Noetherian Rings and Hilbert Basis — Examples

Each ring is a quotient of a polynomial ring in finitely many indeterminates over its base ring: $k[x,y]$ is $k[x_1,x_2]$ after renaming the indeterminates, and $\mathbb{Z}[x]$ is $\mathbb{Z}[x_1]$. Being such a quotient is exactly the condition of being an algebra of finite type over that base ring, so $k[x,y]\mathrm{/}(xy)$ is of finite type over $k$ and $\mathbb{Z}[x]\mathrm{/}(x^2-2)$ is of finite type over $\mathbb{Z}$.

The base rings $k$ and $\mathbb{Z}$ are Noetherian.

An algebra of finite type over a Noetherian commutative ring is a Noetherian ring, so both displayed rings are Noetherian.

The first ring has zero divisors, and the argument above never asked. Let $\pi$ denote the quotient map $k[x,y]\to k[x,y]\mathrm{/}(xy)$. Evaluating at $0$ in the last indeterminate gives a ring homomorphism $\epsilon \text{\hspace{0.17em}⁣}:k[x][y]\to k[x]$ that fixes $k[x]$ and sends $y$ to $0$. Every element of $(xy)$ is $f\cdot xy$ with $f\in k[x,y]$, and $\epsilon (f\cdot xy)=\epsilon (f)\epsilon (x)\epsilon (y)=\epsilon (f)x\cdot 0=0$, whereas $\epsilon (x)=x\mathrm{\ne }0$; so $x\mathrm{\notin }(xy)$ and $\pi (x)\mathrm{\ne }0$. Exchanging the roles of the two indeterminates gives $\pi (y)\mathrm{\ne }0$ in the same way. Yet $\pi (x)\pi (y)=\pi (xy)=0$.

Under the identification of $R_m$ with the constants of $R_{m+1}$, each $R_m$ is a subring of $R_{m+1}$, so the family is increasing and any two of its members are comparable.

The union $R_{\mathrm{\infty }}$ is a commutative ring. Given $f,g\in R_{\mathrm{\infty }}$ there is $m$ with $f,g\in R_m$, by comparability of the two stages containing them; define $f+g$ and $fg$ there. The value does not depend on the stage chosen, because a larger stage contains the smaller as a subring and the operations of a subring are the restrictions of the ambient ones. Each ring axiom involves finitely many elements, which again lie in a common $R_m$, where the axiom holds. The identity is $1_k$ and the zero is $0_k$.

For $m\in \mathbb{N}$ let $I_m:=(x_1,\dots ,x_m)$ be the ideal of $R_{\mathrm{\infty }}$ generated by $x_1,\dots ,x_m$; at $m=0$ the generating set is empty and $I_0=0$. Since the generating sets increase with $m$, so do the ideals: $I_m\subseteq I_{m+1}$.

The inclusions are strict, because $x_{m+1}\mathrm{\notin }{I}_m$. Suppose $x_{m+1}={\sum }_{i=1}^m{f}_i{x}_i$ with $f_i\in R_{\mathrm{\infty }}$; all the $f_i$ lie in a common $R_N$ with $N\ge m+1$, so the equation holds in $R_N=k[x_1,\dots ,x_N]$. Iterating the one-variable universal property along the tower defining $R_N$ produces a $k$-algebra homomorphism $\theta \text{\hspace{0.17em}⁣}:R_N\to k[T]$ with $\theta (x_{m+1})=T$ and $\theta (x_j)=0$ for every $j\le N$ with $j\mathrm{\ne }m+1$. Applying $\theta$ to the supposed equation gives $T={\sum }_{i=1}^m\theta (f_i)\cdot 0=0$ in $k[T]$, which is false. So $x_{m+1}\in I_{m+1}\setminus I_m$ and $I_m⊊I_{m+1}$.

The chain $I_0\subseteq I_1\subseteq \cdots$ is an ascending chain of ideals of $R_{\mathrm{\infty }}$ indexed by $\mathbb{N}$ in which every inclusion is strict, so no index $N$ has $I_n=I_N$ for all $n\ge N$: already $I_{N+1}\mathrm{\ne }{I}_N$. The ascending chain condition therefore fails, and $R_{\mathrm{\infty }}$ is not Noetherian.

$A$ is a subring of $k[x,y]$. It contains $1=1+x\cdot 0$; it is closed under addition and additive inverses because $k$ and $xk[x,y]$ both are; and $(c+xf)(c^{\mathrm{\prime }}+x{f}^{\mathrm{\prime }})=c{c}^{\mathrm{\prime }}+x(c{f}^{\mathrm{\prime }}+c^{\mathrm{\prime }}f+xf{f}^{\mathrm{\prime }})$ lies in $A$.

$A=k[x,xy,x{y}^2,\dots ]$, the smallest subring of $k[x,y]$ containing $k$ and every $x{y}^i$. That subring is contained in $A$ because $A$ is a subring containing $k$ and each $x{y}^i=0+x\cdot y^i$. Conversely every $c+xf$ lies in it: expanding $f$ as a $k$-linear combination of monomials $x^a{y}^b$ gives $xf$ as a $k$-linear combination of $x^{a+1}{y}^b=x^a\cdot (x{y}^b)$, and each of $x=x{y}^0$ and $x{y}^b$ is one of the listed generators.

