# Abstract Nonsense

## 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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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.

July 19, 2011 -