A$-module not an $\mathbb{F}_p$-module.

• Quite impressive! – Leon Nov 22 '13 at 8:55

Let's call $Der(A)$ the set of all your derivations $d:A\to A$. What algebraic structure does it have?

1) $Der(A)$ is an $A$-module via the definition that for $a\in A$ and $d\in Der(A)$ we demand that $(a \bullet d)(b)= a\cdot d(b)$ where $\cdot$ on the right means multiplication in $A$.

non2) $Der(A)$ is not a subalgebra of $\mathcal L_{\mathbb F_p-lin}(A,A),\circ$ in the sense that given two derivations $d,d'\in A$, the composition $d\circ d'$ is not a derivation.
For example $(\frac{\partial}{\partial x} \circ\frac{\partial}{\partial y})(xy)=1\neq x(\frac{\partial}{\partial x} \circ\frac{\partial}{\partial y})(y)+y(\frac{\partial}{\partial x}\circ \frac{\partial}{\partial y})(x)=0$

2) However $Der(A)$ is a Lie algebra under the bracket $[d,d']=d\circ d'-d'\circ d$.

3) The $A$-module of derivations $A\to A$ is the dual module of a very interesting $A$-module $\Omega_A$ (the module of Kähler differentials), i.e. $Der(A)=\mathcal L_{A-lin}(\Omega_A,A)$. So there remains to compute $\Omega_A$.

4) It is easy to see, once you you are acquainted with the easy formalism of Kähler differentials, that $\Omega_A=A^2=A\delta x\oplus A \delta y$, so that $Der(A)=\mathcal L_{A-lin}(\Omega_A,A)= A\frac{\partial}{\partial x}\oplus A\frac{\partial}{\partial y}$

5) Summing up $$Der(\mathbb F_p[x,y])=\ \mathbb F_p[x,y]\frac{\partial}{\partial x}\oplus \mathbb F_p[x,y]\frac{\partial}{\partial y}$$

• Thanks for the informative answer! I regret that I have but one acceptance to give. Do you have any reading suggestions for material on Kahler differentials? I'm interested in knowing more about them. – KReiser Feb 26 '12 at 5:22
• Dear @KReiser, don't worry about acceptance: mt_'s answer is indeed very nice. A good reference for Kähler differentials (and all of commutative algebra) is Matsushima's Commutative Ring Theory, specifically Chapter 9. – Georges Elencwajg Feb 26 '12 at 7:58
• Don't you mean Matsumura? Is the following true: if $K=K[x_1,\ldots,x_n]$, then $Der_KA\cong A^n$ as a $K$-module and $A$-module and Lie $K$-algebra, where the bracket on r.h.s. is trivial $[-,-]=0$? – Leon Nov 22 '13 at 9:04
• Dear @Leon, yes I meant Matsumura! As for your question, it is true that $Der_K A\cong A^n$ but I don't think the Lie algebra is trivial: $y\frac{\partial}{\partial x} \circ (x\frac{\partial}{\partial y})- x\frac{\partial}{\partial y} \circ (y\frac{\partial}{\partial x}) \neq 0$ – Georges Elencwajg Nov 22 '13 at 10:14
• And this is $y\frac{d}{dy}\!+\!yx\frac{d}{dx}\frac{d}{dy}\!-\!x\frac{d}{dx}\!-\!xy\frac{d}{dy}\frac{d}{dx}= y\frac{d}{dy}\!-\!x\frac{d}{dx}\neq0$, right? One more thing: a Kähler derivation is an $R$-linear map $D\!:A\to A$ with $D(xy)=xD(y)+yD(x)$. If $A$ is commutative, then this is precisely an ordinary derivation. Why does commutative algebra use Kähler derivations, if they are ordinary ones? – Leon Nov 22 '13 at 17:43