information:highlighting, structural:expansion
Extension and contraction of an ideal

Let A and B be two commutative rings, and let f : A → B be a ring homomorphism. If a {\displaystyle {\mathfrak {a}}} is an ideal in A, then f (a) {\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 a e {\displaystyle {\mathfrak {a}}^{e}} of a {\displaystyle {\mathfrak {a}}} in B is defined to be the ideal in B generated by f (a) {\displaystyle f({\mathfrak {a}})}. Explicitly,

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

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

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

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

Remark: 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 a = p {\displaystyle {\mathfrak {a}}={\mathfrak {p}}} of A under extension is one of the central problems of algebraic number theory.

The following is sometimes useful: a prime ideal p {\displaystyle {\mathfrak {p}}} is a contraction of a prime ideal if and only if p = p e c {\displaystyle {\mathfrak {p}}={\mathfrak {p}}^{ec}}. (Proof: Assuming the latter, note p e B p = B p ⇒ p e {\displaystyle {\mathfrak {p}}^{e}B_{\mathfrak {p}}=B_{\mathfrak {p}}\Rightarrow {\mathfrak {p}}^{e}} intersects A − p {\displaystyle A-{\mathfrak {p}}}, a contradiction. Now, the prime ideals of B p {\displaystyle B_{\mathfrak {p}}} correspond to those in B that are disjoint from A − p {\displaystyle A-{\mathfrak {p}}}. Hence, there is a prime ideal q {\displaystyle {\mathfrak {q}}} of B, disjoint from A − p {\displaystyle A-{\mathfrak {p}}}, such that q B p {\displaystyle {\mathfrak {q}}B_{\mathfrak {p}}} is a maximal ideal containing p e B p {\displaystyle {\mathfrak {p}}^{e}B_{\mathfrak {p}}}. One then checks that q {\displaystyle {\mathfrak {q}}} lies over p {\displaystyle {\mathfrak {p}}}. The converse is obvious.)