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Zero divisor in $R[x]$
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Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b ne 0$ in $R$ such that $ba_0=ba_1=cdots=ba_n=0$?
abstract-algebra polynomials ring-theory
edited Dec 14 '14 at 12:19
user26857
38.7k123678
asked Nov 17 '11 at 17:47
Mohan
5,72494498
add a comment |Â
up vote
28
down vote
favorite
Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b ne 0$ in $R$ such that $ba_0=ba_1=cdots=ba_n=0$?
abstract-algebra polynomials ring-theory
edited Dec 14 '14 at 12:19
user26857
38.7k123678
asked Nov 17 '11 at 17:47
Mohan
5,72494498
add a comment |Â
up vote
28
down vote
favorite
up vote
28
down vote
favorite
Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b ne 0$ in $R$ such that $ba_0=ba_1=cdots=ba_n=0$?
abstract-algebra polynomials ring-theory
edited Dec 14 '14 at 12:19
user26857
38.7k123678
asked Nov 17 '11 at 17:47
Mohan
5,72494498
Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b ne 0$ in $R$ such that $ba_0=ba_1=cdots=ba_n=0$?
abstract-algebra polynomials ring-theory
edited Dec 14 '14 at 12:19
user26857
38.7k123678
asked Nov 17 '11 at 17:47
Mohan
5,72494498
edited Dec 14 '14 at 12:19
user26857
38.7k123678
edited Dec 14 '14 at 12:19
user26857
38.7k123678
edited Dec 14 '14 at 12:19
user26857
38.7k123678
38.7k123678
asked Nov 17 '11 at 17:47
Mohan
5,72494498
asked Nov 17 '11 at 17:47
Mohan
5,72494498
asked Nov 17 '11 at 17:47
Mohan
5,72494498
5,72494498
add a comment |Â
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
33
down vote
accepted
It is true over any commutative ring, and is sometimes called McCoy's theorem. Below is a proof sketch
from my sci.math post on 5/4/2004:
Theorem $ $ Let $ ,F in R[X]$ be a polynomial over a commutative ring $ ,R.,$
If $ ,F,$ is a zero-divisor then $ ,r,F = 0,$ for some nonzero $ ,r in R.$
Proof $ $ Suppose not. Choose $ ,G ne 0,$ of min degree with $ ,F,G = 0.,$
Write $ ,F =, a +,cdots,+ f X^k +,cdots,+ c X^m $
and $ G = b +,cdots,+ g X^n,,$ where $ ,g ne 0:$ and $ ,f,$ is the highest deg coef of $ ,F,$ with $ ,f,G ne 0,$ (note that such an $ ,f,$ exists else $ ,F,g = 0,$ contra supposition).
Then $ ,F,G = (a +,cdots,+ f X^k) (b +,cdots,+ g X^n) = 0.$
Thus $ ,f,g = 0,$ so $ ,deg(f,G) < n,$ and $ , F,(f,G) = 0,,$ contra minimality of $ ,G. $ QED
Alternatively it follows Gauss's Lemma
(Dedekind-Mertens form) or related results.
answered Nov 17 '11 at 20:12
Number
1
1
Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)?
– mathjacks
Sep 7 '14 at 15:54
1
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $deg(fG)<deg(G)$.
– Martin Brandenburg
Sep 27 '14 at 13:51
1
Bill, what an elegant proof! Thanks for sharing this lovely argument.
– Prism
Jun 26 '15 at 20:43
2
@user95553 While $,fg =X^7 - X = 0,$ as a polynomial function on $,Bbb Z_7,,$ it is not true that $,X^7-X = 0,$ as formal polynomials $in Bbb Z_7[X].,$ Formal vs. functional polynomials is discussed in many prior answers, e.g. this one.
– Number
Aug 23 '15 at 15:07
1
@user386627 e.g. $,Fg = 0 (Rightarrow, gF = 0),,$ and $,F(fG) = f(FG)= 0 $
– Number
Feb 28 '17 at 18:35
 |Â
show 5 more comments
up vote
21
down vote
Assume that $gf=0$ for some $gin R[X]$ and let $c$ be the leading coefficient of $g$. Then $ca_n=0$. Therefore $cf$ is either $0$ (in which case $c$ is your $b$), or $cf$ has degree less than $n$ with $g(cf)=0$. Proceed by induction on $n$. In the end you find that some power of $c$ kills every $a_i$, and $c$ was not nilpotent ...
