## Polynomial Rings (Pt. II)

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

*Degree of a Polynomial Over an Integral Domain*

The degree of a polynomial is of fundamental importance in most of the places polynomials are commonly seen. For example, a fundamental result will tell us that for a polynomial for some field that the number of roots of is less than the degree of . Consequently, it seems desirable to see how the degree of a polynomial reacts in connection with the ring operations. More explicitly, it would seem fruitful to find an expression (or at least a bound) for and in terms of and . While there are bounds for both of these expressions there is a meaningful equality for in terms of , but this requires that we assume that is an integral domain. Namely:

**Theorem: ***Let be a commutative unital ring and , then and . Moreover, if is an integral domain, then this last inequality is an equality.*

**Proof: **The first two inequalities are obvious by inspection. If we assume then that is an integral domain then equality in the second inequality follows since the leading coefficient of is the product of the leading coefficients of and and since this is the product of two non-zero ring elements, we have by assumption the result is non-zero.

Now, it may sound silly but it’s an important observation to note that implies . In particular, we see that if then . With this observation and the previous theorem we get the following result:

**Theorem: ***Let be a commutative unital ring. Then, is an integral domain if and only if is.*

**Proof: **Since embeds into it’s evident that the ‘only if’ part of this statement is true. To prove the converse we consider two cases. Namely, suppose that are non-zero. If these are constant polynomials then their product is non-zero because is an integral domain, if one of them is non-constant then and so as desired.

*Units in and Nilpotent Elements*

What we would now like to do is classify the units in the polynomial ring . This is a deceivingly difficult task, for which we shall need a few lemmas, but before we proceed to these we make a definition. Namely, for a ring we call an element *nilpotent *if for some and call the minimal such the *index of nilpotency *or just *index *when no confusion will arise. With this definition in mind we make the following observation:

**Observation: ***The binomial theorem holds for in an arbitrary commutative ring *

The proof of this is done precisely the same as the case for . Thus:

**Theorem: ***Let be a commutative ring, then the set of all nilpotent elements of forms an ideal of .*

**Proof: **Evidently if then since where is the index of . To see that is a subgroup of we merely note that if and is greater than twice the index of either then since by the binomial theorem we have that is a sum of elements of the form where and since evidently implies we have that each of these terms is zero by assumption.

Moreover, we get the following interesting theorem:

**Theorem: ***Let be a commutative unital ring, and then .*

**Proof: **It evidently suffices to prove this for since , is closed under multiplication, and . So, to prove that is a unit we merely note that by induction one can prove that for all and so in particular, if is the index of then .

Noticing the obvious fact that if implies for every gives us half of the following theorem:

**Theorem: ***Let for some commutative unital ring . Then, if and only if and .*

**Proof: **Evidently this is sufficient since is the sum of nilpotent polynomial and the unit from where the result follows from the previous theorem.

Conversely, we first note that it clearly suffices to prove that since, if so, we have that and so (being the sum of a unit and a nilpotent). To do this we assume that where . We note then that since the coefficient of is we must have . Similarly we must have that . Multiplying this by yields . Similarly we have that and multiplying this by gives . Proceeding in this manner shows that . Now, since the constant term of is we must evidently have that and since we may conclude that . The conclusion follows from previous comment.

Since evidently if is an integral domain then we have the following corollary:

**Theorem: ***Let be an integral domain, then .*

**References:**

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

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

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