Zero-divisors in polynomial rings

Posted: November 1, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Problem. (McCoy) Let $R$ be a commutative ring with identity and let $f(x)= \sum_{i=0}^n a_ix^i \in R[x].$

1) Prove that $f(x)$ is a zero-divisor if and only if there exists some $0 \neq c \in R$ such that $cf(x) = 0.$

2) Prove that if $R$ is reduced and $f(x)g(x)=0$ for some $g(x)=\sum_{i=0}^m b_ix^i \in R[x],$ then $a_ib_j=0$ for all $0 \leq i \leq n$ and $0 \leq j \leq m.$

Solution. 1) If there exists $0 \neq c \in R$ such that $cf(x)=0,$ then clearly $f(x)$ is a zero-divisor of $R[x].$ For the converse, let $g(x)=\sum_{i=0}^m b_ix^i, \ b_m \neq 0,$ be a polynomial with minimum degree such that $f(x)g(x)=0.$ I will show that $m = 0.$ So, suppose to the contrary, that $m \geq 1.$ If $a_jg(x)=0,$ for all $j,$ then $a_jb_m=0,$ for all $j,$ and so $b_mf(x)=0$ contradicting the minimality of $m$ because $\deg b_m = 0 < m$. So we may assume that the set $\{j: \ a_jg(x) \neq 0 \}$ is non-empty and so we can let

$\ell=\max \{j : \ a_jg(x) \neq 0 \}.$

Then

$0=f(x)g(x)=(a_{\ell}x^{\ell} + \cdots + a_0)(b_mx^m + \cdots + b_0).$

Thus

$a_{\ell}b_m=0$ and so $a_{\ell}g(x)=a_{\ell}b_{m-1}x^{m-1} + \cdots + a_{\ell}b_0.$

Hence $\deg a_{\ell}g(x) < m=\deg g.$ But we have $f(x)(a_{\ell}g(x))=a_{\ell}f(x)g(x)=0,$ which is impossible because $g(x)$ was supposed to be a polynomial with minimum degree satisfying $f(x)g(x)=0.$

2) The proof of this part is by induction over $i+j.$ It is obvious from $f(x)g(x)=0$ that $a_0b_0=0.$ Now let $0 < \ell \leq m+n$ and suppose that $a_rb_s=0$ whenever $0 \leq r+s < \ell.$ We need to show that $a_rb_s=0$ whenever $r+s=\ell.$ So suppose that $r+s=\ell.$ The coefficient of $x^{\ell}$ in $f(x)g(x)$ is clearly

$0=\sum_{i < r, \ i+j=\ell}a_ib_j+ a_rb_s + \sum_{i > r, \ i+j=\ell} a_ib_j,$

which after multiplying both sides by $a_rb_s$ gives us

$\sum_{i < r, \ i+j=\ell} a_rb_sa_ib_j+ (a_rb_s)^2 + \sum_{i > r, \ i+j=\ell} a_rb_sa_ib_j=0.$

Call this (1). Now in the first sum in (1), since $i < r,$ we have $i+s < r+s=\ell$ and hence by the induction hypothesis $a_ib_s=0.$ Thus $a_rb_sa_ib_j=0.$ So the first sum in (1) is $0.$ In the second sum in (1), since $i > r$ and $i+j=r+s=\ell,$ we have $j < s.$ Therefore by the induction hypothesis $a_rb_j=0$ and hence $a_rb_sa_ib_j=0.$ So the second sum in (1) is also $0.$ Thus (1) becomes $(a_rb_s)^2=0$ and so, since $R$ is reduced, $a_rb_s=0. \ \Box$

Units in polynomial rings

Posted: September 2, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Lemma. Let $R$ be a commutative ring with 1. If $a \in R$ is nilpotent and $b \in R$ is a unit, then $a+b$ is a unit.

