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The problem of determining a set of generators of the ring of invariants of the group $\textrm{SL}_2$ acting on the complex $n+1$-dimensional vector space of binary $n$-ic forms is known to be very hard and is open for $n > 10$. However we do know thanks to Hilbert that this is a finitely generated ring over $\mathbb{C}$. When $n = 2$ or $3$, the ring of invariants is generated by the discriminant of the binary (quadratic/cubic) form and when $n = 4$, we have two generators $I$ and $J$, both $\mathbb{Z}$- polynomials in the coefficients of the binary quartic form of degrees $2$ and $3$ respectively. This begs the general question -- even if one cannot exhibit a set of generators for the complex ring of invariants, can one show that there exists such a set of integral/rational polynomials (equivalent by scaling) in the coefficients of the binary form? Intuitively, I would guess that the answer is yes, but I don't know if there's a way to see this nor could I find a reference for it.


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up vote 5 down vote accepted

Sure. If you have an algebraic group $G$ defined over $\mathbb Q$ acting on a $\mathbb Q$-algebra $A$, the $\mathbb Q$-algebra of invariants $A^G$ is the equalizer of two usual homomorphisms of algebras $A \to A \otimes_{\mathbb Q}\mathbb Q[G]$. Tensoring with $\mathbb C$ is exact, so $(A \otimes \mathbb C)^{G_{\mathbb C}} = A^G\otimes \mathbb C$.

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Thanks, this is pretty much what I had in mind but "everything should be defined over $\mathbb{Q}$" didn't get me too far. – Ashwath Rabindranath Apr 9 '13 at 10:44

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