- But what does $x, y, z$ even mean? well, they are linear function $x, y, z: \mathbb R^3 \rightarrow \mathbb R$. So $x^2 + y^2 + z^2$ is a "polynomial" of these linear functions.

- If $G$ acts on $V$, how does $G$ act on the polynomial functions $V \rightarrow \mathbb R$?
- In general, if we have a function $f: X \rightarrow Y$ where $g$ acts on $X$ and $Y$(in our case, $G$ acts trivially on $Y=\mathbb R$), what is $g(f)$?
- We define $(gf)(x) \equiv g (f(g^{-1}(x)))$.
- What is the $g^{-1}$? We should write $(gf)(gx) = g(fx)$. This is like $g(ab) = g(a) g(b)$. We want to get $(gf)(x) = (gf)(g(g^{-1}x) = g(f(g^{-1}x))$.
- If we miss out $g^{-1}$ we get a mess. Let's temporarily define $(gf)(x) = f(g(x))$. Consider $(gh)f(x) = f(ghx)$. But we can also take this as $(gh)(f(x)) = g((hf)(x)) = (hf)(gx) = f(hgx)$. This is absurd as it gives $f(ghx) = f(hgx)$.

$\begin{bmatrix}
a & b \\ c & d
\end{bmatrix}$

acting on:
$\begin{bmatrix}
x_1 & y_1 \\ x_2 & y_2
\end{bmatrix}$

This action preserves the polynomial $x_1 y_2 - x_2 y_1$, aka the determinant. anything that
ends with an "-ant" tends to be an "invari-ant" (resultant, discriminant)
- $e_1 \equiv x_1 + x_2 + \dots x_n$
- $e_2 \equiv x_1 x_2 + x_1 x_3 + \dots + x_{n-1} x_n$.
- $e_n \equiv x_1 x_2 \dots x_n$.

- The
says that every invariant polynomial in $x_1, \dots x_n$ is a polynomial in $e_1, \dots, e_n$.*basic theory of symmetric functions*

$\begin{aligned}
P \equiv
&(x_1 + x_2 \dots)^{n_1 - n_2} \\
&\times (x_1 x_2 + x_1 x_2 \dots)^{n_2 - n_3} \\
&\times (x_1 x_2 x_3 + x_1 x_2 x_4 \dots)^{n_3 - n_4} \\
\end{aligned}$

This kills of the biggest monomial in $f$. If $f$ is symmetric,
Then we can order the term we choose such that $n_1 \geq n_2 \geq n_3 \dots$.
We need this to keep the terms $(n_1 - n_2), (n_2 - n_3), \dots$ to be positive.
So we have now killed off the largest term of $f$. Keep doing this to kill of $f$ completely.
This means that the invariants of $S_n$ acting on $\mathbb C^n$ are a finitely generated
algebra over $\mathbb C$. So we have a finite number of generating invariants such that every invariant
can be written as a polynomial of the generating invariants with coefficients in $\mathbb C$. This is
the first non-trivial example of invariants being finitely generated.
The algebra of invariants is a - $z_0 z_2 = x^4y^2 = z_1^2$. So $z_0 z_2 - z_1^2 = 0$.
- $z_1 z_3 x^2y^4 = z_2^2$. So $z_1 z_3 - z_2^2 = 0$.
- $z_0 z_3 = x^3y^3 = z_1 z_2$. So $z_0 z_3 - z_1 z_2 = 0$.

- $p_1 \equiv z_0 z_2 - z_1^2$.
- $p_2 \equiv z_1 z_3 - z_2^2$.
- $p_3 \equiv z_0 z_3 - z_1 z_2$.

$\begin{aligned}
& z_2 p_1 + z_0 p_2 \\
&= z_2 (z_0 z_2 - z_1^2) + z_0 (z_1 z_3 - z_2^2) \\
&= (z_0 z_2^2 - z_2 z_1^2) + (z_0 z_1 z_3 - z_0 z_2^2) \\
&= z_0 z_1 z_3 - z_2 z_1^2 \\
&= z_1(z_0 z_3 - z_1 z_2) \\
&= z_1 p_3
\end{aligned}$

So we have non-trivial relations between $p_1, p_2, p_3$! This is a second order syzygy, a sygyzy between syzygies.
We have a ring $R \equiv k[z_0, z_1, z_2, z_3]$. We have a map $R \rightarrow \texttt{invariants}$.
This has a nontrivial kernel, and this kernel is spanned by $(p_1, p_2, p_3) \simeq R^3$. But this itself
has a kernel $q = z_1 p_1 + z_2 p_2 + z_3 p_3$. So there's an exact sequence:
$\begin{aligned}
0 \rightarrow R^1 \rightarrow R^3 \rightarrow R=k[z_0, z_1, z_2, z_3] \rightarrow \texttt{invariants}
\end{aligned}$

In general, we get an invariant ring of linear maps that are invariant under the group action.
We have polynomials $R \equiv k[z_0, z_1, \dots]$ that map onto the invariant ring. We have
relationships between the $z_0, \dots, z_n$. This gives us a sequence of syzygies. We have many
questions:
- Is $R$ finitely generated as a $k$ algebra? Can we find a finite number of generators?
- Is $R^m$ finitely generated (the syzygies as an $R$-MODULE)? To recall the difference, see that $k[x]$is finitely generated as an ALGEBRA by $(k, x)$ since we can multiply the $x$s. It's not finitely generated as a MODULE as we need to take all powers of $x$: $(x^0, x^1, \dots)$.
- Is this SEQUENCE of sygyzy modulues FINITE?
- Hilbert showed that the answer is YES if $G$ is reductive and $k$ has characteristic zero. We will do a special case of $G$ finite group.