Extension and contraction

Leg $f:A\to B$ be a ring homomorphism. If ${\frak a}$ is an ideal, then $f({\frak a})$ is not necessarily an ideal, since $f$ may maps to only part of the elements of $B$, neglecting some elements to make it an ideal. For example, if $f$ maps $\mathbb{Z}$ to $\mathbb{Q}$, any image of any ideal in $\mathbb{Z}$ is not an ideal in $\mathbb{Q}$. However, we can always construct an ideal generated by $f({\frak a})$, it is called the extension of ideal ${\frak a}$, denoted by ${\frak a}^e$,

\[{\frak a}^e := \{ yf(x) \mid x\in f({\frak a}) \}.\]

If ${\frak b}$ is an ideal of B, then $f^{-1}({\frak b})$ is always an ideal of $A$, called the contraction ${\frak b}^c$ of ${\frak b}$. If ${\frak b}$ is prime, then so is the contraction of it. However, if ${\frak a}\subset A$ is prime, ${\frak a}^e$ need not be prime, take any homomorphism from $\mathbb{Z}$ to $\mathbb{Q}$ for example.

We can factorize $f$ as follows:

\[A \xrightarrow{\text{p}} f(A)\xrightarrow{j}B\]

where $p$ is surjective and $j$ is injective. For $p$ the situation is simple, there is a one-to-one correspondence between the ideals of $A$ containing $\text{ker}(p)$ and ideals of $f(A)$. For $j$, the general situation is quite complicated. The classical example is from algebraic number theory.

Example. Consider $\mathbb{Z}\to \mathbb{Z}[i]$, where $i=\sqrt{ -1 }$. A prime ideal of $\mathbb{Z}$ may not stay a prime ideal when extended to $\mathbb{Z}[i]$. The situation is as follows,

  1. $(2)^e=((1+i)^2)$, the extension of $(2)$ is the ideal generated by the square of a prime ideal in $\mathbb{Z}[i]$,
  2. If $p\equiv {1}(\text{mod }4)$, then $p^e$ is the product of two distinct prime ideals. For example, $(5)^e=(2+i)(2-i)$.
  3. If $p\equiv {3}(\text{mod }4)$, then $p^e$ is prime in $Z[i]$.

Let $f:A\to B$ be a ring homomorphism, $C$ be the set of contracted ideals in $A$ and $E$ the set of extended ideals in $B$. Then

\[C=\{ {\frak a} \mid {\frak a}={\frak a}^{ec} \},\quad D=\{ {\frak b} \mid {\frak b}={\frak b}^{ce} \}\]

and there exists a one-to-one correspondence between ${\frak a}$ and ${\frak a}^e$.

Miscellaneous

Let $x$ be a nilpotent of ring $R$, then $1+x$ is a unit, the inverse is given by $1-x+x^2\dots$.

The polynomial ring $k[x]$ in one indeterminant is a PID. Note that $R[x]$, the polynomials whose coefficients take value from a ring, is in general not a PID, the coefficients have to take value from a field.

Let $R$ be a ring and $R[x]$ be the ring of polynomials in an indeterminant $x$, with coefficients in $R$. Then the Jacobson radical is equal to the nilradical. Recall that the Jacobson radical is the intersection of all the maximal ideal.

Let $R$ be a ring and $R[[x]]$ be the ring of formal power series

\[f=\sum_{n=0}^\infty a_{n} x^n.\]

We have

  • $f$ is a unit iff $a_{0}$ is a unit.
  • If $f$ is nilpotent, then $a_{n}$ is nilpotent for all $n\geq_{0}$.
  • Every prime ideal of $R$ is a contraction of prime ideal of $R[A]$.

Definition. A ring $R$ is Boolean if $x^{2}=x$ for all $x\in R$.

Boolean ring has some peculiar properties:

  • $2x=0$ for all $x$;
  • every prime ideal is maximal, and $A / {\frak p}$ is a field with two elements;
  • every finitely generated ideal is principal.

A local ring (rings with only one maximal ideal) has no idempotents other than $0,1$.

The prime spectrum of a ring

Let $A$ be a ring and let $X$ be the set of all prime ideals of $A$. For each subset $E$ of $A$, let $V(E)$ denote the set of all prime ideals which contain $E$. Convince yourself that

  1. if ${\frak a}$ is the ideal generated by $E$, then $V(E)=V({\frak a})=V(\text{rad}(({\frak a})))$.
  2. $V(0)=X$, $V(1)=\emptyset$.
  3. \[V\left( \bigcup_{i} E_{i} \right) = \bigcap_{i}V(E_{i})\]
  4. $V({\frak a}\cap {\frak b})=V({\frak ab})=V({\frak a})\cup V({\frak b})$.

These results show that the sets $V(E)$ satisfy the axioms for closed sets in a topological space. For example, if define the sets that can be written as $V(E)$ for some $E$ to be closed, then property 2) shows that the full set and empty set are closed. This topology is called Zariski topology. The topological space is called the prime spectrum of ring $A$, and written as $\text{spec}(A)$.

A topological space is said to be irreducible if $X\neq 0$ and every pair of non-empty open sets intersect.

Definition. Let $k$ be an algebraically closed field. The common zero locus of a set of polynomial equations

\[f_{\alpha}(t_{1},\dots,t_{n}) = 0\]

with coefficients in $k$ is called the affine albegraic variety.

On the other hand, given a locus $X$, consider the set of polynomial equations that is satisfied on $X$. Apparent they form an ideal of the polynomial ring, and is denoted by $I(x)$. It is called the ideal of the variety $X$. The quotient ring

\[P(X) := k[t_{1},\dots,t_{n}] / I(X)\]

is the ring of polynomial function on $X$.

Let $\xi_{i}$ be the coordinate function on $X$: for $x\in X$, the $\xi_{i}(x)$ is the $i$-th component of $x$. $P(x)$ is generated as a $k$-algebra (which we explain more in the next papragraph) by the coordinates functions, and is called the coordinate ring of $X$.

Recall that $k$-algebra are essentially a vector field over $k$ together with a ring structure. To be specific, a k-algebra $A$ is a ring with identity that is also a $k$-vector space, such that for $\alpha\in k$ and $a,b\in A$,

\[\alpha(ab) = (\alpha a)b = a (\alpha b).\]

Example of $k$-algebras include $k$ itself, the polynomial ring $k[X,{Y},…]$, the ring of matrices $M_{d}(k)$ under matrix addition and multiplication.