# Polynomial ring zero divisors

Homework Statement:
Let ##R## be a commutative ring with unity. Let ##p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + 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 element ##b \in R## such that ##bp(x) = 0##. Hint: Let ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## be a nonzero polynomial of minimal degree 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 also 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, \dots, n## and show that this implies ##b_mp(x) = 0##.
Relevant Equations:
,
Proof: ##(\Leftarrow)## Suppose there exists non zero ##b \in R## such that ##bp(x) = 0##. Well, ##R \subset R[x]##, and so by definition of zero divisor, ##p(x)## is a zero divisor. (assuming ##p(x) \neq 0##).

##(\Rightarrow)## Suppose ##p(x)## is a zero divisor in ##R[x]##. Then we can choose a nonzero polynomial of minimal degree ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## such that ##g(x)p(x) = 0##. From the reading, the coefficient of ##x^{m+n}## of ##g(x)p(x)## is

$$\sum_{k = 0}^{n+m} a_kb_{n+m-k} = a_nb_m$$

since for ##k > n## we have ##a_k = 0## and for ##k < n## we have ##b_{n+m-k} = 0##. Since ##g(x)p(x) = 0## we can conclude ##a_nb_m = 0##, I think? Now, ##\deg a_ng(x) < \deg g(x)## and ##(a_ng(x))p(x) = a_n(g(x)p(x)) = a_n\cdot 0 = 0##. This implies ##a_ng(x) = 0## since it has degree less than ##g(x)## and thus is not a zero divisor of ##p(x)##.

We will show in general that for all ##i = 0, 1, \dots, n##, ##a_{n-i}g(x) = 0##. We've shown this for ##i = 0## and proceed by induction. I am stuck here. Any advice, please?

Also a question, is ##R[[x]] \subseteq R[x]## where ##R[[x]]## is the ring of formal power series with coefficients from ##R##? By the wikipedia, it seems every polynomial in ##R[x]## has finite degree, but I just wanted to confirm.

fresh_42
Mentor
Homework Statement:: Let ##R## be a commutative ring with unity. Let ##p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + 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 element ##b \in R## such that ##bp(x) = 0##. Hint: Let ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## be a nonzero polynomial of minimal degree 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 also 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, \dots, n## and show that this implies ##b_mp(x) = 0##.
Relevant Equations:: ,

Proof: ##(\Leftarrow)## Suppose there exists non zero ##b \in R## such that ##bp(x) = 0##. Well, ##R \subset R[x]##, and so by definition of zero divisor, ##p(x)## is a zero divisor. (assuming ##p(x) \neq 0##).

##(\Rightarrow)## Suppose ##p(x)## is a zero divisor in ##R[x]##. Then we can choose a nonzero polynomial of minimal degree ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## such that ##g(x)p(x) = 0##. From the reading, the coefficient of ##x^{m+n}## of ##g(x)p(x)## is

$$\sum_{k = 0}^{n+m} a_kb_{n+m-k} = a_nb_m$$

since for ##k > n## we have ##a_k = 0## and for ##k < n## we have ##b_{n+m-k} = 0##. Since ##g(x)p(x) = 0## we can conclude ##a_nb_m = 0##, I think?
This is more than confusing. We have ##g(x)p(x)=b_ma_nx^{n+m}+r(x)=0## with ##\deg r(x)<n+m## and so ##b_ma_n=0##.
Now, ##\deg a_ng(x) < \deg g(x)## and ##(a_ng(x))p(x) = a_n(g(x)p(x)) = a_n\cdot 0 = 0##. This implies ##a_ng(x) = 0## since it has degree less than ##g(x)## and thus is not a zero divisor of ##p(x)##.
Wrong. It is a zero divisor by assumption. The contradiction only requires ##\deg g(x)=0## since the argument needed an ##x-##term. We are done here. I don't understand neither the rest of this post nor the induction hint.

The proof goes:
##p## a zero divisor ##\Longrightarrow## ##g\cdot p=0## for some ##g##. Hence we can choose one of minimal degree. Assume ##\deg g > 0 ## then contradiction, ergo ##\deg g = 0##. Done. The other direction is trivial.
We will show in general that for all ##i = 0, 1, \dots, n##, ##a_{n-i}g(x) = 0##. We've shown this for ##i = 0## and proceed by induction. I am stuck here. Any advice, please?

