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Prime Polynomials and Irreducible Polynomials

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Dummit and Foote on page 284 give the following definitions of irreducible and prime for integral domains. (I have some issues/problems with the definitions - see below)

------------------------------------------------------------------------------------------------------------------------------- "Definition:

Let R be an integral domain.

(1) Suppose \(\displaystyle r \in R \) is non-zero and not a unit. Then r is called irreducible in R if whenever r = ab with [TEX] a, b \in R [/TEX] at least one of a or b must be a unit in R. Otherwise r is said to be reducible.

(2) The non-zero element [TEX] p \in R [/TEX] is called prime in R if the ideal (p) generated by p is a prime ideal. In other words, a non-zero element p is a prime if it is not a unit and whenever p|ab for any [TEX] a,b \in R [/TEX], then either p|a or p|b."

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So where a ring of polynomials is an integral domain we have a definition of prime and irreducible elements (polynomials).

However, most algebra books I have referenced just talk about irreducible polynomials - and indeed define them in ways that look like D&F's definition of a prime element in an integral domain - so maybe for polynomials (for some reason???) irreducible and prime are the same thing? Can someone clarify this point?

Indeed, in the book Elements of Modern Algebra by Gilbert and Gilbert we find the following definition of an irreducible polynomial:

"A polynomial f(x) in F[x] is irreducible (or prime) over F (a field?) if f(x) has positive degree and f(x) cannot be expressed as a product f(x) = g(x)h(x) with both g(x) and h(x) of positive degree in F[x]."

Gilbert and Gilbert then follow this with the following theorem: (which seems to follow D&F definition of a prime in an integral domain - also - why is G&G using field and not an integral domain?)

"If p(x) is an irreducible polynomial over the field F and p(x) divides f(x)g(x), then either p(x) | f(x) or p(x) | g(x) in F[x] How does G&G's definition square with D&F's definitions. Can someone please clarify?

Another point is that I am unsure why D&F restrict these definitions to an integral domain thus leaving the terms undefined for general rings that are not integral domains. Can someone clarify?

Yet another problem I have with the above definitions by D&F is the following: D&F write: "In other words, a non-zero element p is a prime if it is not a unit and whenever p|ab for any [TEX] a,b \in R [/TEX], then either p|a or p|b." - How does this follow from (p) being a prime ideal.

Peter


Note: D&F's definition of prime ideal is on page 255 and is as follows:
Definition: Assume R is commutative. An ideal P is called a prime ideal if [TEX] P \ne R [/TEX] and whenever the product of two elements [TEX] a,b \in R [/TEX] is an element of P, then at least on of a and b is an element of P.
 
Last edited:

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Dummit and Foote on page 284 give the following definitions of irreducible and prime for integral domains. (I have some issues/problems with the definitions - see below)

------------------------------------------------------------------------------------------------------------------------------- "Definition:

Let R be an integral domain.

(1) Suppose \(\displaystyle r \in R \) is non-zero and not a unit. Then r is called irreducible in R if whenever r = ab with [TEX] a, b \in R [/TEX] at least one of a or b must be a unit in R. Otherwise r is said to be reducible.

(2) The non-zero element [TEX] p \in R [/TEX] is called prime in R if the ideal (p) generated by p is a prime ideal. In other words, a non-zero element p is a prime if it is not a unit and whenever p|ab for any [TEX] a,b \in R [/TEX], then either p|a or p|b."

--------------------------------------------------------------------------------------------------------------------------------

So where a ring of polynomials is an integral domain we have a definition of prime and irreducible elements (polynomials).

However, most algebra books I have referenced just talk about irreducible polynomials - and indeed define them in ways that look like D&F's definition of a prime element in an integral domain - so maybe for polynomials (for some reason???) irreducible and prime are the same thing? Can someone clarify this point?

Indeed, in the book Elements of Modern Algebra by Gilbert and Gilbert we find the following definition of an irreducible polynomial:

"A polynomial f(x) in F[x] is irreducible (or prime) over F (a field?) if f(x) has positive degree and f(x) cannot be expressed as a product f(x) = g(x)h(x) with both g(x) and h(x) of positive degree in F[x]."

Gilbert and Gilbert then follow this with the following theorem: (which seems to follow D&F definition of a prime in an integral domain - also - why is G&G using field and not an integral domain?)

"If p(x) is an irreducible polynomial over the field F and p(x) divides f(x)g(x), then either p(x) | f(x) or p(x) | g(x) in F[x] How does G&G's definition square with D&F's definitions. Can someone please clarify?

Another point is that I am unsure why D&F restrict these definitions to an integral domain thus leaving the terms undefined for general rings that are not integral domains. Can someone clarify?

Yet another problem I have with the above definitions by D&F is the following: D&F write: "In other words, a non-zero element p is a prime if it is not a unit and whenever p|ab for any [TEX] a,b \in R [/TEX], then either p|a or p|b." - How does this follow from (p) being a prime ideal.

Peter


Note: D&F's definition of prime ideal is on page 255 and is as follows:
Definition: Assume R is commutative. An ideal P is called a prime ideal if [TEX] P \ne R [/TEX] and whenever the product of two elements [TEX] a,b \in R [/TEX] is an element of P, then at least on of a and b is an element of P.
Hi Peter, :)

I might not be able to answer all your questions but let me try to answer some of them. :)

Every prime element in an integral domain is irreducible. But the converse is true only within an Unique factorization domain(UFD). It could be shown that if \(F\) is a UFD so is \(F[x]\) (the ring of polynomials with coefficients in \(F\)). From what I have read in most books the polynomial ring (say \(F[x]\)) is defined in terms of a field \(F\). So you see the point of G&G using a field to define the polynomial ring. In this situation irreducibility an primeness are the same. Had he used an integral domain he would be forced to distinguish between irreducibility and primeness.

I hope my explanation helped you to clarify at least some of your doubts. :)