# 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 $\displaystyle \forall a,b\in R$ such that $\displaystyle a,b\neq 0$, $\displaystyle ab\neq 0$.

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

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

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

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

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

Originally Posted by Siron
Is it not easier to say, if $\displaystyle a,b$ are no zero divisors, then $\displaystyle \forall a,b \in R$: $\displaystyle a.b=0$ such $\displaystyle a=0$ or $\displaystyle b=0$.
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 $\displaystyle R[X]$ is a polynomial ring, his coefficients are $\displaystyle \in R$.

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 $\displaystyle \forall a,b\in R$ such that $\displaystyle a,b\neq 0$, $\displaystyle ab\neq 0$.

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

$\displaystyle R[ab]=R[a]R[b]\neq 0$
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 $\displaystyle \in R$?
And I think dwsmith means with:
$\displaystyle R[a]=\sum_{i=0}^{n} a_iX^{i}$
$\displaystyle R[b]=\sum_{i=0}^{n} b_iX^{i}$

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 $\displaystyle \forall a,b\in R$ such that $\displaystyle a,b\neq 0$, $\displaystyle ab\neq 0$.

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

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

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

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

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

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

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

But we know that the only way for $\displaystyle PQ$ to be the zero polynomial is if all its coefficients are $\displaystyle 0$, and since all the coefficients are zero then $\displaystyle a_{k_1}b_{i_1} = 0$, but $\displaystyle a_{k_1},b_{i_1} \neq 0$ so $\displaystyle a_{k_1}$ and $\displaystyle b_{i_1}$ 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 $\displaystyle \forall a,b\in R$ such that $\displaystyle a,b\neq 0$, $\displaystyle ab\neq 0$.

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

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

Perhaps simpler to say, is just look at the leading coefficient of $\displaystyle pq$ 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 :

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

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

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

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

QED.

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