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By Francois Dumas

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**Extra info for Noncommutative invariants**

**Example text**

We give here a short independent proof. Let z be a normal element of kq [x, y]. We have in particular zy = uz and zx = vz for some u, v ∈ kq [x, y]. Considering degx in the first equality, we have u ∈ k[y]. Denoting z = m fm (y)xm , relation zy = uz implies m fm (y)(q m y − u(y))xm = 0; since q is not a root of one, there exists one nonnegative integer i such that z = fi (y)xi . From the second equality fi (y)xi+1 = vz, it is easy to deduce that z = αy j xi for some nonnegative integer j and some α ∈ k.

1 Deformations of Poisson algebras General deformation process We fix a non necessary commutative k-algebra B. We suppose that there exists some element h of B which is central in B not invertible and not a zero divisor in B, such that A := B/hB is a commutative k-algebra. A being commutative, any u, v ∈ B satisfy (u+hB)(v +hB) = (v +hB)(u+hB) and then [u, v] := uv −vu ∈ hB. We denote by γ(u, v) the element of B defined by [u, v] = hγ(u, v). We set: {u, v} = γ(u, v) for any u, v ∈ A. This is independent of the choice of u, v.

Thus each vector of M lies in the Lie subalgebra g of S := C[x1 , . . , xn , y1 , . . , yn ] generated by H and T . Moreover, (ad X) acts as ni=1 ∂yi and (ad (−Y )) as ni=1 ∂xi then we have for 1 ≤ i ≤ n: (ad (−Y ))i (Xi ) = 12 (i + 2)! x2i and (adX (Y ))i (Yi ) = 12 (i + 2)! yi2 , which imply that x21 , . . , x2n , y12 , . . , yn2 ∈ g. Applying again (ad (−Y )) and (ad X), we deduce that g contains x1 , . . , xn , y1 , . . , yn . Hence g contains the homogeneous component of degree one S1 = V = Cx1 ⊕ · · · ⊕ Cxn ⊕ Cy1 ⊕ · · · ⊕ Cyn .