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In abstract algebra, a non-zero non-unit element in an integral domain is said to be irreducible if it is not a product of two non-units.

Relationship with prime elements

Irreducible elements should not be confused with prime elements. (A non-zero non-unit element a in a commutative ring R is called prime if, whenever a | bc for some b and c in R, then a|b or a|c.) In an integral domain, every prime element is irreducible,[1][2] but the converse is not true in general. The converse is true for unique factorization domains[2] (or, more generally, GCD domains.)

Moreover, while an ideal generated by a prime element is a prime ideal, it is not true in general that an ideal generated by an irreducible element is an irreducible ideal. However, if D is a GCD domain, and x is an irreducible element of D, then the ideal generated by x is a prime ideal of D.[3]
Example

In the quadratic integer ring \( \mathbf{Z}[\sqrt{-5}] \), it can be shown using norm arguments that the number 3 is irreducible. However, it is not a prime element in this ring since, for example,

\( 3 \mid \left(2 + \sqrt{-5}\right)\left(2 - \sqrt{-5}\right)=9, \)

but 3 does not divide either of the two factors.[4]
See also

Irreducible polynomial

References

Consider p a prime that is reducible: p=ab. Then p | ab \Rightarrow p | a or p | b. Say p | a \Rightarrow a = pc, then we have p=ab=pcb \Rightarrow p(1-cb)=0. Because R is an integral domain we have cb=1. So b is a unit and p is irreducible.
Sharpe (1987) p.54
http://planetmath.org/encyclopedia/IrreducibleIdeal.html

William W. Adams and Larry Joel Goldstein (1976), Introduction to Number Theory, p. 250, Prentice-Hall, Inc., ISBN 0-13-491282-9

Sharpe, David (1987). Rings and factorization. Cambridge University Press. ISBN 0-521-33718-6. Zbl 0674.13008.

Any non-trivial idempotent a is a zero divisor (because ab = 0 with neither a nor b being zero, where b = 1 − a). This shows that integral domains and division rings don't have such idempotents. Local rings also don't have such idempotents, but for a different reason. The only idempotent contained in the Jacobson radical of a ring is 0.

For associative algebras or Jordan algebras over a field, the Peirce decomposition is a decomposition of an algebra as a sum of eigenspaces of commuting idempotent elements.

Another method to obtain a field from a commutative ring R is taking the quotient R / m, where m is any maximal ideal of R. The above construction of F = E[X] / (p(X)), is an example, because the irreducibility of the polynomial p(X) is equivalent to the maximality of the ideal generated by this polynomial. Another example are the finite fields Fp = Z / pZ.

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