Proof. So $a^n = 0$ for some integer $n \geq 1$ and $bc = 1$ for some $c \in R.$ Let

$u = (b^{n-1}- ab^{n-2} + \ldots + (-1)^{n-2}a^{n-2}b + (-1)^{n-1}a^{n-1})c^n$

and see that $(a+b)u=1. \ \Box$

Problem. Let $R$ be a commutative ring with 1. Let $p(x) = \sum_{j=0}^n a_j x^j, \ a_j \in R,$ be an element of the polynomial ring $R[x].$ Prove that $p(x)$ is a unit if and only if $a_0$ is a unit and all $a_j, \ j \geq 1,$ are nilpotent.

First Solution. ($\Longrightarrow$) Suppose that $a_1, \cdots , a_n$ are nilpotent and $a_0$ is a unit. Then clearly $p(x)-a_0$ is nilpotent and thus $p(x)=p(x)-a_0 + a_0$ is a unit, by the lemma.

($\Longleftarrow$) We’ll use induction on $n,$ the degree of $p(x).$ It’s clear for $n = 0.$ So suppose that the claim is true for any polynomial which is a unit and has degree less than $n.$ Let $p(x) = \sum_{j=0}^n a_jx^j, \ n \geq 1,$ be a unit. So there exists some $q(x)=\sum_{j=0}^m b_jx^j \in R[x]$ such that $p(x)q(x)=1.$ Then $a_0b_0=1$ and so $b_0$ is a unit. We also have

$a_nb_m = 0, \ a_nb_{m-1}+ a_{n-1}b_m = 0, \ \cdots , a_nb_0 +a_{n-1}b_1 + \cdots = 0.$

So $AX=0,$ where

$A=\begin{pmatrix}a_n & 0 & 0 & . & . & . & 0 \\ a_{n-1} & a_n & 0 & . & . & . & 0 \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ . & . & . & & . & . & . \\ * & * & * & . & . & . & a_n \end{pmatrix}, \ \ X=\begin{pmatrix}b_m \\ b_{m-1} \\ . \\ . \\ . \\ b_0 \end{pmatrix}.$

Thus $a_n^{m+1}X =(\det A)X = \text{adj}(A)A X = 0.$ Therefore $a_n^{m+1}b_0=0$ and hence $a_n^{m+1} = 0$ because $b_0$ is a unit. Thus $a_n,$ and so $-a_nx^n,$ is nilpotent. So $p_1(x)=p(x) -a_nx^n$ is a unit, by the lemma. Finally, since $\deg p_1(x) < n,$ we can apply the induction hypothesis to finish the proof. $\Box$

Second Solution. ($\Longrightarrow$) This part is the same as the first solution.

($\Longleftarrow$) Let $p(x) = \sum_{j=0}^n a_jx^j, \ a_n \neq 0,$ be a unit of $R[x]$ and let $q(x)=\sum_{i=0}^m b_i x^i \in R[x], \ b_m \neq 0,$ be such that $p(x)q(x)=1.$ Then $a_0b_0=1$ and so $a_0$ is a unit. To prove that $a_j$ is nilpotent for all $j \geq 1,$ we consider two cases:

Case 1 . $R$ is an integral domain. Suppose that $n > 0.$ Then from $p(x) q(x)=1$ we get $a_n b_m = 0,$ which is impossible because both $a_n$ and $b_m$ are non-zero and $R$ is an integral domain. So $n=0$ and we are done.

Case 2 . $R$ is arbitrary. Let $P$ be any prime ideal of $R$ and let $\overline{R}=R/P.$ For every $r \in R$ let $\overline{r}=r+P.$ Let

$\overline{p(x)}=\sum_{j=0}^n \overline{a_j}x^j, \ \ \overline{q(x)}=\sum_{i=0}^m \overline{b_i}x^i.$

Then clearly $\overline{p(x)} \cdot \overline{q(x)}=\overline{1}$ in $\overline{R}[x]$ and thus, since $\overline{R}$ is an integral domain, $\overline{a_j}=\overline{0}$ for all $j \geq 1,$ by case 1. Hence $a_j \in P$ for all $j \geq 1.$ So $a_j, \ j \geq 1,$ is in every prime ideal of $R$ and thus $a_j$ is nilpotent. $\Box$

