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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

True. An irreducible ideal $I$ is a proper ideal such that $I=J\cap K$ forces $I=J$ or $I=K$. Passing to the quotient ring $A=R/I$ turns this into the assertion that zero cannot be the intersection of two nonzero ideals. Also $A$ is Noetherian: ascending chains of ideals lift to ascending chains of ideals containing $I$ in $R$.

Suppose $ab=0$ in $A$, and suppose $b$ is not a nilpotent element. The ascending chain condition makes the chain of annihilatorsstabilize. Choose $n\ge 1$ with $\mathrm{Ann}(b^n)=\mathrm{Ann}(b^{n+1})$. We claimIf $z=ua=v{b}^n$ lies in this intersection, then $bz=uab=0$, so $v{b}^{n+1}=0$. The equality of annihilators implies $v{b}^n=0$, hence $z=0$. Since $b$ is not nilpotent, $(b^n)\ne 0$. Irreducibility of zero therefore forces $(a)=0$, or $a=0$.

We have proved that $ab=0$ and $a\ne 0$ force $b$ to be a nilpotent element. This says precisely that zero is a primary ideal in $A$, and therefore that $I$ is a primary ideal in $R$.

If its radical of an ideal is to be denoted by a prime ideal, this requires no extra hypothesis: for any primary ideal $I$, $uv\in \sqrt{I}$ and $u\notin \sqrt{I}$ imply $u^n{v}^n\in I$ with $u^n\notin I$, so some power of $v$ lies in $I$. Thus $v\in \sqrt{I}$.

False: existence holds, but uniqueness fails. First, the Lasker–Noether theorem supplies existence; here is an argument from the preceding part. If some proper ideal in a Noetherian ring were not a finite intersection of irreducible ideals, the ascending chain condition would let us choose such an ideal $I$ maximal under inclusion. It is not irreducible, so $I=J\cap K$ with both $J$ and $K$ strictly larger than $I$. By maximality, both are finite intersections of irreducible ideals, and therefore so is $I$, a contradiction. Part (d) makes each factor a primary ideal. The whole ring is represented by the empty intersection if that convention is needed.

For a counterexample to uniqueness, work in $R=k[x,y]$ over a field $k$ and put $I=(x^2,xy)$. The Hilbert basis theorem makes $R$ a Noetherian ring. There are two distinct minimal primary decompositions:The ideal $(x)$ is a prime ideal, hence a primary ideal. Each of the other factors has quotient ring isomorphic to $k[x]/(x^2)$, by substituting respectively $y=0$ and $y=x$. As in part (b), each is $(x,y)$-primary.

To check both intersections, let $Q_c=(x^2,y-cx)$, for $c=0,1$. Substitution identifies $R/Q_c$ with $k[x]/(x^2)$. The class of $xg(x,y)$ is zero exactly when $g(0,0)=0$, equivalently when $g\in (x,y)$. Therefore $(Q_c:x)=(x,y)$ andThe two $Q_c$ are different, since $y$ belongs to $Q_0$ but has nonzero class $x$ modulo $Q_1$. Both decompositions are irredundant: $x\notin Q_c$, while $y-cx\notin (x)$. Applying the radical of an ideal operation to the two components in either decomposition gives $(x)$ and $(x,y)$, respectively.

The second uniqueness theorem for primary decomposition explains the distinction: the component at an isolated prime of a primary decomposition is unique, but embedded primary components need not be. The counterexample varies an embedded component.

For a point $p\in {\mathbb{A}}_k^1$, let $a$ be the image of $t$ in its residue field $\kappa (p)$. The scheme-theoretic fiber isIf $p\ne (t)$, then $a\ne 0$. The quadratic polynomial $xy-a{z}^2$ is irreducible. Indeed, after setting $z=0$, any hypothetical linear factors must restrict, up to nonzero scalars, to $x$ and $y$; comparing the $xz$ and $yz$ coefficients then forces both $z$ coefficients to vanish, contradicting the nonzero $z^2$ coefficient. Hence its homogeneous coordinate ring is an integral domain, so $Y_p$ is an integral scheme.

At the origin $p=(t)$, the fiber is $V_{+}(xy)$, the union of the two distinct projective lines $x=0$ and $y=0$, and is therefore not irreducible. It is nevertheless a reduced scheme because the ideal $(xy)=(x)\cap (y)$ equals its radical. Every other fiber is integral and hence reduced. Thus the fiber is integral exactly away from the origin, and it is reduced at every point of ${\mathbb{A}}_k^1$.

For an ideal $I\subseteq k[T_1,\dots ,T_n]$, defineFor $X\subseteq k^n$, defineand recall that $\sqrt{I}=\{f:f^r\in I\text{ for some }r\ge 1\}$ is the radical of an ideal.