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
add a comment |Â
up vote
6
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Apparently, Bill Dubuque's argument is not really about polynomial rings. Here is a generalization.
Let $A$ be a commutative $mathbbN$-graded ring. Let $f in A$ be a zero divisor. Then there is some $0 neq a in A$ homogeneous such that $a f = 0$.
Proof. Choose some $0 neq g in A$ of minimal total degree with $fg=0$. Let $f=f_0+f_1+cdots$ and $g=g_0+g_1+cdots+g_d$ be the homogeneous decompositions with $g_d neq 0$. If $f g_d = 0$, we are done. Otherwise, we have $f_i g_d neq 0$ for some $i$, and hence $f_i g neq 0$. Choose $i$ maximal with $f_i g neq 0$. Then $0=fg=(f_0+cdots+f_i)g=(f_0+cdots+f_i)(g_0+cdots+g_d)$ implies $f_i g_d = 0$. Then $f_i g$ has smaller degree than $g$, but still satisfies $f(f_i g)=0$ and $f_i g neq 0$, a contradiction. $square$
This may be applied to $A=R[x]$ with the usual grading. Hence, any zero divisor in $R[x]$ is killed by some element of the form $r x^n$ ($r in R setminus 0$) and then also by $r$.
edited Aug 2 '16 at 15:06
user26857
38.7k123678
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
3
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)?
– Pham Hung Quy
Dec 14 '14 at 14:11
1
I've asked about the correctness of this proof here
– Jay
Jul 31 '16 at 21:04
1
@Jay The above proof can be easily fixed if one considers $g$ as having the minimal number of non-zero homogeneous components among the polynomials with the property $fg=0$. (As you can notice, the answerer wanted to mimic the proof given for polynomials in the accepted answer, and this is an easy task.)
– user26857
Aug 2 '16 at 15:01
@user26857: Yeah that makes sense. That's a neat trick, looking at the number of homogeneous components instead of the degree. Good to know this proof can be fixed!
– Jay
Aug 2 '16 at 22:03
Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above.
– Dietrich Burde
Nov 6 '17 at 15:54
add a comment |Â
up vote
4
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This is the case of Armendariz Rings, which I studied in last summer briefly. It is an interesting topic.
A ring $R$ is called Armendariz if whenever $f(x)=sum_i=0^ma_ix^i, g(x)=sum_j=0^nb_jx^j in R[x]$ such that $f(x)g(x)=0$, then $a_ib_j=0 forall i,j$.
In his paper "A NOTE ON EXTENSIONS OF BAER AND P. P. -RINGS" in 1973, Armendariz proved that Reduced rings are Armendariz which is a nice generalization of your result.
Proof- Let $fg=0$ and assuming $m=n$ is sufficient. We then have $$a_0b_0=0,\ a_1b_0+a_0b_1=0 ,\ vdots \a_nb_0+dots +a_0b_n=0$$
Since $R$ is reduced, $a_0b_0=0implies (b_0a_0)^2=b_0(a_0b_0)a_0=0 implies b_0a_0=0$.
Now left multiplying $a_1b_0+b_1a_0=0$ by $b_0$ we get $b_0a_1b_0=-b_0a_0b_1=0 implies (a_1b_0)^2=a_1(b_0a_1b_0)=0 implies a_1b_0=0$.
Similarly we get $a_ib_0=0 forall 1leq i leq n$.
Now original equations reduces to
$$a_0b_1=0\ a_1b_1+a_0b_2=0\ vdots\ a_n-1b_1+dots +a_0b_n=0$$ and then by same process first we will get that $a_0b_1=0$ and then multiplying on left of $a_1b_1+a_0b_2=0$ by $b_1$ we get $a_1b_1=0$, and so on we will get, $a_ib_1=0 forall 1le i le n$.
Similarly, Repeating it we get $a_ib_j=0 forall 1 leq i,j leq n$. $hspace5.5cmblacksquare$
Some other examples of Armendariz Rings are:
- $BbbZ/nBbbZ forall n$.
- All domains and fields (which are reduced).
