Abstract Nonsense

Crushing one theorem at a time

Polynomial Rings (Pt. II)

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

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Degree of a Polynomial Over an Integral Domain

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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 p(x)\in k[x] for some field k that the number of roots of p(x) is less than the degree of p(x). 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 \deg(p(x)+q(x)) and \deg(p(x)q(x)) in terms of \deg(p(x)) and \deg(q(x)). While there are bounds for both of these expressions there is a meaningful equality for \deg(p(x)q(x)) in terms of \deg(p(x)),\deg(q(x)), but this requires that we assume that R is an integral domain. Namely:

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Theorem: Let R be a commutative unital ring and p(x),q(x)\in R[x], then \deg(p(x)+q(x))\leqslant \max\{\deg(p(x)),\deg(q(x))\} and \deg(p(x)q(x))\leqslant \deg(p(x))+\deg(q(x)). Moreover, if R 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 R is an integral domain then equality in the second inequality follows since the leading coefficient of p(x)q(x) is the product of the leading coefficients of p(x) and q(x) and since this is the product of two non-zero ring elements, we have by assumption the result is non-zero. \blacksquare

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Now, it may sound silly but it’s an important observation to note that \deg(p(x))\ne\deg(q(x)) implies p(x)\ne q(x). In particular, we see that if \deg(p(x))>0 then p(x)\ne 0. With this observation and the previous theorem we get the following result:

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Theorem: Let R be a commutative unital ring. Then, R[x] is an integral domain if and only if R is.

Proof: Since R embeds into R[x] it’s evident that the ‘only if’ part of this statement is true. To prove the converse we consider two cases. Namely, suppose that p(x),q(x)\in R[x] are non-zero. If these are constant polynomials then their product is non-zero because R is an integral domain, if one of them is non-constant then \deg(p(x)q(x))=\deg(p(x))+\deg(q(x))>0 and so p(x)q(x)\ne 0 as desired. \blacksquare

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Units in R[x] and Nilpotent Elements

What we would now like to do is classify the units in the polynomial ring R[x]. 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 R we call an element r\in R nilpotent if r^n=0 for some n\in\mathbb{N} and call the minimal such n the index of nilpotency or just index when no confusion will arise. With this definition in mind we make the following observation:

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Observation: The binomial theorem \displaystyle (a+b)^n=\sum_{j=0}^{n}{n\choose j}a^j b^{n-j} holds for a,b in an arbitrary commutative ring R

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The proof of this is done precisely the same as the case for \mathbb{R}. Thus:

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Theorem: Let R be a commutative ring, then the set \mathfrak{N}(R) of all nilpotent elements of R forms an ideal of R.

Proof: Evidently if a\in\mathfrak{N}(R) then a\in\mathfrak{N}(R) since (ab)^m=a^mb^m=0b^m=0 where m is the index of a. To see that \mathfrak{N}(R) is a subgroup of R we merely note that if a,b\in\mathfrak{N}(R) and m is greater than twice the index of either a,b then (a+b)^m=0 since by the binomial theorem we have that (a+b)^m is a sum of elements of the form a^j b^{m-j} where j\in\{0,\cdots,m\} and since evidently \displaystyle j<\frac{m}{2}  implies \displaystyle m-j>\frac{m}{2} we have that each of these terms is zero by assumption. \blacksquare

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Moreover, we get the following interesting theorem:

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Theorem: Let R be a commutative unital ring, r\in\mathfrak{N}(R) and u\in U(R) then r+u\in U(R).

Proof: It evidently suffices to prove this for u=1 since r+u=u(1+u^{-1}r), U(R) is closed under multiplication, and  u^{-1}r\in\mathfrak{N}(R). So, to prove that r+1 is a unit we merely note that by induction one can prove that (1+x)(1-x+x^2-x^3+\cdots+(-1)^{n-1} x^{n-1})=1-(-1)^nx^n for all x\in R and so in particular, if m is the index of r then (1+r)(1-r+r^2-\cdots+(-1)^{m-1}r^{m-1})=1+(-1)^m r^m=1. \blacksquare

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Noticing the obvious fact that if r\in \mathfrak{N}(R) implies rx^n\in\mathfrak{N}(R[x]) for every n\in\mathbb{N} gives us half of the following theorem:

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Theorem: Let p(x)=a_0+\cdots+a_nx^n\in R[x] for some commutative unital ring R. Then, p(x)\in U(R[x]) if and only if a_0\in U(R) and a_1,\cdots,a_n\in \mathfrak{N}(R).

Proof: Evidently this is sufficient since p(x) is the sum of nilpotent polynomial a_1x+\cdots+a_n x^n and the unit a_0 from where the result follows from the previous theorem.

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Conversely, we first note that it clearly suffices to prove that a_n\in\mathfrak{N}(R) since, if so, we have that a_n x^n\in\mathfrak{N}(R[x]) and so a_{n-1}x^{n-1}+\cdots+a_0=p(x)-a_nx^n\in U(R[x]) (being the sum of a unit and a nilpotent). To do this we assume that p(x)q(x)=1 where q(x)=b_0+\cdots+b_mx^m. We note then that since the coefficient of x^{n+m} is a_nb_m we must have a_nb_m=0. Similarly we must have that a_nb_{m-1}+a_{n-1}b_m=0. Multiplying this by a_n yields a_n^2 b_{m-1}=0. Similarly we have that a_nb_{m-2}+a_{n-1}b_{m-1}+a_{n-1}b_{m}=0 and multiplying this by a_n^2 gives a_n^3 b_{m-2}=0. Proceeding in this manner shows that b_0 a_{n}^{m+1}=0. Now, since the constant term of p(x)q(x) is a_0b_0 we must evidently have that b_0\in U(R) and since b_0a_n^{m+1}=0 we may conclude that a_n^{m+1}=0. The conclusion follows from previous comment. \blacksquare

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Since evidently if R is an integral domain then \mathfrak{N}(R)=\{0\} we have the following corollary:

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Theorem: Let R be an integral domain, then U(R[x])=U(R).

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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.


July 19, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , , , , ,


  1. […] Point of Post: This post is a continuation of this one. […]

    Pingback by Polynomial Rings (Pt. III) « Abstract Nonsense | July 19, 2011 | Reply

  2. […] know  is an integral domain. To see that for each we merely note that the kernel of the evaluation homomorphism  is (this follows from the classic property of complex polynomials [as we shall see more […]

    Pingback by Prime Ideals (Pt. II) « Abstract Nonsense | August 17, 2011 | Reply

  3. […] is maximal (as discussed before) since . Conversely, let be a maximal ideal, and suppose that for each there existed a such that […]

    Pingback by Maximal Ideals (Pt. II) « Abstract Nonsense | September 6, 2011 | Reply

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