The ideal $\mathfrak{a}$ of $A$ generated by $\{x{y}^i:i\in \mathbb{N}\}$ equals $xk[x,y]$. An element of $\mathfrak{a}$ is a finite sum ${\sum }_j{a}_{j}x{y}^{n_j}$ with $a_j\in A$, which lies in $xk[x,y]$; conversely, for $f\in k[x,y]$ expanded as a $k$-linear combination of monomials $x^a{y}^b$, the element $xf$ is the corresponding combination of $x^a\cdot (x{y}^b)$ with $x^a\in A$, hence lies in $\mathfrak{a}$.

Suppose $\mathfrak{a}=(f_1,\dots ,f_m)$ as an ideal of $A$, with $m\in \mathbb{N}$ and $f_j\in \mathfrak{a}$. Since $x=x{y}^0\in \mathfrak{a}$ and $x\mathrm{\ne }0$, the ideal $\mathfrak{a}$ is nonzero, so at least one $f_j$ is nonzero; let $N\in \mathbb{N}$ be an index beyond which every $f_j$ has vanishing $y$-coefficients, that is, the $y^b$-coefficient of every $f_j$ is $0$ for $b>N$. Any element of $(f_1,\dots ,f_m)$ is ${\sum }_j{a}_j{f}_j$ with $a_j\in A$; writing $a_j=c_j+x{h}_j$ with $c_j\in k$ and $h_j\in k[x,y]$, and using $f_j\in xk[x,y]$ from step 2.1, this element equals ${\sum }_j{c}_j{f}_j+x^{2}g$ for some $g\in k[x,y]$. Its $y^{N+1}$-coefficient is therefore the sum of ${\sum }_j{c}_j\cdot 0=0$ and an element of $x^{2}k[x]$, hence lies in $x^{2}k[x]$.

But $x{y}^{N+1}$ lies in $\mathfrak{a}$ and its $y^{N+1}$-coefficient is $x$, which is not in $x^{2}k[x]$: every element of $x^{2}k[x]$ has vanishing coefficient at $x^1$, whereas $x$ has coefficient $1\mathrm{\ne }0$ there. So $x{y}^{N+1}\mathrm{\notin }(f_1,\dots ,f_m)$, contradicting $\mathfrak{a}=(f_1,\dots ,f_m)$. Hence $\mathfrak{a}$ is not finitely generated and $A$ is not Noetherian.

The ambient ring $k[x,y]$ is Noetherian, since $k$ is a field and a polynomial ring in finitely many variables over a Noetherian ring is Noetherian. So the Noetherian ring $k[x,y]$ has the non-Noetherian subring $A$. And $A$ is not of finite type over $k$: an algebra of finite type over the Noetherian ring $k$ would be Noetherian, which $A$ is not, while $k[x,y]$ itself is of finite type over $k$.

A field is an integral domain: it is a commutative ring with $1\mathrm{\ne }0$, and if $ab=0$ with $a\mathrm{\ne }0$ then $b=a^{-1}ab=0$. Hence $k[x,y]$ is an integral domain, its field of fractions $k(x,y)$ is a field, and $k[x,y]$ embeds in it.

Let $R$ be the intersection of all subrings of $k(x,y)$ containing $k$, $x$, $y$ and every $x\mathrm{/}{y}^i$ with $i\ge 1$; this family is nonempty because $k(x,y)$ itself belongs to it, and an intersection of subrings is a subring, so $R$ is the smallest such subring.

$R$ is the set of $k$-linear combinations of the elements $x^c{y}^d$ with $c\ge 1$ and $d\in \mathbb{Z}$, together with the elements $y^d$ with $d\ge 0$. Call that set $S$. It is a subring: it contains $1=y^0$, is visibly closed under addition and additive inverses, and is closed under multiplication because $(x^c{y}^d)(x^{c^{\mathrm{\prime }}}{y}^{d^{\mathrm{\prime }}})=x^{c+c^{\mathrm{\prime }}}{y}^{d+d^{\mathrm{\prime }}}$, where $c+c^{\mathrm{\prime }}\ge 1$ as soon as one of $c,c^{\mathrm{\prime }}$ is, and where $d+d^{\mathrm{\prime }}\ge 0$ when $c=c^{\mathrm{\prime }}=0$. It contains the listed generators, since $x=x^1{y}^0$, $y=y^1$ and $x\mathrm{/}{y}^i=x^1{y}^{-i}$. Conversely every element of $S$ lies in any subring containing the generators, because $x^c{y}^d$ with $c\ge 1$ is $x^{c-1}$ times $x{y}^d$, and $x{y}^d$ is $x\cdot y^d$ when $d\ge 0$ and is the generator $x\mathrm{/}{y}^{-d}$ when $d<0$. So $S=R$.

