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True, without a Noetherian hypothesis. By definition, the symbolic power is the contraction of an idealwhere the intersection notation means inverse image under $R\to R_P$, even if this map is not injective. In the local ring $A=R_P$, put $\mathfrak{m}=P{R}_P$. Then $P^n{R}_P={\mathfrak{m}}^n$, and $\sqrt{{\mathfrak{m}}^n}=\mathfrak{m}$: every element of $\mathfrak{m}$ has its $n$th power in ${\mathfrak{m}}^n$, whereas a unit cannot have a power in this proper ideal.

The ideal ${\mathfrak{m}}^n$ is $\mathfrak{m}$-primary. Indeed, if $uv\in {\mathfrak{m}}^n$ and $v\notin \mathfrak{m}$, then $v$ is a unit, so $u\in {\mathfrak{m}}^n$. Its contraction of an ideal is therefore $P$-primary: its radical of an ideal contracts to $P$, and the same implication applies to the images of any two elements of $R$. Thus $P^{(n)}$ is always $P$-primary, proving the reverse implication immediately when $P^n=P^{(n)}$.

For the other implication, suppose $P^n$ is $P$-primary. Membership in the contraction can be expressed by clearing denominators:For completeness, if $r/1=a/t$ with $a\in P^n$ and $t\notin P$, equality of fractions gives $u(tr-a)=0$ for some $u\notin P$, so $(ut)r=ua\in P^n$; the converse follows by inverting $s$. Since $s\notin \sqrt{P^n}=P$, the primary ideal property forces $r\in P^n$. The inclusion $P^n\subseteq P^{(n)}$ always holds, giving

For a discrete valuation ring $A$,Let $\pi$ be a uniformizer, so its unique nonzero maximal ideal is $\mathfrak{m}=(\pi )$. We use the defining description of a discrete valuation ring: it is a non-field local ring that is a principal ideal domain, and every nonzero element is $u{\pi }^n$ for a unit $u$ and an integer $n\ge 0$.

Every nonzero proper ideal is therefore $({\pi }^n)$ for some $n\ge 1$. Such an ideal is a prime ideal only when $n=1$: for $n\ge 2$, $\pi {\pi }^{n-1}\in ({\pi }^n)$ while neither factor belongs to $({\pi }^n)$. The ideal $(\pi )$ is prime because it is maximal. Also $(0)$ is prime because $A$ is an integral domain. Thusand the longest strict chain has length one.

The original PDF asks about a discrete valuation ring; the local TeX transcription's phrase “discrete variable $X$” is an extraction error.

Put $a=(123)$ and $t=(12)$. Conjugation by $a$ fixes the three given idempotents of $k⟨a⟩$, while $ta{t}^{-1}=a^{-1}$ fixes $e_1$ and interchanges $e_2,e_3$. Their orbits are therefore $\boxed{\{e_1\},\{e_2,e_3\}}$, with central orbit sumsusing $\omega +{\omega }^2=1$ in characteristic $2$.

To ensure these sums really are primitive central idempotents, examine the two ideals. The first has basis $b_0,b_{0}t$, with $b_{0}a=b_0$, so $b_{0}kG\cong k{C}_2$, a local ring. The second has dimension $4$: $b_{1}k⟨a⟩=k{e}_2\oplus k{e}_3$ has dimension $2$, and the two cosets of $⟨a⟩$ double it. In the two-dimensional representationThese matrices satisfy $a^3=t^2=1$ and $ta{t}^{-1}=a^{-1}$, so they define a representation of $S_3$. In it, $b_1$ acts as identity and $b_0$ as zero. The distinct diagonal entries supply the two diagonal matrix units, and multiplying by the swap matrix supplies the off-diagonal units. The induced map $b_{1}kG\to M_2(k)$ is thus surjective and, by dimension, an isomorphism. Both summands have no nontrivial central idempotents, so $b_0,b_1$ are exactly the 2-modular blocks of S3.

The Brauer morphism at the trivial subgroup is the identity. For $H=⟨t⟩$, $C_G(H)=H$, and neither $a$ nor $a^2$ lies in this centralizer. ThusIndeed $H$ is Sylow, and all nontrivial $2$-subgroups are conjugate to it; these images therefore determine maximality in the definition of a defect group of a block. The original PDF provides the $S_3,{\mathbb{F}}_4,N,H$ setup missing from the TeX transcription.

Write the affine scheme as $U=\mathrm{Spec}B$. A point $x\in U$ corresponds to a prime ideal $\mathfrak{p}\subset B$, and its local ring is ${\mathcal{O}}_{X,x}\cong B_{\mathfrak{p}}$ with maximal ideal $\mathfrak{p}{B}_{\mathfrak{p}}$. HenceThe last condition defines the principal open subscheme $D(\bar{f})$, soEvery point of the scheme $X$ has such an affine neighbourhood. Thus $X_f$ is locally open, and hence is open in the Zariski topology.

The Jacobson radical $J=J(R)$ is the intersection of all maximal right ideals, equivalently the largest ideal annihilating every simple right module. The Artin–Wedderburn theorem givesbecause $k$ is algebraically closed.