- If A and B are Armendariz , then $A(+)B$ in which multiplication is defined by $(a,b)(a',b')=(aa',ab'+a'b)$ is Armendariz.
- Direct Product of Armendariz rings.
edited May 4 '17 at 19:28
Xam
4,45451445
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
33
down vote
accepted
It is true over any commutative ring, and is sometimes called McCoy's theorem. Below is a proof sketch
from my sci.math post on 5/4/2004:
Theorem $ $ Let $ ,F in R[X]$ be a polynomial over a commutative ring $ ,R.,$
If $ ,F,$ is a zero-divisor then $ ,r,F = 0,$ for some nonzero $ ,r in R.$
Proof $ $ Suppose not. Choose $ ,G ne 0,$ of min degree with $ ,F,G = 0.,$
Write $ ,F =, a +,cdots,+ f X^k +,cdots,+ c X^m $
and $ G = b +,cdots,+ g X^n,,$ where $ ,g ne 0:$ and $ ,f,$ is the highest deg coef of $ ,F,$ with $ ,f,G ne 0,$ (note that such an $ ,f,$ exists else $ ,F,g = 0,$ contra supposition).
Then $ ,F,G = (a +,cdots,+ f X^k) (b +,cdots,+ g X^n) = 0.$
Thus $ ,f,g = 0,$ so $ ,deg(f,G) < n,$ and $ , F,(f,G) = 0,,$ contra minimality of $ ,G. $ QED
Alternatively it follows Gauss's Lemma
(Dedekind-Mertens form) or related results.
answered Nov 17 '11 at 20:12
Number
1
1
Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)?
– mathjacks
Sep 7 '14 at 15:54
1
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $deg(fG)<deg(G)$.
– Martin Brandenburg
Sep 27 '14 at 13:51
1
Bill, what an elegant proof! Thanks for sharing this lovely argument.
– Prism
Jun 26 '15 at 20:43
2
@user95553 While $,fg =X^7 - X = 0,$ as a polynomial function on $,Bbb Z_7,,$ it is not true that $,X^7-X = 0,$ as formal polynomials $in Bbb Z_7[X].,$ Formal vs. functional polynomials is discussed in many prior answers, e.g. this one.
– Number
Aug 23 '15 at 15:07
1
@user386627 e.g. $,Fg = 0 (Rightarrow, gF = 0),,$ and $,F(fG) = f(FG)= 0 $
– Number
Feb 28 '17 at 18:35
 |Â
show 5 more comments
up vote
33
down vote
accepted
It is true over any commutative ring, and is sometimes called McCoy's theorem. Below is a proof sketch
from my sci.math post on 5/4/2004:
Theorem $ $ Let $ ,F in R[X]$ be a polynomial over a commutative ring $ ,R.,$
If $ ,F,$ is a zero-divisor then $ ,r,F = 0,$ for some nonzero $ ,r in R.$
Proof $ $ Suppose not. Choose $ ,G ne 0,$ of min degree with $ ,F,G = 0.,$
Write $ ,F =, a +,cdots,+ f X^k +,cdots,+ c X^m $
and $ G = b +,cdots,+ g X^n,,$ where $ ,g ne 0:$ and $ ,f,$ is the highest deg coef of $ ,F,$ with $ ,f,G ne 0,$ (note that such an $ ,f,$ exists else $ ,F,g = 0,$ contra supposition).
Then $ ,F,G = (a +,cdots,+ f X^k) (b +,cdots,+ g X^n) = 0.$
Thus $ ,f,g = 0,$ so $ ,deg(f,G) < n,$ and $ , F,(f,G) = 0,,$ contra minimality of $ ,G. $ QED
Alternatively it follows Gauss's Lemma
(Dedekind-Mertens form) or related results.
answered Nov 17 '11 at 20:12
Number
1
1
Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)?
– mathjacks
Sep 7 '14 at 15:54
1
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $deg(fG)<deg(G)$.
– Martin Brandenburg
Sep 27 '14 at 13:51
1
Bill, what an elegant proof! Thanks for sharing this lovely argument.
– Prism
Jun 26 '15 at 20:43
2
@user95553 While $,fg =X^7 - X = 0,$ as a polynomial function on $,Bbb Z_7,,$ it is not true that $,X^7-X = 0,$ as formal polynomials $in Bbb Z_7[X].,$ Formal vs. functional polynomials is discussed in many prior answers, e.g. this one.