Representation of polynomials

Posted: May 23, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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This is a generalization of the ordinary representation of polynomials:

Problem. Let $R$ be a commutative ring with $1$ and $A \in R[x]$ have degree $n \geq 0$ and let $B \in R[x]$ have degree at least $1$. Prove that if the leading coefficient of $B$ is a unit of $R$, then there exist unique polynomials $Q_0,Q_1,...,Q_n \in R[x]$ such that $\deg Q_i < \deg B,$ for all $i$, and $A = Q_0+Q_1B+...+Q_nB^n$

SolutionUniqueness of the representation : Since the leading coefficient of $B$ is a unit, for any $C \in R[x]$ we have $\deg (BC)=\deg B + \deg C.$ Now suppose that $Q_0 + Q_1B + \cdots + Q_nB^n = 0,$ with $Q_n \neq 0.$ Let $\alpha, \ \beta$ be the leading coefficients of $Q_n$ and $B$ repectively. Then the leading coefficient of $Q_0 + Q_1B + \cdots +Q_nB^n$ is $\alpha \beta^n.$ Thus $\alpha \beta^n = 0.$ Since $\beta$ is a unit, we’ll get $\alpha = 0,$ which contradicts $Q_n \neq 0.$ Therefore $Q_0 = Q_1= \cdots = Q_n=0.$

Existence of the representation : We only need to prove the claim for $A=x^n.$ The proof is by induction over $n.$ It is clear for $n = 0,$ Suppose that the claim is true for any $k < n.$ If $n < \deg B,$ then choose $A=Q_0$ and $Q_1 = \cdots = Q_n=0.$ So we may assume that $n \geq \deg B.$ Let $B=b_mx^m + b_{m-1}x^{m-1}+ \cdots + b_0.$ Therefore, since $b_m$ is a unit, we will have $x^m=b_m^{-1}B-b_m^{-1}b_{m-1}x^{m-1} - \cdots - b_m^{-1}b_0,$ which will give us $x^n = b_m^{-1}x^{n-m}B - b_m^{-1} b_{m-1}x^{n-1} - \cdots - b_m^{-1}b_0 x^{n-m}.$ Now apply the induction hypothesis to each term $x^{n-k}, \ 1 \leq k \leq m,$ to finish the proof.

Module-finiteness of Laurent polynomial rings

Posted: March 11, 2010 in Elementary Algebra; Problems & Solutions, Rings and Modules
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Let $R$ be a commutative ring with identity and $S=R[x,x^{-1}],$ the ring of Laurent polynomials with coefficients in $R.$ Obviously $S$ is not a finitely generated $R$-module but we can prove this:

Problem. There exists $f \in S$ such that $S$ is a finitely generated $R[f]$-module.

Solution. Let $f=x+x^{-1}.$ Then $x=f - x^{-1}$ and $x^{-1}=f-x.$ Now an easy induction shows that $x^n \in xR[f]+R[f]$ for all $n \in \mathbb{Z}.$ Hence $S=xR[f] + R[f]. \ \Box$

x being reducible in R[x]

Posted: December 1, 2009 in Elementary Algebra; Problems & Solutions, Rings and Modules
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There are many commutative rings $R$ satisfying this property that the indeterminate $x$ is reducible in the polynomial ring $R[x]$. Here are two examples: in $(\mathbb{Z}/6\mathbb{Z})[x]$ we have $x=(3x+4)(4x+3)$ and in $(\mathbb{Z}/10\mathbb{Z})[x]$ we have $x=(5x+4)(4x+5).$ In general, if a commutative ring $R$  has an idempotent $e \neq 0,1,$ then

$x=(ex + 1-e)((1-e)x + e).$

So if an integer $n > 1$ has at least two distinct prime divisors, then $x$ will be reducible in $(\mathbb{Z}/n\mathbb{Z})[x].$