The descending chain $J\supseteq J^2\supseteq \cdots$ stabilizes since $R$ is finite-dimensional. If $J^m=J^{m+1}$, Nakayama lemma applied to the finite right module $J^m$ gives $J^m=0$. Thus $J$ is nilpotent.

For $f\in {\mathrm{End}}_R(P_i)$, the Fitting lemma givesfor large $n$. Indecomposability makes one summand zero, so $f$ is either invertible or nilpotent. In the latter case $1-f$ is invertible. This is the criterion that ${\mathrm{End}}_R(P_i)$ is a local ring.

Let $M={⨁}_i{P}_i$. Reduction modulo $MJ$ definesSince $M/MJ={⨁}_i{S}_i$ and the $S_i$ are pairwise nonisomorphic simples,by Schur lemma. Arbitrary scalars on the direct summands lift to scalar identity maps on the $P_i$, so $\theta$ is surjective.

If $f\in \ker \theta$, then $f(M)\subseteq MJ$. A product of $r$ such maps sends $M$ into $M{J}^r$, so $\ker \theta$ is nilpotent. A nilpotent ideal lies in the Jacobson radical, while the semisimplicity of the quotient $k^n$ gives the reverse inclusion. HenceThis is exactly the definition of a basic algebra.

For every $t\in T_N$, the torus action mapis an automorphism of $X_{\Sigma }$. It induces isomorphisms of local rings, so it carries smooth points to smooth points and singular points to singular points. The singular locus is consequently invariant under $T_N$. If it contains one point $x$, it contains the entire orbit $T_N\cdot x$; hence the torus-invariance of the singular locus of a toric variety proves that $X_{\Sigma }^{\mathrm{sing}}$ is a union of the orbits in the orbit-cone correspondence.

Assume tensor products of two nonzero modules never vanish. If $\mathfrak{m}\ne \mathfrak{n}$ were distinct maximal ideals, then $\mathfrak{m}+\mathfrak{n}=R$ andalthough both residue fields are nonzero. Hence $R$ has one maximal ideal $\mathfrak{m}$ and is a local ring.

For every module $M$,If $\mathfrak{m}M=M$, this tensor product vanishes. Since $R/\mathfrak{m}\ne 0$, the assumed property forces $M=0$. Thus condition (a) implies condition (b).

For an irreducible variety which is an affine variety of dimension $d$ and a nonzero nonunit regular function $f$, each irreducible component of $V(f)$ has dimension $d-1$. Choose a closed point on just the component in question. In its local ring $R$, $\sqrt{(f)}$ is that component's prime ideal. A system of parameters of $R/(f)$, lifted and supplemented by $f$, generates a maximal-primary ideal; the Krull height theorem gives $\dim R/(f)\ge d-1$. Extending any chain above $(f)$ by the zero prime of the domain gives the reverse inequality. This proof does not identify dimension with transcendence degree.

On a normal integral Noetherian scheme, a nonzero rational function defines the displayed Weil divisor, with the sum over codimension-one prime divisors. The local ring at each such divisor is a discrete valuation ring, so the order is well defined. The divisor class group is the group of Weil divisors modulo principal Weil divisors. On a smooth curve this agrees with a principal divisor on an algebraic curve.

For a local ring $A$, every map from $\mathrm{Spec}A$ into projective space lands in one standard chart: the inverse image of a chart containing the closed point is the whole spectrum. This gives homogeneous coordinates with at least one unit entry, unique up to a common unit. For a nonlocal ring, coordinates may instead merely generate the unit ideal, and general maps can involve a nontrivial invertible sheaf quotient.

A finite locally free sheaf of ideals on a reduced scheme has rank at most one at every point. On a nonempty affine open subscheme where its rank is $r$, localize its inclusion into the structure sheaf at a minimal prime ideal. The resulting local ring is a field $K$, giving an injection $K^r\hookrightarrow K$, hence $r\le 1$. This works without a Noetherian assumption. The empty scheme is a vacuous exception to claims phrased as nonexistence of a sheaf of a prescribed rank.

A scheme is reduced when every local ring ${\mathcal{O}}_{X,x}$ is a reduced ring, equivalently when its structure sheaf has no nonzero nilpotent element. The affine scheme $\mathrm{Spec}A$ is reduced exactly when $A$ is reduced.

The closed subset of points where an algebraic variety is not smooth. Over a perfect field it is equivalently the locus whose local rings are not regular local rings. A normal variety has no singular points in codimension zero or one.

A smooth algebraic curve is a one-dimensional algebraic variety whose local ring at every point is regular.

Over a perfect field, a point of an algebraic variety is smooth exactly when its local ring is a regular local ring. The smooth locus of a variety is open; its closed complement is the singular locus.

Over a perfect field, a point of a variety is smooth when its local ring is a regular local ring. For an irreducible variety of dimension $n$ at a closed point, this is equivalent to its Zariski tangent space having dimension $n$. The Jacobian criterion expresses the equality as a matrix-rank condition.

The structure sheaf ${\mathcal{O}}_X$ records the regular functions on every open subset of a scheme $X$. Its stalk ${\mathcal{O}}_{X,x}$ is the local ring of functions defined near $x$.

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 $D$.