## Polynomial Rings in Relation to Euclidean Domains, PIDs, and UFDs (Pt. II)

**Point of Post: **This is a continuation of this post.

*Polynomial Rings as UFDs*

So, we have stated earlier that a polynomial ring is a UFD if and only if its underlying ring is a UFD. But, why is this true? Certainly if the polynomial ring is a UFD then the underlying ring must be a UFD (if this isn’t obvious we shall prove it shortly) it is the other direction that is difficult. There seems to be an obvious way one might try to approach this. Namely, if is a UFD we know that is a UFD and since we can factor any element of into irreducibles in . Hopefully, we can pass this factorization down. This is the content of Gauss’s Lemma:

**Lemma(Gauss): ***Let be a UFD and let . Let . If is reducible in then is reducible in .*

**Proof: **Since is reducible in we can write with . Multiplying through by an appropriate constant we have that where now . If then and we’re done, so assume that . Since is a UFD and is a non-zero non-unit we can write it as a product of irreducibles, say . Since is irreducible (and so prime) we have that . We know then by previous discussion that is a prime ideal in . and is an integral domain. But, note that and since is an integral domain we may conclude that or is in which implies, if we say that , that . So, . Continuing in this way allows us to write with (of the form and where run over some two subsets of ).

Note that in all reality the only difference between factorization in and come from units in the coefficients. With this in mind we define the *content *of a polynomial , denoted , to be the ideal . With this idea and Gauss’s lemma it becomes clear that:

**Theorem: ***Let be a UFD and let . If then is irreducible in if and only if is irreducible in .*

We now wish to prove what we set out to do:

**Theorem: ***A ring is a UFD if and only if is a UFD.*

**Proof: **Suppose first that is a UFD and let . Write as where is a greatest common divisor of the coefficients of (i.e. a generator of ). We know that can be factored uniquely into irreducibles in and so it suffices to show that can be factored uniquely. From this it’s clear that in general, we can assume that has content (i.e. the coefficients are coprime) and that they are non-constant.

So, let be non-unit non-zero polynomial in with content . Since is non-zero and non-unit in we know that can be factored into irreducibles in . But, we know that each of the factors in this factorizations are just irreducibles in times a unit in . But, since our original polynomial has content one we have by the previous theorem that each of the factors are irreducible in . Thus, every element of can be factored into irreducibles.

It now remains to show that every irreducible element of is prime. This is mostly just busy work, and can be found in [2] on page 308-309.

So, now let’s formally prove that being a UFD implies that is a UFD. We already know that being an integral domain implies that is an integral domain and so it suffices to prove that every irreducible in is prime and that every element of can be written as a product of irreducibles. Now, suppose that is irreducible, then clearly is irreducible in (since the only way to factor it is as constant terms etc.) and so by assumption is prime in . Thus, is an integral domain, but and so is an integral domain. Thus, is an integral domain and so is prime in . Now, let be arbitrary. Since is a UFD we can factor as a product of irreducibles in , but note that being irreducible in implies being irreducible in , and since is an integral domain (so that degree is additive) we know that . Thus, we can factor any as the product of irreducibles.

**References:**

[1] Dummit, David Steven., and Richard M. Foote. *Abstract Algebra*. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. *Advanced Modern Algebra*. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. *Basic Abstract Algebra*. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

[…] Suppose that was reducible in , then Gauss’s lemma would tell us that is reducible in , say where , and since there is a natural homomorphism one […]

Pingback by Some Polynomial Irreducibility Criteria « Abstract Nonsense | October 26, 2011 |

I am interested in doing my Research work on Euclidean Near Rings containing Noetherian Regular delta Near Rings by following thoroughly Reading this article.

CH.Padma

Asst.Prof (Marhs)

LBRCE, Mylavaram

Comment by Chpadma | January 30, 2012 |

[…] first note that the polynomial is irreducible over . Indeed, by Gauss’s lemma we need merely check that it is irreducible in but and this is clearly irreducible in the latter […]

Pingback by Automorphisms of k(t)/k « Abstract Nonsense | February 28, 2012 |