# Extension and contraction of ideals

(Redirected from Extension of an ideal)

In commutative algebra, the extension and contraction of ideals are operations performed on sets of ideals.

## Extension of an ideal

Let A and B be two commutative rings with unity, and let f : AB be a (unital) ring homomorphism. If ${\displaystyle {\mathfrak {a}}}$ is an ideal in A, then ${\displaystyle f({\mathfrak {a}})}$ need not be an ideal in B (e.g. take f to be the inclusion of the ring of integers Z into the field of rationals Q). The extension ${\displaystyle {\mathfrak {a}}^{e}}$ of ${\displaystyle {\mathfrak {a}}}$ in B is defined to be the ideal in B generated by ${\displaystyle f({\mathfrak {a}})}$. Explicitly,

${\displaystyle {\mathfrak {a}}^{e}={\Big \{}\sum y_{i}f(x_{i}):x_{i}\in {\mathfrak {a}},y_{i}\in B{\Big \}}}$

## Contraction of an ideal

If ${\displaystyle {\mathfrak {b}}}$ is an ideal of B, then ${\displaystyle f^{-1}({\mathfrak {b}})}$ is always an ideal of A, called the contraction ${\displaystyle {\mathfrak {b}}^{c}}$ of ${\displaystyle {\mathfrak {b}}}$ to A.

## Properties

Assuming f : AB is a unital ring homomorphism, ${\displaystyle {\mathfrak {a}}}$ is an ideal in A, ${\displaystyle {\mathfrak {b}}}$ is an ideal in B, then:

• ${\displaystyle {\mathfrak {b}}}$ is prime in B ${\displaystyle \Rightarrow }$ ${\displaystyle {\mathfrak {b}}^{c}}$ is prime in A.
• ${\displaystyle {\mathfrak {a}}^{ec}\supseteq {\mathfrak {a}}}$
• ${\displaystyle {\mathfrak {b}}^{ce}\subseteq {\mathfrak {b}}}$

It is false, in general, that ${\displaystyle {\mathfrak {a}}}$ being prime (or maximal) in A implies that ${\displaystyle {\mathfrak {a}}^{e}}$ is prime (or maximal) in B. Many classic examples of this stem from algebraic number theory. For example, embedding ${\displaystyle \mathbb {Z} \to \mathbb {Z} \left\lbrack i\right\rbrack }$. In ${\displaystyle B=\mathbb {Z} \left\lbrack i\right\rbrack }$, the element 2 factors as ${\displaystyle 2=(1+i)(1-i)}$ where (one can show) neither of ${\displaystyle 1+i,1-i}$ are units in B. So ${\displaystyle (2)^{e}}$ is not prime in B (and therefore not maximal, as well). Indeed, ${\displaystyle (1\pm i)^{2}=\pm 2i}$ shows that ${\displaystyle (1+i)=((1-i)-(1-i)^{2})}$, ${\displaystyle (1-i)=((1+i)-(1+i)^{2})}$, and therefore ${\displaystyle (2)^{e}=(1+i)^{2}}$.

On the other hand, if f is surjective and ${\displaystyle {\mathfrak {a}}\supseteq {\mathop {\mathrm {ker} }}f}$ then:

• ${\displaystyle {\mathfrak {a}}^{ec}={\mathfrak {a}}}$ and ${\displaystyle {\mathfrak {b}}^{ce}={\mathfrak {b}}}$.
• ${\displaystyle {\mathfrak {a}}}$ is a prime ideal in A ${\displaystyle \Leftrightarrow }$ ${\displaystyle {\mathfrak {a}}^{e}}$ is a prime ideal in B.
• ${\displaystyle {\mathfrak {a}}}$ is a maximal ideal in A ${\displaystyle \Leftrightarrow }$ ${\displaystyle {\mathfrak {a}}^{e}}$ is a maximal ideal in B.

## Extension of prime ideals in number theory

Let K be a field extension of L, and let B and A be the rings of integers of K and L, respectively. Then B is an integral extension of A, and we let f be the inclusion map from A to B. The behaviour of a prime ideal ${\displaystyle {\mathfrak {a}}={\mathfrak {p}}}$ of A under extension is one of the central problems of algebraic number theory.