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Every commutative Artinian ring is Noetherian. Its nilradical $N$ is nilpotent by the nilradical of a commutative Artinian ring, and the Chinese remainder theorem identifies the reduced quotient $R/N$ with the finite product of the residue fields at the finitely many maximal ideals. Each $N^i/N^{i+1}$ is an Artinian module over that product of fields and hence a finite-dimensional vector space. The finite filtration by these layers makes every ideal finitely generated.

Every finite-dimensional vector space subspace of a normed vector space is closed. Coordinates in a finite basis identify it homeomorphically with a finite-dimensional scalar space; it is therefore complete, and every complete subspace of a metric space is closed.

On a finite-dimensional vector space equipped with a norm, the weak topology and norm topology coincide. The weak topology is no finer because every element of the dual is norm-continuous. Conversely, finitely many coordinate functionals in a basis control the norm, so every sufficiently small basic weak neighbourhood lies in a prescribed norm ball.

Two norms $N_1,N_2$ are equivalent norms when constants $c,C>0$ satisfyfor every $v$.

For $x,y\in {\mathbb{R}}^n$, the reverse triangle inequality givesby the Cauchy-Schwarz inequality. Hence $g$ is Lipschitz continuous, and therefore continuous.

On the compact Euclidean unit sphere, $g$ is positive and continuous, so the extreme value theorem gives $0<m\le g\le M<\infty$. Homogeneity then yieldsThus every norm on a finite-dimensional vector space is equivalent to the Euclidean norm; comparing two such bounds proves any two norms on $V$ are Lipschitz equivalent.

Let $P_n$ be the finite-dimensional vector space of real polynomials of degree at most $n$. For fixed $a>0$,is a norm: if it vanishes, the polynomial vanishes on an interval and hence is the zero polynomial. Alsois a norm on $P_n$. By finite-dimensional equivalence of norms, there is $C>0$, depending only on $n$ and $a$, such that $M(p)\le CN(p)$. ThereforeThe extreme value theorem supplies $y\in [0,a]$ at which the last supremum is attained, and hence

Choose a basis of the finite-dimensional vector space and let $|x{|}_{\infty }$ be the maximum absolute coordinate. For any norm $N$, the triangle inequality givesso $N$ is continuous in the coordinate topology. On the compact set $\{x:|x{|}_{\infty }=1\}$, the positive continuous function $N$ attains a positive minimum $m$ and a finite maximum $M$. By absolute homogeneity,This proves the equivalence of norms in finite dimensions: every norm is equivalent to the maximum norm, and transitivity of these inequalities proves that any two norms on a finite-dimensional vector space are Lipschitz equivalent.