– Number
Aug 23 '15 at 15:07
1
@user386627 e.g. $,Fg = 0 (Rightarrow, gF = 0),,$ and $,F(fG) = f(FG)= 0 $
– Number
Feb 28 '17 at 18:35
 |Â
show 5 more comments
up vote
33
down vote
accepted
up vote
33
down vote
accepted
It is true over any commutative ring, and is sometimes called McCoy's theorem. Below is a proof sketch
from my sci.math post on 5/4/2004:
Theorem $ $ Let $ ,F in R[X]$ be a polynomial over a commutative ring $ ,R.,$
If $ ,F,$ is a zero-divisor then $ ,r,F = 0,$ for some nonzero $ ,r in R.$
Proof $ $ Suppose not. Choose $ ,G ne 0,$ of min degree with $ ,F,G = 0.,$
Write $ ,F =, a +,cdots,+ f X^k +,cdots,+ c X^m $
and $ G = b +,cdots,+ g X^n,,$ where $ ,g ne 0:$ and $ ,f,$ is the highest deg coef of $ ,F,$ with $ ,f,G ne 0,$ (note that such an $ ,f,$ exists else $ ,F,g = 0,$ contra supposition).
Then $ ,F,G = (a +,cdots,+ f X^k) (b +,cdots,+ g X^n) = 0.$
Thus $ ,f,g = 0,$ so $ ,deg(f,G) < n,$ and $ , F,(f,G) = 0,,$ contra minimality of $ ,G. $ QED
Alternatively it follows Gauss's Lemma
(Dedekind-Mertens form) or related results.
answered Nov 17 '11 at 20:12
Number
1
It is true over any commutative ring, and is sometimes called McCoy's theorem. Below is a proof sketch
from my sci.math post on 5/4/2004:
Theorem $ $ Let $ ,F in R[X]$ be a polynomial over a commutative ring $ ,R.,$
If $ ,F,$ is a zero-divisor then $ ,r,F = 0,$ for some nonzero $ ,r in R.$
Proof $ $ Suppose not. Choose $ ,G ne 0,$ of min degree with $ ,F,G = 0.,$
Write $ ,F =, a +,cdots,+ f X^k +,cdots,+ c X^m $
and $ G = b +,cdots,+ g X^n,,$ where $ ,g ne 0:$ and $ ,f,$ is the highest deg coef of $ ,F,$ with $ ,f,G ne 0,$ (note that such an $ ,f,$ exists else $ ,F,g = 0,$ contra supposition).
Then $ ,F,G = (a +,cdots,+ f X^k) (b +,cdots,+ g X^n) = 0.$
Thus $ ,f,g = 0,$ so $ ,deg(f,G) < n,$ and $ , F,(f,G) = 0,,$ contra minimality of $ ,G. $ QED
Alternatively it follows Gauss's Lemma
(Dedekind-Mertens form) or related results.
answered Nov 17 '11 at 20:12
Number
1
edited Jul 17 '16 at 3:33
answered Nov 17 '11 at 20:12
Number
1
answered Nov 17 '11 at 20:12
Number
1
answered Nov 17 '11 at 20:12
Number
1
1
1
Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)?
– mathjacks
Sep 7 '14 at 15:54
1
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $deg(fG)<deg(G)$.
– Martin Brandenburg
Sep 27 '14 at 13:51
1
Bill, what an elegant proof! Thanks for sharing this lovely argument.
– Prism
Jun 26 '15 at 20:43
2
@user95553 While $,fg =X^7 - X = 0,$ as a polynomial function on $,Bbb Z_7,,$ it is not true that $,X^7-X = 0,$ as formal polynomials $in Bbb Z_7[X].,$ Formal vs. functional polynomials is discussed in many prior answers, e.g. this one.
– Number
Aug 23 '15 at 15:07
1
@user386627 e.g. $,Fg = 0 (Rightarrow, gF = 0),,$ and $,F(fG) = f(FG)= 0 $
– Number
Feb 28 '17 at 18:35
 |Â
show 5 more comments
1
Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)?
– mathjacks
Sep 7 '14 at 15:54
1
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $deg(fG)<deg(G)$.
