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True. This is the Artinian commutative ring is Noetherian theorem. One proof uses the nilpotent nilradical $N$ of an Artinian ring $A$. The quotient $A/N$ is a finite product of fields. Each quotient $N^j/N^{j+1}$ is an Artinian module over the semisimple ring $A/N$, hence has finite length and is Noetherian. The finite filtrationthen makes $A$ a Noetherian module over itself, which is exactly the ascending chain condition on its ideals.

False. The integers $\mathbb{Z}$ form a Noetherian ring, since every ideal is principal, but this ring is not an Artinian ring: the strictly descending chainnever stabilizes.

The claim fails for a general extension. If $k$ is an infinite field, all the infinitely many maximal ideals $(x-a)$ of $k[x]$ contract to $(0)$ in $k$.

It still fails for an integral extension. Take a finite field $k={\mathbb{F}}_q$ andEvery $b\in B$ satisfies the monic equation $b^q-b=0$, so $B$ is integral over the diagonal copy of $k$. The coordinate kernels are infinitely many distinct maximal ideals, all lying over $(0)$.

The claim is true for a module-finite ring extension. The primes above $\mathfrak{p}$ correspond to the primes of the fiber ringThis is a finite-dimensional algebra over the residue field $\kappa (\mathfrak{p})$, hence an Artinian ring, and an Artinian ring has only finitely many prime ideals.

Form the finite-dimensional $k$-algebraBecause a finite extension is integral, the Lying-over theorem supplies a prime of $B$ above $\ker f$; after localization and extension of the residue field to $k$, this shows $C\ne 0$. Therefore $C$ has at least one maximal ideal.

As an Artinian ring, $C$ has only finitely many maximal ideals. For each such ideal $\mathfrak{n}$, the quotient $C/\mathfrak{n}$ is a finite field extension of the algebraically closed field $k$, so it equals $k$. Consequently the quotient maps $C\to k$ are in bijection with the required extensions $g:B\to k$. The set of extensions is therefore finite and nonempty.

A commutative Artinian ring has finitely many prime ideals, and every one is a maximal ideal. If $\mathfrak{p}$ is prime, the Artinian domain $R/\mathfrak{p}$ is a field: stabilization of $(x)\supseteq (x^2)\supseteq \cdots$ gives $x^n=x^{n+1}a$, and cancellation gives $xa=1$. Infinitely many distinct maximal ideals ${\mathfrak{m}}_i$ would give a strictly descending chainbecause distinct maximal ideals are comaximal.

A nonzero central idempotent is primitive central if it is not a sum of two nonzero orthogonal central idempotents. For an Artinian ring, it determines a block of an Artinian algebra. This differs from a primitive idempotent in the whole ring: for a matrix algebra of size greater than one over a field, $1$ is primitive central but is a sum of diagonal idempotents.

For a $p$-subgroup $D$ of a finite group $G$, set $I_D={\mathrm{Tr}}_D^G((kG{)}^D)$, a transfer ideal of conjugation-fixed elements in $Z(kG)$. A block of a group algebra idempotent $b$ lies in $I_D$ if and only if a defect group of a block of $b$ is conjugate to a subgroup of $D$. In particular, defect groups can equivalently be defined as minimal subgroups $D$ with $b\in I_D$.

Here is a proof using the Brauer morphism. A subgroup of smallest order with $b\in I_D$ exists, since a Sylow subgroup $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_D^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_D(b)=0$, then $ba$ belongs to the kernel of the Brauer morphism, which is the sum of proper-subgroup relative traces. Hence $b\in {\sum }_{Q<D}{I}_Q$. The algebra $bZ(kG)$ is a commutative Artinian ring and a local ring, so multiplying this equality by $b$ forces some ideal $b{I}_Q$ to contain its identity $b$. This contradicts minimality. On the other hand, ${\mathrm{Br}}_E(I_D)=0$ unless $E$ is conjugate into $D$: the $E$-action on $G/D$ has no fixed coset otherwise, and nonfixed orbit sizes vanish in characteristic $p$. Thus these minimal subgroups are precisely the maximal subgroups with nonzero Brauer image. They are all conjugate. Trace transitivity proves the stated criterion for larger $D$.