The elements $x^c{y}^d$ of $k(x,y)$, for $c\ge 0$ and $d\in \mathbb{Z}$, are $k$-linearly independent, and consequently $1\mathrm{/}y\mathrm{\notin }R$. For the independence, a finite relation $\sum {\lambda }_{c,d}{x}^c{y}^d=0$ becomes, after multiplication by $y^M$ for $M$ large enough that every $d+M$ occurring is at least $0$, a relation $\sum {\lambda }_{c,d}{x}^c{y}^{d+M}=0$ among distinct monomials of $k[x,y]$; a polynomial is a finitely supported family of coefficients, so it vanishes exactly when every coefficient does, and the embedding of $k[x,y]$ is injective, so all ${\lambda }_{c,d}=0$. Now $1\mathrm{/}y=x^0{y}^{-1}$, whose index pair has $c=0$ and $d=-1<0$; by step 3.1 and the independence just proved, $1\mathrm{/}y$ is not among the $k$-linear combinations making up $R$.

For every $n\in \mathbb{N}$, $(x\mathrm{/}{y}^n)\subseteq (x\mathrm{/}{y}^{n+1})$ and the inclusion is strict. The inclusion holds because $y\in R$ and $x\mathrm{/}{y}^n=y\cdot (x\mathrm{/}{y}^{n+1})$, so the generator of the left ideal lies in the right one. If the two were equal then $x\mathrm{/}{y}^{n+1}=r\cdot x\mathrm{/}{y}^n$ for some $r\in R$; multiplying by $y^n\mathrm{/}x$ in the field $k(x,y)$, which is legitimate because $x\mathrm{\ne }0$ and $y\mathrm{\ne }0$ there, gives $r=1\mathrm{/}y$, contradicting step 4.1. At $n=0$ this reads $(x)⊊(x\mathrm{/}y)$.

The chain $(x)\subseteq (x\mathrm{/}y)\subseteq (x\mathrm{/}{y}^2)\subseteq \cdots$ is an ascending chain of ideals of $R$ indexed by $\mathbb{N}$ with every inclusion strict, so for no index $N$ is it constant from $N$ onwards. The ascending chain condition fails and $R$ is not Noetherian.

Every element of $\mathfrak{a}$ is $a\cdot 2x+b\cdot 3x^2+c\cdot x^3$ with $a,b,c\in \mathbb{Z}[x]$. Its constant coefficient is $0$, since each of $2x$, $3x^2$, $x^3$ has support in the indices $\ge 1$; and its coefficient at $x^1$ is $2a_0$, since the second and third products have support in the indices $\ge 2$. So $\mathfrak{a}$ contains no nonzero constant, giving ${\mathfrak{a}}_0=0$, and every element of ${\mathfrak{a}}_1$ lies in $(2)$.

The element $2x$ lies in $\mathfrak{a}$, is nonzero of degree $1$ and has leading coefficient $2$, so $2\in {\mathfrak{a}}_1$; with step 1.1 this gives ${\mathfrak{a}}_1=(2)$. The element $x^2=3x^2-x\cdot (2x)$ lies in $\mathfrak{a}$, is nonzero of degree $2$ and has leading coefficient $1$, so $1\in {\mathfrak{a}}_2$ and ${\mathfrak{a}}_2=\mathbb{Z}$.

The stages ascend, so $\mathbb{Z}={\mathfrak{a}}_2\subseteq {\mathfrak{a}}_n\subseteq \mathbb{Z}$ and ${\mathfrak{a}}_n=\mathbb{Z}$ for every $n\ge 2$. The chain reads $0⊊(2)⊊\mathbb{Z}=\mathbb{Z}=\cdots$ and stabilises at $N=2$; it does not stabilise at $N=1$, since ${\mathfrak{a}}_1=(2)\mathrm{\ne }\mathbb{Z}={\mathfrak{a}}_2$.

Run the finite-generation lemma with $N=2$. Stage $0$ has ${\mathfrak{a}}_0=0$ and contributes no realiser; stage $1$ has the single generator $2$ of $(2)$, realised by $2x$; stage $2$ has the single generator $1$ of $\mathbb{Z}$, realised by $x^2$. So the lemma returns $\mathfrak{a}=(2x,x^2)$. This is checkable by hand: $2x$ and $x^2$ lie in $\mathfrak{a}$ by steps 1.1 and 2.1, while $3x^2=3\cdot x^2$ and $x^3=x\cdot x^2$ lie in $(2x,x^2)$.

Take $f=5x^4+2x$, which lies in $\mathfrak{a}$ because $5x^4=5x^2\cdot x^2$ and $2x$ do. It is nonzero of degree $d=4$ with $\mathrm{lc}(f)=5\in \mathbb{Z}={\mathfrak{a}}_2$, so the cancellation lemma applies at $n=\min (4,2)=2$ with the single realiser $x^2$ and coefficient $r_1=5$: it forms $h=5x^{4-2}\cdot x^2=5x^4$ and leaves $f-h=2x$, nonzero of degree $1<4$. Applying it again to $2x$ at $n=1$, with realiser $2x$ and coefficient $r_1=1$, forms $h=1\cdot x^0\cdot 2x=2x$ and leaves $0$. Adding the two corrections back recovers $f=5x^4+2x$ as an element of $(2x,x^2)$.