Also a question, is ##R[[x]] \subseteq R[x]## where ##R[[x]]## is the ring of formal power series with coefficients from ##R##? By the wikipedia, it seems every polynomial in ##R[x]## has finite degree, but I just wanted to confirm.

This is more than confusing. We have ##g(x)p(x)=b_ma_nx^{n+m}+r(x)=0## with ##\deg r(x)<n+m## and so ##b_ma_n=0##.

Wrong. It is a zero divisor by assumption. The contradiction only requires ##\deg g(x)=0## since the argument needed an ##x-##term. We are done here. I don't understand neither the rest of this post nor the induction hint.

The proof goes:
##p## a zero divisor ##\Longrightarrow## ##g\cdot p=0## for some ##g##. Hence we can choose one of minimal degree. Assume ##\deg g > 0 ## then contradiction, ergo ##\deg g = 0##. Done. The other direction is trivial.
Thank you for the response and the corrections. I'm sorry but i'm confused about the contradiction part.

Proof: ##(\Rightarrow)## Assume ##p## is a zero divisor. Then we can choose a non zero ##g \in R[x]## of minimal degree such that ##gp = 0##. Assume by contradiction that ##\deg g > 0##. We have
$$p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$ and
$$g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0$$

and
$$g(x)p(x) = a_nb_mx^{n+m} + r(x) = 0$$
where ##\deg r(x) < n+m##. This implies ##a_nb_m = 0##. Why is ##a_ng(x)## is a zero divisor, if I understand your above post? I think ##a_ng(x) = 0## because we have ##\deg g(x) > \deg a_ng(x)##. My question is sincere, I understand I am wrong.

fresh_42
Mentor
Thank you for the response and the corrections. I'm sorry but i'm confused about the contradiction part.

Proof: ##(\Rightarrow)## Assume ##p## is a zero divisor. Then we can choose a non zero ##g \in R[x]## of minimal degree such that ##gp = 0##. Assume by contradiction that ##\deg g > 0##.
You can either say "Assume ##\deg g > 0##" (nothing with contradiction here) or distinguish cases: If ##g=b_0## we are done, otherwise there is a ##g## of positive degree. Out of all those we choose one of minimal degree. (That's important. And since the natural numbers - the possible degrees - are bounded from below, such a polynomial exists.)
We have
$$p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$ and
$$g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0$$

and
$$g(x)p(x) = a_nb_mx^{n+m} + r(x) = 0$$
where ##\deg r(x) < n+m##. This implies ##a_nb_m = 0##. Why is ##a_ng(x)## is a zero divisor, if I understand your above post?
The way you have proven it in the original post. I just was lazy and didn't repeat it. Maybe I should have used dots. ##g## is a zero divisor per assumption. Then ##b_ma_n=0## as the highest degree of the zero polynomial ##0=gp##. Now as before: ##\deg (a_ng) < \deg g## and ##(a_ng)p=0## contradicting the minimality of ##\deg g##. Hence ##\deg g> 0 ## leads to a contradiction, or in case of prove by distinction of cases: such a case does not exist.
I think ##a_ng(x) = 0## because we have ##\deg g(x) > \deg a_ng(x)##. My question is sincere, I understand I am wrong.
No, you were right in your argument, except that it was too long and had redundancies. Wrong was the phrase
... and thus [g(x)] is not a zero divisor of ##p(x)##.
It is, since we have chosen it as such. You cannot say it isn't. The contradiction is the minimal degree, not the zero dividing.

fishturtle1
You can either say "Assume ##\deg g > 0##" (nothing with contradiction here) or distinguish cases: If ##g=b_0## we are done, otherwise there is a ##g## of positive degree. Out of all those we choose one of minimal degree. (That's important. And since the natural numbers - the possible degrees - are bounded from below, such a polynomial exists.)

The way you have proven it in the original post. I just was lazy and didn't repeat it. Maybe I should have used dots. ##g## is a zero divisor per assumption. Then ##b_ma_n=0## as the highest degree of the zero polynomial ##0=gp##. Now as before: ##\deg (a_ng) < \deg g## and ##(a_ng)p=0## contradicting the minimality of ##\deg g##. Hence ##\deg g> 0 ## leads to a contradiction, or in case of prove by distinction of cases: such a case does not exist.