For an algebraically closed field $k$, the Weak Hilbert Nullstellensatz says that every maximal ideal of $k[T_1,\dots ,T_n]$ isfor a unique $a\in k^n$, equivalently every proper ideal has a common zero. The Strong Hilbert Nullstellensatz says

To prove the weak form, let $\mathfrak{m}$ be maximal. The residue fieldis a field finitely generated as a $k$-algebra. By the Zariski lemma, $K/k$ is finite algebraic; algebraic closedness gives $K=k$. If $a_i$ is the image of $T_i$, the quotient map is evaluation at $a=(a_1,\dots ,a_n)$ and its kernel is $(T_1-a_1,\dots ,T_n-a_n)$. This proves the assertion.

Use the quotient filtration induced from the order filtration of a Weyl algebra. The key step is to identify the entire associated graded module of the left ideal $I$, rather than merely take symbols of its displayed generators.

For every member $x^{\alpha }{y}^{\beta }$ of the ordered monomial basis of a Weyl algebra, right multiplication by the pure $y$-monomials givesNo commuting past an $x_i$ is necessary. It follows that $I$ is exactly the complex linear span of the basis monomials satisfying eitherIntersecting this linear span with each filtration piece and then taking symbols therefore yields the monomial idealBy b,

To determine its Krull dimension, compute the radical of an ideal:Indeed, $K$ lies in the displayed intersection, an intersection of prime ideals. Conversely, $({\xi }_1{\xi }_2{)}^3\in K$ and $({\xi }_1{\xi }_3{)}^3\in K$, so both generators of that intersection lie in $\sqrt{K}$. The equality of the intersection with $({\xi }_1{\xi }_2,{\xi }_1{\xi }_3)$ follows by factoring out ${\xi }_1$ and reducing modulo $({\xi }_2,{\xi }_3)$.

The order characteristic variety of a Weyl algebra module in this case has two irreducible components:They are coordinate affine spaces of dimensions $2n-1$ and $2n-2$, respectively, in the $2n$ coordinates $x_1,\dots ,x_n,{\xi }_1,\dots ,{\xi }_n$. The Krull dimension is the larger of these component dimensions. ThusIn particular, two displayed generators do not force a drop of two in dimension: their common factor creates the component ${\xi }_1=0$ of dimension $2n-1$.

For an ideal $I$ in a commutative ring $R$, put $I^0=R$ and let $I^n$ for $n\ge 1$ be the product of ideals with $n$ factors equal to $I$. It consists of finite sums of products of $n$ elements of $I$. These ideals form a descending chain $R\supseteq I\supseteq I^2\supseteq \cdots$.

Every positive power has the same radical of an ideal as $I$:One inclusion follows from $I^n\subseteq I$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, giving the other inclusion. Different ideals can therefore define the same radical ideal, even when their successive powers are all distinct.

In a commutative ring, the radical of an ideal is monotone and idempotent: inclusion $I\subseteq J$ gives $\sqrt{I}\subseteq \sqrt{J}$, and $a^n\in \sqrt{I}$ means $a^{nm}\in I$ for some $m$, giving $\sqrt{\sqrt{I}}=\sqrt{I}$. Since $\sqrt{I+J}$ contains both $\sqrt{I}$ and $\sqrt{J}$, it contains their sum of ideals. Taking radicals of ideals gives $\sqrt{\sqrt{I}+\sqrt{J}}\subseteq \sqrt{I+J}$. The reverse inclusion follows from $I+J\subseteq \sqrt{I}+\sqrt{J}$.

In particular, $\sqrt{I}+\sqrt{J}=R$ implies $I+J=R$. One can also see this directly: write $1=a+b$ with $a^m\in I$ and $b^n\in J$, then expand $(a+b{)}^{m+n-1}$ by the binomial theorem. Each term belongs to $I$ or $J$, so $1\in I+J$.

Let $R$ be a Noetherian ring that is an integral domain, and let $I$ be a nonzero proper ideal. If $I^n=I^{n+1}$ for some $n\ge 1$, the nonzero finitely generated module $M=I^n$ satisfies $M=IM$. Choose generators $u_1,\dots ,u_r$ and write $u_i={\sum }_j{a}_{ij}{u}_j$ with $a_{ij}\in I$. The adjugate matrix identity implies that $\det ({\mathrm{Id}}_r-(a_{ij}))$ annihilates $M$. This determinant is $1-b$ for some $b\in I$, since every term except the constant term contains an entry from $I$.

The determinant trick has thus supplied $(1-b)M=0$. A nonzero element of $M$ and multiplicative cancellation in an integral domain force $b=1$, contradicting properness of $I$. Therefore all positive powers of an ideal are distinct in this setting. Since they have the same radical of an ideal, there are infinitely many ideals with radical $\sqrt{I}$.

In $k[u,v]$, the ideal $I=(u^2,uv)$ has radical of an ideal $(u)$, a prime ideal, and hence a unique minimal prime ideal. Nevertheless $uv\in I$, $u\notin I$, and $v^j\notin I$ for all $j\ge 1$, so $I$ is not a primary ideal. The same example works after adjoining inverses of $1+u$ and $1+v$; thus it also gives an example in a two-variable Laurent polynomial ring.