Fields and $\mathbb{Z}$ are Noetherian, and so are their polynomial rings in finitely many variables

Given: A field $K$, the ring $\mathbb{Z}$ of integers, and $n\in \mathbb{N}$.

A field is a set $F$ with two operations and distinguished elements $0\mathrm{\ne }1$ such that $(F,+)$ is an abelian group, multiplication is associative and commutative on all of $F$ with $x\cdot 1=x$, every $x\mathrm{\ne }0$ has a multiplicative inverse, and multiplication distributes over addition (Field).

Every subgroup $H\le (\mathbb{Z},+)$ equals $n\mathbb{Z}=⟨n⟩$ for exactly one natural number $n$; in particular every subgroup of $(\mathbb{Z},+)$ is cyclic (Every subgroup of $(\mathbb{Z},+)$ is $⟨n⟩=n\mathbb{Z}$ for exactly one natural number $n$).

For $S\subseteq R$, $(S)$ is the intersection of all two-sided ideals containing $S$; $(\{a\})$ is written $(a)$ and is called principal (The ideal generated by a subset and principal ideals).

An ideal of a ring is in particular an additive subgroup of it (Left, right and two-sided ideals).

In a commutative ring, $(S)$ consists of finite sums $\sum r_i{s}_i$, and $(a)=Ra$ (In a commutative ring, $(S)$ consists of finite sums $\sum r_i{s}_i$, and $(a)=Ra$).

For a commutative ring, being Noetherian is equivalent to every ideal being finitely generated (A commutative ring is Noetherian exactly when every ideal is finitely generated, exactly when its ideals satisfy the ascending chain condition, and exactly when every nonempty set of ideals has a maximal member).

If $R$ is a Noetherian commutative ring then $R[x_1,\dots ,x_n]$ is Noetherian for every $n\in \mathbb{N}$ (If $R$ is Noetherian then $R[x_1,\dots ,x_n]$ is Noetherian for every $n\in \mathbb{N}$).