No, you were right in your argument, except that it was too long and had redundancies. Wrong was the phrase

It is, since we have chosen it as such. You cannot say it isn't. The contradiction is the minimal degree, not the zero dividing.
Sorry for the late response.

Proof: ##(\Rightarrow)## Assume ##p## is a zero divisor. Then we can choose nonzero ##g \in R[x]## of minimal degree such that ##gp = 0##. If ##\deg g = 0##, then we are done. Else ##\deg g > 0##. We have
$$p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$
$$g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0$$
and so ##g(x)p(x) = b_ma_nx^{m+n} + r(x) = 0## where ##\deg r < m + n##. This shows ##b_ma_n = 0##. We can conclude ##(a_ng)p = 0## and ##\deg a_ng < \deg g## and ##a_ng \neq 0##. This contradicts the minimality of the degree of ##g##.

We can conclude that ##\deg g = 0##.

##(\Leftarrow)## Suppose there exists nonzero ##b \in R## such that ##bp = 0##. Well ##b## is a nonzero element in ##R[x]##. We can conclude ##p## is a zero divisor in ##R[x]##. []

Is this correct?

fresh_42
Mentor
Close. I think I start to understand the rest of the hint. ##a_ng=0## cannot be ruled out, which you correctly listed as necessary. What if ##a_n## kills all coefficients of ##g##?

Sorry, my bad. I haven't seen that. You can repair this by looking at the next term at ##x^{n+m-1}## and iterate.

fishturtle1
member 587159
Also a question, is ##R[[x]] \subseteq R[x]## where ##R[[x]]## is the ring of formal power series with coefficients from ##R##? By the wikipedia, it seems every polynomial in ##R[x]## has finite degree, but I just wanted to confirm.

Nope ##1+X+X^2+X^3+\dots \in R[[X]]\setminus R[X]##. The other inclusion holds though.

fishturtle1
I'm not sure.. We have to consider the case where ##a_ng = 0##. We want to show ##a_{n-i}g= 0## for all ##i = 0, 1, \dots, n##. This holds for ##i = 0##. We proceed by induction on ##i##.

Well,

$$(a_{n-(i+1)}g)p = (a_{n-i-1}g)p = 0$$

and then I think I have to look at the coefficient of ##x^{n-i-1}##?

Hi, I'm still having trouble with the induction step. This is what I have:

Proof: ##(\Rightarrow)## Suppose ##p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0## is a zero divisor in ##R[x]##. Then we can choose a polynomial of minimal degree ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## in ##R[x]## such that ##pg = 0##. If ##\deg g = 0##, we're done. Otherwise, we consider that ##pg = 0## implies ##a_nb_m = 0##. So, ##(a_ng)p = 0## and ##\deg a_ng < \deg g##. If ##a_ng \neq 0##, then we've reached contradiction.

Otherwise ##a_ng = 0## and we need to show ##a_{n-1}g = 0##? I'm not sure how to.

Edit: From the textbook,

##g(x)p(x) = \sum_{l=0}^{n+m} c_lx^l## where ##c_l = \sum_{k=0}^{l} a_kb_{l-k}##

fresh_42
Mentor
Otherwise ##a_ng = 0## and we need to show ##a_{n-1}g = 0##? I'm not sure how to.
Assume ##a_ng=a_nb_mx^m+a_nb_{m-1}x^{m-1}+\ldots+a_nb_1x+a_nb_0=0##. We also know that ##pg=a_nb_mx^{n+m}+(a_nb_{m-1}+a_{n-1}b_m)x^{n+m-1}+\ldots =0##.

The first equation gives us ##a_nb_{m-1}=0## and the second ##(a_nb_{m-1}+a_{n-1}b_m)=0##. Combining both yields ##a_{n-1}b_m=0##. Now consider ##a_{n-1}g=a_{n-1}b_mx^m+a_{n-1}b_{m-1}x^{m-1}+\ldots = a_{n-1}b_{m-1}x^{m-1}+\ldots## which is again a zero divisor of ##p## of less degree than ##g##. If it is different from zero, we are done.

