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The induced map on the spectrum of a commutative ring is contraction of an ideal:To see that it is well defined, let $\mathfrak{q}$ be a prime ideal of $S$. Its inverse image is an ideal of $R$, and it is proper because $\phi (1)=1\notin \mathfrak{q}$. If $ab\in {\phi }^{-1}(\mathfrak{q})$, then $\phi (a)\phi (b)\in \mathfrak{q}$, whence $\phi (a)\in \mathfrak{q}$ or $\phi (b)\in \mathfrak{q}$. This proves that the inverse image is a prime ideal.

For any ideal $I\subseteq R$, its extension of an ideal to $S$ is $IS=\phi (I)S$. We haveThus inverse images of Zariski-closed sets are Zariski-closed sets, which proves that $f$ is a continuous map.

Here the coefficient fields are $\mathbb{R}$ and $\mathbb{C}$. A polynomial ring in one variable over a field is a Euclidean domain, hence a principal ideal domain; its nonzero prime ideals are generated by irreducible polynomials. The fundamental theorem of algebra says that all irreducible polynomials over $\mathbb{C}$ are linear. Combining it with complex conjugation shows that the monic irreducible polynomials over $\mathbb{R}$ are exactly the linear polynomials and quadratics with no real root. ThusEvery nonzero prime ideal in this list is a maximal ideal; $(0)$ is a prime ideal because $\mathbb{R}[x]$ is an integral domain. Likewise,The map is contraction of an ideal along the inclusion. It sends $(0)$ to $(0)$. For $z=a+ib$, the contraction of $(x-z)$ consists of the real polynomials vanishing at $z$. If $b=0$, the factor theorem gives the ideal $(x-a)$. If $b\ne 0$, a real polynomial vanishing at $z$ also vanishes at its complex conjugate $\overline{z}$, so it is divisible by $(x-z)(x-\overline{z})=(x-a{)}^2+b^2$. Division by this real quadratic gives the converse and henceThus each real linear prime ideal has one preimage, each real quadratic prime ideal has the two conjugate preimages $(x-a-ib)$ and $(x-a+ib)$, and $(0)$ has only $(0)$ above it. In particular, $f$ is surjective.

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

We use three theorems for an integral extension $S\subseteq R$, none requiring Noetherian hypotheses. The Lying-over theorem says that each prime ideal $\mathfrak{p}$ of $S$ is $\mathfrak{q}\cap S$ for some prime ideal $\mathfrak{q}$ of $R$. The Going-up theorem says that if $\mathfrak{p}\subseteq {\mathfrak{p}}^{\prime }$ in $S$ and $\mathfrak{q}\cap S=\mathfrak{p}$, there is ${\mathfrak{q}}^{\prime }\supseteq \mathfrak{q}$ with ${\mathfrak{q}}^{\prime }\cap S={\mathfrak{p}}^{\prime }$. The incomparability theorem for integral extensions says that comparable prime ideals of $R$ with the same contraction to $S$ are equal.

Take any strict chainin $R$. The contraction of an ideal operation gives ${\mathfrak{p}}_j={\mathfrak{q}}_j\cap S$, which are prime ideals of $S$, by the proof in Question 1(b), and remain strictly increasing by the incomparability theorem for integral extensions. Thus every chain length in $R$ occurs in $S$, giving $\dim R\le \dim S$.

Conversely, take any strict chain ${\mathfrak{p}}_0⊊\cdots ⊊{\mathfrak{p}}_n$ in $S$. Use the Lying-over theorem to choose ${\mathfrak{q}}_0$ over ${\mathfrak{p}}_0$, then repeatedly use the Going-up theorem to obtain ${\mathfrak{q}}_0\subseteq \cdots \subseteq {\mathfrak{q}}_n$ with the prescribed contractions. Each inclusion must be strict, since its contractions are distinct. Hence every chain length in $S$ occurs in $R$, giving $\dim S\le \dim R$.

Taking suprema proves integral extensions preserve Krull dimension:The argument works equally well when the dimensions are infinite, since it compares all finite chain lengths.

The $n$th symbolic power of a prime ideal $P\subset R$ iswhere the intersection denotes contraction of an ideal. It is always a $P$-primary ideal. The ordinary power $P^n$ equals $P^{(n)}$ precisely when $P^n$ is $P$-primary; this assertion does not require $R$ to be Noetherian.