– Martin Brandenburg
Sep 27 '14 at 13:51
1
Bill, what an elegant proof! Thanks for sharing this lovely argument.
– Prism
Jun 26 '15 at 20:43
2
@user95553 While $,fg =X^7 - X = 0,$ as a polynomial function on $,Bbb Z_7,,$ it is not true that $,X^7-X = 0,$ as formal polynomials $in Bbb Z_7[X].,$ Formal vs. functional polynomials is discussed in many prior answers, e.g. this one.
– Number
Aug 23 '15 at 15:07
1
@user386627 e.g. $,Fg = 0 (Rightarrow, gF = 0),,$ and $,F(fG) = f(FG)= 0 $
– Number
Feb 28 '17 at 18:35
1
1
Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)?
– mathjacks
Sep 7 '14 at 15:54
Why does this imply that F(fG)=0? I understand that FG = 0 but why does this say something about the effect of F on the polynomial fG with deg(fG) < deg(G)?
– mathjacks
Sep 7 '14 at 15:54
1
1
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $deg(fG)<deg(G)$.
– Martin Brandenburg
Sep 27 '14 at 13:51
$F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $deg(fG)<deg(G)$.
– Martin Brandenburg
Sep 27 '14 at 13:51
1
1
Bill, what an elegant proof! Thanks for sharing this lovely argument.
– Prism
Jun 26 '15 at 20:43
Bill, what an elegant proof! Thanks for sharing this lovely argument.
– Prism
Jun 26 '15 at 20:43
2
2
@user95553 While $,fg =X^7 - X = 0,$ as a polynomial function on $,Bbb Z_7,,$ it is not true that $,X^7-X = 0,$ as formal polynomials $in Bbb Z_7[X].,$ Formal vs. functional polynomials is discussed in many prior answers, e.g. this one.
– Number
Aug 23 '15 at 15:07
@user95553 While $,fg =X^7 - X = 0,$ as a polynomial function on $,Bbb Z_7,,$ it is not true that $,X^7-X = 0,$ as formal polynomials $in Bbb Z_7[X].,$ Formal vs. functional polynomials is discussed in many prior answers, e.g. this one.
– Number
Aug 23 '15 at 15:07
1
1
@user386627 e.g. $,Fg = 0 (Rightarrow, gF = 0),,$ and $,F(fG) = f(FG)= 0 $
– Number
Feb 28 '17 at 18:35
@user386627 e.g. $,Fg = 0 (Rightarrow, gF = 0),,$ and $,F(fG) = f(FG)= 0 $
– Number
Feb 28 '17 at 18:35
 |Â
show 5 more comments
up vote
21
down vote
Assume that $gf=0$ for some $gin R[X]$ and let $c$ be the leading coefficient of $g$. Then $ca_n=0$. Therefore $cf$ is either $0$ (in which case $c$ is your $b$), or $cf$ has degree less than $n$ with $g(cf)=0$. Proceed by induction on $n$. In the end you find that some power of $c$ kills every $a_i$, and $c$ was not nilpotent ...
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
add a comment |Â
up vote
21
down vote
Assume that $gf=0$ for some $gin R[X]$ and let $c$ be the leading coefficient of $g$. Then $ca_n=0$. Therefore $cf$ is either $0$ (in which case $c$ is your $b$), or $cf$ has degree less than $n$ with $g(cf)=0$. Proceed by induction on $n$. In the end you find that some power of $c$ kills every $a_i$, and $c$ was not nilpotent ...
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
add a comment |Â
up vote
21
down vote
up vote
21
down vote
Assume that $gf=0$ for some $gin R[X]$ and let $c$ be the leading coefficient of $g$. Then $ca_n=0$. Therefore $cf$ is either $0$ (in which case $c$ is your $b$), or $cf$ has degree less than $n$ with $g(cf)=0$. Proceed by induction on $n$. In the end you find that some power of $c$ kills every $a_i$, and $c$ was not nilpotent ...
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
Assume that $gf=0$ for some $gin R[X]$ and let $c$ be the leading coefficient of $g$. Then $ca_n=0$. Therefore $cf$ is either $0$ (in which case $c$ is your $b$), or $cf$ has degree less than $n$ with $g(cf)=0$. Proceed by induction on $n$. In the end you find that some power of $c$ kills every $a_i$, and $c$ was not nilpotent ...
