Noether Normalisation and Nullstellensatz - Examples

The set $\{s,t\}$ is algebraically independent over $k$ by construction, so $k(s,t)$ is a rational function field in two variables.

The remaining generator $u$ satisfies the polynomial $X^2-(s+t)\in k(s,t)[X]$, so $u$ is algebraic over $k(s,t)$. Since $K=k(s,t,u)$, the whole field $K$ is algebraic over $k(s,t)$.

Therefore [L1] shows that $\{s,t\}$ is a transcendence basis of $K$ over $k$.

Introduce the triangular coordinate $u=x-y$, so $x=u+y$. Then $(u+y{)}^2+y^2=u^2+2uy+2y^2\mathrm{.}$ Multiplying by $2^{-1}$ gives the monic polynomial $y^2+uy+\frac{1}{2}{u}^2$ in the variable $y$. This is the concrete instance of [L1].

In the quotient algebra, $\bar{u}=\bar{x}-\bar{y}$ and the class $\bar{y}$ satisfies ${\bar{y}}^2+\bar{u}\bar{y}+\frac{1}{2}{\bar{u}}^2=0$ over $k[\bar{u}]$. By [L2], $\bar{y}$ is integral over $k[\bar{u}]$. Since $A=k[\bar{u},\bar{y}]$, the algebra $A$ is module-finite over $k[\bar{u}]$.

The highest homogeneous part of $f$ is $f$ itself. For any $a\in {\mathbb{F}}_q$, $f(a,1)=a^q-a=0,$ so the infinite-field linear-change argument cannot choose a scalar $a\in {\mathbb{F}}_q$ with nonzero leading coefficient.

Now substitute $x\mapsto x+y^N$ with $N>q$. Because $q$ is a power of the characteristic of ${\mathbb{F}}_q$, the Frobenius identity gives $(x+y^N{)}^q=x^q+y^{Nq}$. Hence $f(x+y^N,y)=x^{q}y+y^{Nq+1}-x{y}^q-y^{N+q}\mathrm{.}$ Because $Nq+1>N+q$, the term $y^{Nq+1}$ is the unique highest power of $y$. Therefore the transformed polynomial is already monic in $y$, exactly as [L1] predicts.

Every element of $B$ can be written as $h\mathrm{/}{t}^m$ with $h\in k[t]$ and $m\ge 0$. Thus only powers of $t$ occur in the denominator.

If $1\mathrm{/}(t-1)$ belonged to $B$, then $1\mathrm{/}(t-1)=h\mathrm{/}{t}^m$ for some $h\in k[t]$. Cross-multiplication would give $t^m=h(t-1)$, so $t-1$ would divide $t^m$. But substituting $t=1$ gives $1=0$, impossible. This is the concrete obstruction behind [L1].

The quotient $\mathbb{R}[x]\mathrm{/}(x^2+1)$ is isomorphic to $\mathbb{C}$ by sending the class of $x$ to $i$. Since $\mathbb{C}$ is a field, $(x^2+1)$ is maximal.

For every $a\in \mathbb{R}$, [L1] says the evaluation ideal at $a$ is $(x-a)$, and no such ideal equals $(x^2+1)$ because $x^2+1$ has no real root. So weak Nullstellensatz fails in point form over $\mathbb{R}$. The residue field extension here is $\mathbb{C}\mathrm{/}\mathbb{R}$, which is finite of degree $2$, as [L2] allows.

A point $(a,b)\in k^2$ lies in $V(I)$ exactly when $a^2=0$ and $ab=0$. Since $k$ is a field, this means $a=0$ and $b$ is arbitrary. Hence $V(I)=\{(0,b):b\in k\}$.

The vanishing ideal of the $y$-axis is $(x)$. Also $\sqrt{I}=(x)$ because $x^2\in I$ and every element of $I$ is divisible by $x$. Therefore $I(V(I))=(x)=\sqrt{I}$, in agreement with [L1].

The zero locus of $I=(x,y)$ is the single point $(0,0)$, and $f(0,0)=0$. So the Rabinowitsch hypothesis holds. The displayed formula is already an explicit unit-ideal identity in the auxiliary ideal.

In the localization where $x+y$ is invertible, substitute $u=1\mathrm{/}(x+y)$ into $1=ux+uy+(1-u(x+y))$ to get $1=\frac{x}{x+y}+\frac{y}{x+y}\mathrm{.}$ The zero-denominator case is excluded precisely because this localization inverts $x+y$. Multiplying by $x+y$ yields $x+y=x+y\in I$. This is the denominator-clearing step of [L3] with $N=1$.