# Thread: A commutative Ring R has no zero divisors

1. ## A commutative Ring R has no zero divisors

Show that if R has no zero divisors, then R[x] has no zero divisors.

Since R has no zero divisors, then $\forall a,b\in R$ such that $a,b\neq 0$, $ab\neq 0$.

Is this next part correct (not sure what to do next)?

$R[ab]=R[a]R[b]\neq 0$

2. ## Re: A commutative Ring R has no zero divisors

Is it not easier to say, if $a,b$ are no zero divisors, then $\forall a,b \in R$: $a.b=0$ such $a=0$ or $b=0$.

3. ## Re: A commutative Ring R has no zero divisors

Originally Posted by Siron
Is it not easier to say, if $a,b$ are no zero divisors, then $\forall a,b \in R$: $a.b=0$ such $a=0$ or $b=0$.
My question isn't how to define a zero divisor. Regardless of how you or I want to define it, how can I use that fact to do the proof?

4. ## Re: A commutative Ring R has no zero divisors

Yes indeed, I agree with your proof. Because if $R[X]$ is a polynomial ring, his coefficients are $\in R$.

5. ## Re: A commutative Ring R has no zero divisors

Originally Posted by dwsmith
Show that if R has no zero divisors, then R[x] has no zero divisors.

Since R has no zero divisors, then $\forall a,b\in R$ such that $a,b\neq 0$, $ab\neq 0$.

Is this next part correct (not sure what to do next)?

$R[ab]=R[a]R[b]\neq 0$
This doesn't even make sense. R[x] is the ring of polynomials with variable x, with coeffcients in x. I have no idea what you mean by "R[a]" where a is a member of R. The best I can think of is that it is just a complicated way of saying "R"!

6. ## Re: A commutative Ring R has no zero divisors

@HallsofIvy:
You're absolutely right, but is it not enough to say that the coefficients of the polynomial ring R[X] are $\in R$?
And I think dwsmith means with:
$R[a]=\sum_{i=0}^{n} a_iX^{i}$
$R[b]=\sum_{i=0}^{n} b_iX^{i}$

That's how I interpreted it.

7. ## Re: A commutative Ring R has no zero divisors

Originally Posted by dwsmith
Show that if R has no zero divisors, then R[x] has no zero divisors.

Since R has no zero divisors, then $\forall a,b\in R$ such that $a,b\neq 0$, $ab\neq 0$.

Is this next part correct (not sure what to do next)?

$R[ab]=R[a]R[b]\neq 0$

The first thing that comes to my mind is to prove the contrapositive statement. That is I'm going to prove that if $R[X]$ has any zero divisors the so does $R$. Here goes:

Let $P,Q \in R[X]$ such that $P,Q \neq 0$ and $PQ = 0$ (the zero polynomial). We will write

$P = \sum_{k=0}^{n} a_k x^k$ and $Q= \sum_{i=0}^{m} b_i x^i.$.

Now since $P$ and $Q$ are nonzero then there exist some coefficients $a_{k_1}$ and $b_{i_1}$ such that $a_{k_1},b_{i_1} \neq 0$ (these coefficients are in $R$). Now we multiply $P$ and $Q$ to get

$PQ = \sum_{k=0}^{n} \sum_{i=0}^{m} a_k b_i x^{k+i}=0$.

But we know that the only way for $PQ$ to be the zero polynomial is if all its coefficients are $0$, and since all the coefficients are zero then $a_{k_1}b_{i_1} = 0$, but $a_{k_1},b_{i_1} \neq 0$ so $a_{k_1}$ and $b_{i_1}$ must be zero divisors. qed

8. ## Re: A commutative Ring R has no zero divisors

Originally Posted by dwsmith
Show that if R has no zero divisors, then R[x] has no zero divisors.

Since R has no zero divisors, then $\forall a,b\in R$ such that $a,b\neq 0$, $ab\neq 0$.

Is this next part correct (not sure what to do next)?

$R[ab]=R[a]R[b]\neq 0$
The simple way to do this is to note that if $R$ is an integral domain then $\deg(pq)=\deg(p)+\deg(q)$. So, break the problem in two cases, either $p,q\in R$ or $\max\{\deg(p),\deg(q)\}\geqslant 1$ from where it follows that $\deg(pq)\geqslant 1$ and so $pq\ne 0$.

Perhaps simpler to say, is just look at the leading coefficient of $pq$ and you tell me why the conclusion follows.

9. ## Re: A commutative Ring R has no zero divisors

A little simpler than obd2 but it's the same idea :

$P,Q \in \mathcal R [X]\backslash\{0\}$ such that $PQ=0$

Let $k_P = \max \{k\in\mathbb N / X^k | P\}$ and $k_Q = \max \{k\in\mathbb N / X^k | Q\}$ (those exist because of the non nullity of $P$ and $Q$). Let now $R = \frac P {X^{k_P}}, S = \frac Q {X^{k_Q}}$.

One have $0 = PQ = X^{k_P}X^{k_Q}RS$.
So $r_0s_0$ is the coefficient of $X^{k_P+k_Q}$. And so it is null : $r_0s_0 = 0$. So is $r_0 = 0$ or $s_0 = 0$.

If $r_0 = 0$, it's a contradiction with the definition of $k_P$. If $s_0 = 0$, it's a contradiction with the definition of $k_Q$.

QED.

P.S. : I'm new on this forum, and I love the $\LaTeX$ integration. It's really more efficient and beautiful than in other forums.