Invariants of binary forms under a $\begin{pmatrix} 1& 1 \\ 0& 1 \end{pmatrix}$ action The special linear group $\text{SL}_2(\mathbb{Z})$ of $2\times 2$ invertible matrices in $\mathbb{Z}$ acts on binary cubic forms $\{ax^3 + bx^2y + cxy^2 + dy^3\}$ by acting on the vector $(x,y)^T$. This action descends to an action of the group $\Gamma_\infty := \left\{ \begin{pmatrix} 1 & n \\ 0 & 1 \end{pmatrix} : n \in \mathbb{Z}\right\}$ on binary cubic forms. It happens to be that $\Gamma_\infty$ is generated by $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. So understanding the action of this one element allows us to understand the whole $\Gamma_\infty$ action. (This is the action that sends $(x,y) \to (x+y, y)$)
I am interested in the invariants of binary forms under this action. In this (the binary cubic forms) case, the invariants form a 3-dimensional algebra, and I know the invariants. This action makes sense on other binary forms as well. It happens to be that the invariants of the binary quadratic case form a 2-dimensional algebra, which I also know. But I found these through the "brute force" method.
I am not really familiar with invariant theory at all, so I don't know if this is an easy or a hard question. But I'm wondering if there's some nice form for the invariants of binary n-degree forms under $\Gamma_\infty$? Alternately, perhaps there's a nice form for the invariants under $\text{SL}_2(\mathbb{Z})$ that one might perhaps be able to modify?
 A: Let $f(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0$ (I'm not sure how to frame the following with homogeneized polynomials, so I will stick with this together with the action induced by translations)
The translation by the opposite of the unique root of $f^{(n-1)}(x)$ gives you a canonical way to pick a representative in each equivalence class :
$I_n = a_n$ is always an invariant of the action induced by translation, and if $x$ is the root of $f^{(n-1)}$,
then $I_0 = f(x), I_1 = f'(x), I_2 = f''(x),\ldots, I_{n-2} = f^{(n-2)}(x)$ are all invariant. This gives you $n-1$ invariants. 
Conversely, given $I_0, I_1, \ldots I_{n-2}$ and $I_n \in \Bbb K$ (a field with characteristic $0$) , the corresponding unique class of polynomials modulo translation by an element of $\Bbb K$ with those invariants is the equivalence class of $I_n x^n + I_{n-2} \frac {x^{n-2}}{(n-2)!} + \ldots + I_1 x + I_0$
For example with $f(x) = ax^3+bx^2+cx+d$, 
The only root of $f''$ is the inflexion point of $f$, $x = - \frac b{3a}$, and by evaluating $f$ and $f'$ at $x$, you get the other two invariants $3ac-b^2$ and $2b^3-9abc+27a^2d$. 

Note that if a rational function $g \in \Bbb K(a_0,a_1,\ldots,a_n)$ is invariant by translation-by-one, then it is invariant by all the translations with parameter in $\Bbb K$ (I need this because I'm translating by $-a_{n-1}/na_n)$ :
Here the action on the degree $1$ part of $\Bbb K[a_0,a_1,\ldots,a_n]$ is a linear transformation. In fact there is a particular matrix $T \in GL_{n+1}(\Bbb Z[X])$ such that the action of the translation by $q$ is the linear transformation given by $T(q)$.
Therefore, for each $g$ there is an element $\tilde{g} \in \Bbb K(a_0,a_1,\ldots,a_n,X)$ such that forall $q \in \Bbb K$, $g^{T(q)}(a_0,a_1,\ldots,a_n) = g(T(q)(a_0,a_1,\ldots,a_n)) = \tilde{g}(a_0,a_1,\ldots,a_n,q)$.
Now, if $g$ is invariant by $T(1)$, then forall $q \in \Bbb N$, $\tilde{g}(a_0,a_1,\ldots,a_n,q) = \tilde{g}(a_0,a_1,\ldots,a_n,0)$, which implies that $g(a_0,a_1,\ldots,a_n,X) - g(a_0,a_1,\ldots,a_n)$ is a rational function of $X$ with infinitely many roots, so it is zero, and so $g$ is invariant by every translation $T(q)$ for $q \in \Bbb K$
