Irreducible ideal explained

In mathematics, a proper ideal of a commutative ring is said to be irreducible if it cannot be written as the intersection of two strictly larger ideals.[1]

Examples

J

and

K

be ideals of a commutative ring

R

, with neither one contained in the other. Then there exist

a\inJ\setminusK

and

b\inK\setminusJ

, where neither is in

J\capK

but the product is. This proves that a reducible ideal is not prime. A concrete example of this are the ideals

2Z

and

3Z

contained in

Z

. The intersection is

6Z

, and

6Z

is not a prime ideal.

Properties

An element of an integral domain is prime if and only if the ideal generated by it is a non-zero prime ideal. This is not true for irreducible ideals; an irreducible ideal may be generated by an element that is not an irreducible element, as is the case in

Z

for the ideal

4Z

since it is not the intersection of two strictly greater ideals.

In algebraic geometry, if an ideal

I

of a ring

R

is irreducible, then

V(I)

is an irreducible subset in the Zariski topology on the spectrum

\operatorname{Spec}R

. The converse does not hold; for example the ideal

(x2,xy,y2)

in

C[x,y]

defines the irreducible variety consisting of just the origin, but it is not an irreducible ideal as

(x2,xy,y2)=(x2,y)\cap(x,y2)

.

See also

Notes and References

  1. .
  2. .
  3. Book: Dummit. David S.. Foote. Richard M.. Abstract Algebra. 2004. John Wiley & Sons, Inc.. Hoboken, NJ. 0-471-43334-9. 683–685. Third.
  4. . Theorem 1, p. 3.