Linked Questions

13
votes
1answer
5k views

Zero divisors in polynomial rings [duplicate]

The following is an exercise in Hungerford (Ch. III, ex. 5.6). Let $R$ be a commutative ring with identity. If $f=a_nx^n+\dots+a_0$ is a zero divisor in $R[x]$, then there exists a nonzero $b$ in ...
2
votes
0answers
81 views

Zero divisor polynomial [duplicate]

Let $f(x)\in R[x]$ be a zero divisor. How to prove that there is an element $0\neq a\in R$ such that $af(x) = 0$? If $R$ has no nilpotent elements, it is easy. What about the general case? Can ...
0
votes
0answers
45 views

Let $A$ be a commutative ring and $f\in A[X]$ with $f \not\equiv 0$. Show that if $f|0$ in $A[X]$ then $\exists a \in A$ st $a \neq 0$ and $af = 0$ [duplicate]

For example, let $f= \sum_{i=0}^n a_ix^i$. If every $a_i\not\in U(A)$, then $\exists b_i\in A$ with $b_i \neq 0$ for every $i$ such that $a_ib_i=0$, so I can take $a = \prod b_i$ reaches what I need. ...
2
votes
1answer
44 views

Zero divisor for polynomial ring [duplicate]

I am having trouble with how to begin with this problem from Abstract Algebra by Dummit and Foote (2nd ed): Let $R$ be a commutative ring with 1. Let $p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be ...
0
votes
0answers
33 views

Zero divisors in polynomial in a commutative ring [duplicate]

There's an exercise in Herstein marked as very hard Let $R$ be a commutative ring. If $q(x) \in R[x]$ be a zero divisor in $R[x]$, then if $\displaystyle q(x) = \sum_{0 \leq i \leq k} a_i x^i$, ...
0
votes
0answers
20 views

The coefficients of a polynomial over a commutative ring without nilpotent elements [duplicate]

Claim: Let $R$ be a commutative ring with no nonzero nilpotent elements. If $0 \not = f(x) = a_0 + a_1x + \dots + a_mx^m \in R[x]$ is a zero-divisor, then there is a nonzero $b \in R$ such that $ba_0 =...
24
votes
5answers
2k views

Why are polynomials defined to be “formal”?

Despite the fact that $\forall n, n^3 + 2n \equiv 0 \pmod 3$, I understand that $n^3 + 2n$ (considered as a polynomial with coefficients in $\mathbb Z/3\mathbb Z$) is not equal to the zero polynomial. ...
0
votes
3answers
396 views

Ring without zero divisor [closed]

Does there exist a ring $R[x]$ without a zero divisor but the ring R is having a zero divisor.
5
votes
3answers
1k views

Zero-divisors and units in $\mathbb Z_4[x]$

Consider the ring $\mathbb Z_4[x]$. Clearly the elements of the form $2f(x)$ are zero divisors. 1. Is it true that they are all the zero divisors? I mean is it true that if $p(x)$ is a zero ...
3
votes
4answers
739 views

Prove that $R$ is an integral domain $\Leftrightarrow$ $R[x]$ is an integral domain

Here is an exercise(p.129, ex.1.15) from Algebra: Chapter 0 by P.Aluffi. Prove that $R[x]$ is an integral domain if and only if $R$ is an integral domain. The implication part makes no problems, ...
5
votes
1answer
404 views

Zero divisors in $A[x_1,x_2,\dots,x_r]$

Let $A$ be a commutative ring. I am trying to show that if $f(x_1,x_2,\ldots, x_r) \in A[x_1,x_2,\ldots, x_r]$ is a zero divisor then there exists $a$ in $A-\{0\}$ such that $af=0$ in $A[x_1,x_2,\...
3
votes
3answers
157 views

Prove that: $R[x]$ has a zero divisor $\Rightarrow$ $R$ has a zero divisor

The problem I've been trying to solve the above problem. There seems to already be some work regarding zero divisors in polynomial rings over here, but I'm not sure it is applicable to me, because it ...
0
votes
1answer
801 views

If $p$ is a prime ideal then $p[X]$ is a prime ideal

If $Z$ is a ring and $p$ is a prime ideal of $Z$ then $p[X]$ is a prime ideal of $Z[X]$. Is it true or false? I believe that it is true and I try to prove it like that: Take $f(x)\in p[X]$ and ...
0
votes
2answers
634 views

General questions about Polynomial Rings [closed]

I'm learning about polynomial rings in my class. My instructor and book are both spectacularly unhelpful and didn't even bother to define most of the terms in my homework. So I have some general ...
1
vote
1answer
864 views

$R$ has nonzero nilpotent elements…

Let $R$ be a commutative ring with 1. a) Suppose $R$ has no nonzero nilpotent elements (that is, $a^n=0$ implies $a=0$). If $f(X)=a_0+a_1X+\cdots+a_nX^n$ in $R\left[X\right]$ is a zero-divisor, ...

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