Assume ##a_{n-1}g=a_{n-1}b_mx^m+a_{n-1}b_{m-1}x^{m-1}+\ldots+a_{n-1}b_1x+a_{n-1}b_0=0##. We also know that ##pg=a_nb_mx^{n+m}+(a_nb_{m-1}+a_{n-1}b_m)x^{n+m-1}+(a_nb_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_m)x^{n+m-2}+\ldots =0##. Now we look at ##a_nb_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_m = a_{n-2}b_m=0## and ##a_{n-2}g## etc.

This is the idea. I'm sure it can be written in a better, less algorithmic way.

fishturtle1
Assume ##a_ng=a_nb_mx^m+a_nb_{m-1}x^{m-1}+\ldots+a_nb_1x+a_nb_0=0##. We also know that ##pg=a_nb_mx^{n+m}+(a_nb_{m-1}+a_{n-1}b_m)x^{n+m-1}+\ldots =0##.

The first equation gives us ##a_nb_{m-1}=0## and the second ##(a_nb_{m-1}+a_{n-1}b_m)=0##. Combining both yields ##a_{n-1}b_m=0##. Now consider ##a_{n-1}g=a_{n-1}b_mx^m+a_{n-1}b_{m-1}x^{m-1}+\ldots = a_{n-1}b_{m-1}x^{m-1}+\ldots## which is again a zero divisor of ##p## of less degree than ##g##. If it is different from zero, we are done.

Assume ##a_{n-1}g=a_{n-1}b_mx^m+a_{n-1}b_{m-1}x^{m-1}+\ldots+a_{n-1}b_1x+a_{n-1}b_0=0##. We also know that ##pg=a_nb_mx^{n+m}+(a_nb_{m-1}+a_{n-1}b_m)x^{n+m-1}+(a_nb_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_m)x^{n+m-2}+\ldots =0##. Now we look at ##a_nb_{m-2}+a_{n-1}b_{m-1}+a_{n-2}b_m = a_{n-2}b_m=0## and ##a_{n-2}g## etc.

This is the idea. I'm sure it can be written in a better, less algorithmic way.
Thank you. We have

$$f(x)g(x) = \sum_{l=0}^{n+m}c_l$$

where $$c_k = \sum_{l=0}^{k}a_lb_{l-k}$$

We've shown ##a_{n-i}g = 0## for ##i = 0##. We want to show this for ##i = 0, 1, 2, \dots, n##. We proceed by induction on ##i##.

Well, $$c_{n+m-i} = \sum_{k=0}^{n+m-i}a_kb_{n+m-i-k} = \sum_{k=n-i}^{n}a_kb_{n+m-i-k} = a_{n-i}b_m + a_{n-i+1}b_{m-1} + \dots + a_{n-1}b_{m-i+1} + a_nb_{m-i} = a_{n-i}b_m= 0$$

This completes the induction. This implies ##a_{n-i}b_m = 0## for all ##i## and so ##b_mp(x) = 0##. []

i'm not super confident what i've written is correct, but this is what I have so far.

fresh_42
Mentor
I would rewrite it from scratch.

Let ##gp=0## with ##g=b_mx^m+\ldots+b_0 \neq 0## of minimal degree with these properties.

Statement: ##b_0p=0##.

If ##\deg g = 0## then there is nothing to show.

Now let's test the prototype for the induction step.

If ##\deg g = 1## then ##0=b_1xp+b_0p##, i.e. ##b_1a_n=0##. Then ##a_ngp=0## and ##\deg (a_ng)<\deg g##. By minimality of ##\deg g## we get ##a_ng=0## and so ##a_nb_1=a_nb_0=0##.
...

How would you go on? If you can prove it for this ##g##, then assume it for all ##g## with ##\deg = k <m## and show it for ##\deg g = m##. It is a bit tricky, as more than one method of proof is involved. That's why I would start from the beginning. Otherwise there is a chance to build in flaws.

Thank you for the response.

Can we possibly avoid doing a proof by contradiction like this?

