3
$\begingroup$

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, because $R$ is a subring of $R[x]$. For the ''if'' part, however..

Let $R$ be an integral domain. Now let's review pairs $f(x)=\sum \nolimits a_ix^i,g(x)= \sum \nolimits b_ix_i \in R[x]$ such that $f(x)g(x) = 0$ So, we have $\sum \limits_{k=0}^{\infty} \sum \nolimits_{i+j=k} a_ib_jx^{i+j} = 0$.

Now, I'm not sure how to deduce that $f(x) = 0 \vee g(x) = 0$

If I look at $f(x),g(x)$ such that, for example, $deg((f(x))=3, deg((g(x))=2)$, it makes sense. I begin by something like that "if $f(x)g(x)=0$ then $a_0b_0=0$ then $a_0 = 0 \vee b_0 = 0$. If $a_0 =0 $, then $a_0b_1 = a_0b_2 = 0$" and so on.

As I understand, it comes down to proving the following implication: $(\forall k \in \mathbb{N} \sum \nolimits_{i+j=k} a_ib_j = 0)(1) \Rightarrow ((\forall n \in \mathbb{N} \ \ a_n = 0) \vee (\forall m \in \mathbb{N} \ \ b_m = 0))(2)$

We can say that $(1)$ is a system of equations in $R$. And $(2)$ is it solution.

$\endgroup$
4
$\begingroup$

Suppose that neither $f$ nor $g$ is the zero polynomial. Then there exist non-negative integers $k$ and $l$ and ring elements $a_0,a_1,\dots, a_k$, with $a_k\ne 0$, and $b_0,b_1,\dots,b_l$, with $b_l\ne 0$, such that $$f=a_0+a_1x+\cdots+a_kx^k \quad\text{and}\quad g=b_0+b_1x+\cdots+b_lx^l.$$

The coefficient of $x^{k+l}$ in the product $fg$ is $a_kb_l$. Since $R$ is an integral domain, we have $a_kb_l\ne 0$, and therefore $fg$ is not the zero polynomial.

$\endgroup$
1
$\begingroup$

This is a bit extreme, but this is also a consequence of the famed McCoy's Theorem. If $f(x)\in R[x]$ is a zero divisor, then there is a non-zero $r\in R$ such that $r\cdot f(x)=0$. Thus if $R[x]$ were not a domain, there would be an $f(x)$ which is non-zero and a zero-divisor. But that would give you an $r\in R$ that would kill all of the coefficients of $f(x)$, so $R$ would not be a domain. Bill Dubuque sketches a proof here: Zero divisor in $R[x]$

$\endgroup$
0
$\begingroup$

A simple observation: assume $f,g \in R[x] \setminus \{0\}$ then $$f(x) = \sum_{n=0}^N a_n x^n \ g(x) = \sum_{n=0}^M b_n x^n$$ where $a_N,b_M \in R \setminus \{0\}$.

The leading coefficient of $fg$, i.e. the coefficient of degree $N+M$ is $a_Nb_M$ which is not null since $a_N,b_M \ne 0$ and $R$ is a domain. As a consequence $fg$ as degree at least $N+M$ so it can not be the null polynomial.

$\endgroup$
0
$\begingroup$

A proof by contradiction:

If $R$ is an integral domain, but $R[x]$ is not, there must exist some minimal counter-example, that is some $f$ of minimal $\text{deg}\ n \geq 0$, for which there exists $g \neq 0$ such that $fg = 0$.

If $f(x) = a_0 + a_1x +\cdots + a_nx^n$ and: $g(x) = b_0 + b_1x +\cdots + b_mx^m$

it follows from $fg = 0$ that $a_nb_m = 0$.

But $g \neq 0 \implies b_m \neq 0 \implies a_n = 0$ (since $R$ is an integral domain), contradicting the minimality of $n$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.