# Zero divisor for polynomial rings

• MHB
• cbarker1
In summary: Apply this argument to $a_{n-i}g(x)$ for $i=0,1,\cdots,n$ by induction and show that implies $b_mp(x)=0$.In summary, the problem from Abstract Algebra by Dummit and Foote (2nd ed) discusses proving that an element in the polynomial ring $R[x]$ is a zero divisor if and only if there exists a nonzero element $b\in R$ such that $bp(x)=0$. The hint suggests using a nonzero polynomial $g(x)$ of minimal degree such that $g(x)p(x)=0$ and showing that $b_ma_n=0$ and $a_ng(x)=0$. This can be applied to $a_{n-i cbarker1 Gold Member MHB Dear Everybody,I am having trouble with how to begin with this problem from Abstract Algebra by Dummit and Foote (2nd ed): Let$R$be a commutative ring with 1. Let$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$be an element of the polynomial ring$R[x]$. Prove that$p(x)$is a zero divisor in$R[x]$if and only if there is a nonzero$b\in R$such that$bp(x)=0$. Hint: Let$g(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0$be a nonzero polynomial of minimal degree of such that$g(x)p(x)=0$. Show that$b_ma_n=0$and so$a_ng(x)$is a polynomial of degree than$m$that gives 0 when multiplied by$p(x)$. Conclude that$a_ng(x)=0$. Apply a similar argument to show by induction on$i$that$a_{n-i}g(x)=0$for$i=0,1,\cdots,n$and show that implies$b_mp(x)=0$. Thanks CBarker1 Cbarker1 said: Dear Everybody,I am having trouble with how to begin with this problem from Abstract Algebra by Dummit and Foote (2nd ed): Let$R$be a commutative ring with 1. Let$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$be an element of the polynomial ring$R[x]$. Prove that$p(x)$is a zero divisor in$R[x]$if and only if there is a nonzero$b\in R$such that$bp(x)=0$. Hint: Let$g(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0$be a nonzero polynomial of minimal degree of such that$g(x)p(x)=0$. Show that$b_ma_n=0$and so$a_ng(x)$is a polynomial of degree less than$m$that gives 0 when multiplied by$p(x)$. Conclude that$a_ng(x)=0$. Apply a similar argument to show by induction on$i$that$a_{n-i}g(x)=0$for$i=0,1,\cdots,n$and show that implies$b_mp(x)=0$. Thanks CBarker1 My attempt at least the implication, but not the converse: Suppose$p(x)$is a zero divisor. Let$g(x)=b_mx^m+b_{m-1}x^{m-1}+\cdots+b_0$be a nonzero polynomial of minimal degree of such that$g(x)p(x)=0$. Then$g(x)p(x)=(b_mx^m+\cdots+b_0)(a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0)=b_ma_nx^(m+n)+\cdots+b_0a_0=0$by the assumption. so$a_ng(x)$is a polynomial of degree less than$m$that gives 0 when multiplied by$p(x)$. Thus,$a_ng(x)=0$because$R\$ is commutative then the polynomial ring is, too.

## What is a zero divisor for polynomial rings?

A zero divisor for polynomial rings is an element in the ring that, when multiplied by another element, results in the product being equal to zero. In other words, it is an element that has a non-zero product with another element, but the product is equal to zero.

## Why are zero divisors important in polynomial rings?

Zero divisors are important in polynomial rings because they can help identify the factors of a polynomial. If a polynomial has a zero divisor, it can be factored into smaller polynomials, making it easier to solve. Additionally, zero divisors can also provide insight into the structure and properties of the polynomial ring.

## Can every polynomial ring have zero divisors?

No, not every polynomial ring has zero divisors. For a polynomial ring to have zero divisors, it must have a non-zero element that, when multiplied by another non-zero element, results in the product being equal to zero. Some polynomial rings, such as integral domains, do not have any zero divisors.

## How are zero divisors related to the zero element in polynomial rings?

The zero element in polynomial rings is the element that, when added to any other element, results in the element being unchanged. Zero divisors are related to the zero element because they both have the property of producing a zero product when multiplied by another element. However, the zero element is unique and does not have a multiplicative inverse, while zero divisors can have multiplicative inverses.

## What is the connection between zero divisors and irreducible polynomials?

Irreducible polynomials are polynomials that cannot be factored into smaller polynomials over a given field. Zero divisors can help identify irreducible polynomials because if a polynomial has a zero divisor, it is not irreducible. This is because if a polynomial can be factored into smaller polynomials, it cannot be irreducible. Therefore, the absence of zero divisors in a polynomial ring can indicate the presence of irreducible polynomials.

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