Beck's monadicity theorem

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 ${\mathfrak {a}}$ is an ideal in A, then $f({\mathfrak {a}})$ 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 ${\mathfrak {a}}^{e}$ of ${\mathfrak {a}}$ in B is defined to be the ideal in B generated by $f({\mathfrak {a}})$. Explicitly,

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

Contraction of an ideal

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

Properties

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

• ${\mathfrak {b}}$ is prime in B $\Rightarrow$ ${\mathfrak {b}}^{c}$ is prime in A.

• ${\mathfrak {a}}^{ec}\supseteq {\mathfrak {a}}$

• ${\mathfrak {b}}^{ce}\subseteq {\mathfrak {b}}$

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

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

• ${\mathfrak {a}}^{ec}={\mathfrak {a}}$ and ${\mathfrak {b}}^{ce}={\mathfrak {b}}$.

• ${\mathfrak {a}}$ is a prime ideal in A $\Leftrightarrow$ ${\mathfrak {a}}^{e}$ is a prime ideal in B.

• ${\mathfrak {a}}$ is a maximal ideal in A $\Leftrightarrow$ ${\mathfrak {a}}^{e}$ 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 ${\mathfrak {a}}={\mathfrak {p}}$ of A under extension is one of the central problems of algebraic number theory.

See also

• Splitting of prime ideals in Galois extensions

References

• Atiyah, M. F. and Macdonald, I. G., Introduction to Commutative Algebra, Perseus Books, 1969, ISBN 0-201-00361-9