# Noether normalization for $k_{x}$

According to the Noether normalization theorem, there exists a $k[t]$ where $t$ is an indeterminate and $k[t]\subseteq k[x]_{x}$ is a $k$-algebra extension so that $k[x]_{x}$ is a finitely generated $k[t]$-module. Note: I am using the notation where $k[x]_{x}$ denotes localization at the multiplicative set $S=\{1,x,x^2\,\cdots\}$.

I cannot think of a good candidate for $t$ inside $k[x]_{x}$. Obviously, letting $t=x$ would not work since $k[x]_{x}$ would not be finite, nor would $t = 1/x$. Any hints would be greatly appreciated!

#### Solutions Collecting From Web of "Noether normalization for $k_{x}$"

Notice that the proof of Noether Normalization is constructive.

Define $t = x – x^{-1}$. Then $xt = x^2 – 1$, hence $x$ is integral over $k[t]$. Also $t$ is algebraically independent and $k[x,x^{-1}]=k[t][x]$.

You can also interpret this geometrically: $k[x,x^{-1}]$ is the coordinate ring of the hyperbola $V(xy-1) \subseteq \mathbb{A}^2$. We project it onto the line $V(y+x) \subseteq \mathbb{A}^2$. This has clearly degree $2$. Using $V(y+x) \cong \mathbb{A}^1$, the projection becomes $V(xy-1) \to \mathbb{A}^1$, $(x,y) \mapsto x-y$. This is our Noether normalization.

$x+ 1/x = x+ 1/x$ and $x \cdot 1/x = 1$ so

$x$, $1/x$ are roots of

$$\lambda^2 – (x+1/x) \lambda + 1$$

and so $k[x,1/x]$ is integral (and finite ) over $k[x+1/x]$.

$t = x + 1/x$ ( or $t = ax + b/x$ except when $ab=0$).