Extension and contraction of ideals

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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 : A → B be a (unital) ring homomorphism. If [math]\displaystyle{ \mathfrak{a} }[/math] is an ideal in A, then [math]\displaystyle{ f(\mathfrak{a}) }[/math] 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 [math]\displaystyle{ \mathfrak{a}^e }[/math] of [math]\displaystyle{ \mathfrak{a} }[/math] in B is defined to be the ideal in B generated by [math]\displaystyle{ f(\mathfrak{a}) }[/math]. Explicitly,

[math]\displaystyle{ \mathfrak{a}^e = \Big\{ \sum y_if(x_i) : x_i \in \mathfrak{a}, y_i \in B \Big\} }[/math]

Contraction of an ideal

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

Properties

Assuming f : A → B is a unital ring homomorphism, [math]\displaystyle{ \mathfrak{a} }[/math] is an ideal in A, [math]\displaystyle{ \mathfrak{b} }[/math] is an ideal in B, then:

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

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

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

  • [math]\displaystyle{ \mathfrak{a}^{ec}=\mathfrak{a} }[/math] and [math]\displaystyle{ \mathfrak{b}^{ce}=\mathfrak{b} }[/math].
  • [math]\displaystyle{ \mathfrak{a} }[/math] is a prime ideal in A [math]\displaystyle{ \Leftrightarrow }[/math] [math]\displaystyle{ \mathfrak{a}^e }[/math] is a prime ideal in B.
  • [math]\displaystyle{ \mathfrak{a} }[/math] is a maximal ideal in A [math]\displaystyle{ \Leftrightarrow }[/math] [math]\displaystyle{ \mathfrak{a}^e }[/math] 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 [math]\displaystyle{ \mathfrak{a} = \mathfrak{p} }[/math] of A under extension is one of the central problems of algebraic number theory.

See also

References




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