Proof: ##(\Rightarrow)## Let ##p(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 \in R[x]## and suppose ##p## is a zero divisor. We can choose an element ##g(x) = b_mx^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0## of minimal degree such that ##gp = 0##. This implies ##a_nb_m = 0##. So ##\deg a_ng < \deg g## and ##(a_ng)p = 0##. Thus, ##a_ng = 0##. In particular ##a_nb_m = a_nb_{m-1} = \dots = a_nb_{1} = a_nb_0 = 0##. We'd like to show ##a_{n-i}g = 0## for ##i = 0, 1, \dots, n##. Suppose ##a_{n-j}g = 0## is true up to some ##j = i-1 \ge 0##. We'll show ##a_{n-i}g = 0##. Observe,

$$p(x)g(x) = \sum_{k=0}^{n+m}c_kx^k$$
where $$c_k = \sum_{l=0}^k a_lb_{k-l}$$

Since ##pg = 0## we have

\begin{align*}
c_{n+m-i} &= 0 \\
\sum_{k=0}^{n+m-i}a_kb_{n+m-i-k} &= 0 \\
\sum_{k=n-i}^{n}a_kb_{n+m-i-k} &= 0 \\
a_{n-i}b_m + 0 + \dots + 0 &= 0 \\
a_{n-i}b_m &= 0 \\
\end{align*}

This implies ##(a_{n-i}g)p = 0## and ##\deg a_{n-i}g < \deg g##. Therefore ##a_{n-i}g = 0##. In particular ##a_{n-i}b_m = a_{n-i}b_{m-1} = \dots = a_{n-i}b_1 = a_{n-i}b_0 = 0##. This completes the induction.

Observe, ##b_mp = b_ma_nx^n + b_ma_{n-1}x^{n-1} + \dots + b_ma_1x + b_ma_0 = 0 + 0 + \dots + 0 + 0 = 0##, as desired.

##(\Leftarrow)## Suppose there is nonzero ##b \in R## such that ##bp = 0##. By definition, ##bp## is a zero divisor in ##R[x]##.

fresh_42
Mentor
You have a strange wording: up to some ## i -1##. You cannot use a running index as a certain border: either it is running, or it is fixed - not both at a time. I therefore do not see how your ##c_k## look like, i.e. how you are allowed to write them as you did.

I wouldn't use an induction, because we have a finite number of terms which all can be checked. We can write it as an and-so-on, but the best way is likely to define ##k:=\min\{\,0\leq j \leq n\,\,|\,a_nb_m=\ldots =a_jb_m=0\}##. With ##a_nb_m=0## we know that such a minimum exists, i.e. ##0\leq k\leq n.##

If we now could show that this implies ##a_{k-1}b_m=0## then all ##a_jb_m=0## and ##b_mp=0\,.##

(I am not sure whether it is sufficient to define ##k## with ##b_m##. Perhaps it is necessary or easier to use ##g##, but then the way back to ##b_m## is ugly.)

If you insist on an induction, then you have to go over ##\deg g = m##. But already the case ##m=1## looks like:

\begin{align*}
gp & =0 =\underbrace{ b_1a_n}_{=0}x^{n+1} + (b_1a_{n-1}+ \underbrace{b_0a_n}_{=0}) x^n+ (b_1a_{n-2}+b_0a_{n-1})x^{n-1}+ \ldots \\
& \Longrightarrow \\
b_1a_{n-1}&=0\\
&\Longrightarrow \\
a_{n-1}gp &= 0 \;\wedge \;\deg (a_{n-1}g) < \deg g \\
&\Longrightarrow \\
b_1a_{n-2} &=0 \\
&\Longrightarrow \\
&\vdots \\
&\Longrightarrow \\
b_1a_1&=0\\
&\Longrightarrow \\
gp&=0=(b_1x+b_0)p=b_1a_0x+b_0p\\
&\Longrightarrow \\
b_1a_0=&0\\
&\Longrightarrow \\
b_0p&=0
\end{align*}

The conclusion ##deg (c\cdot g) < \deg g## has to appear somewhere! I tried to write it as the and-so-on version, but I think the version with the minimal index ##k## where all higher coefficients ##a_kb_m=a_{k+1}b_m=\ldots =a_nb_m=0## is the best. The idea: if ##k## then also ##k-1## which means ##k## is already the lowest possible, i.e. ##k=0##.