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
answered Nov 17 '11 at 18:01
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
up vote
6
down vote
Apparently, Bill Dubuque's argument is not really about polynomial rings. Here is a generalization.
Let $A$ be a commutative $mathbbN$-graded ring. Let $f in A$ be a zero divisor. Then there is some $0 neq a in A$ homogeneous such that $a f = 0$.
Proof. Choose some $0 neq g in A$ of minimal total degree with $fg=0$. Let $f=f_0+f_1+cdots$ and $g=g_0+g_1+cdots+g_d$ be the homogeneous decompositions with $g_d neq 0$. If $f g_d = 0$, we are done. Otherwise, we have $f_i g_d neq 0$ for some $i$, and hence $f_i g neq 0$. Choose $i$ maximal with $f_i g neq 0$. Then $0=fg=(f_0+cdots+f_i)g=(f_0+cdots+f_i)(g_0+cdots+g_d)$ implies $f_i g_d = 0$. Then $f_i g$ has smaller degree than $g$, but still satisfies $f(f_i g)=0$ and $f_i g neq 0$, a contradiction. $square$
This may be applied to $A=R[x]$ with the usual grading. Hence, any zero divisor in $R[x]$ is killed by some element of the form $r x^n$ ($r in R setminus 0$) and then also by $r$.
edited Aug 2 '16 at 15:06
user26857
38.7k123678
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
3
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)?
– Pham Hung Quy
Dec 14 '14 at 14:11
1
I've asked about the correctness of this proof here
– Jay
Jul 31 '16 at 21:04
1
@Jay The above proof can be easily fixed if one considers $g$ as having the minimal number of non-zero homogeneous components among the polynomials with the property $fg=0$. (As you can notice, the answerer wanted to mimic the proof given for polynomials in the accepted answer, and this is an easy task.)
– user26857
Aug 2 '16 at 15:01
@user26857: Yeah that makes sense. That's a neat trick, looking at the number of homogeneous components instead of the degree. Good to know this proof can be fixed!
– Jay
Aug 2 '16 at 22:03
Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above.
– Dietrich Burde
Nov 6 '17 at 15:54
add a comment |Â
up vote
6
down vote
Apparently, Bill Dubuque's argument is not really about polynomial rings. Here is a generalization.
Let $A$ be a commutative $mathbbN$-graded ring. Let $f in A$ be a zero divisor. Then there is some $0 neq a in A$ homogeneous such that $a f = 0$.
Proof. Choose some $0 neq g in A$ of minimal total degree with $fg=0$. Let $f=f_0+f_1+cdots$ and $g=g_0+g_1+cdots+g_d$ be the homogeneous decompositions with $g_d neq 0$. If $f g_d = 0$, we are done. Otherwise, we have $f_i g_d neq 0$ for some $i$, and hence $f_i g neq 0$. Choose $i$ maximal with $f_i g neq 0$. Then $0=fg=(f_0+cdots+f_i)g=(f_0+cdots+f_i)(g_0+cdots+g_d)$ implies $f_i g_d = 0$. Then $f_i g$ has smaller degree than $g$, but still satisfies $f(f_i g)=0$ and $f_i g neq 0$, a contradiction. $square$
This may be applied to $A=R[x]$ with the usual grading. Hence, any zero divisor in $R[x]$ is killed by some element of the form $r x^n$ ($r in R setminus 0$) and then also by $r$.
edited Aug 2 '16 at 15:06
user26857
38.7k123678
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
3
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)?
– Pham Hung Quy
Dec 14 '14 at 14:11
1
I've asked about the correctness of this proof here
– Jay
Jul 31 '16 at 21:04
1
@Jay The above proof can be easily fixed if one considers $g$ as having the minimal number of non-zero homogeneous components among the polynomials with the property $fg=0$. (As you can notice, the answerer wanted to mimic the proof given for polynomials in the accepted answer, and this is an easy task.)
– user26857
Aug 2 '16 at 15:01
@user26857: Yeah that makes sense. That's a neat trick, looking at the number of homogeneous components instead of the degree. Good to know this proof can be fixed!
– Jay
Aug 2 '16 at 22:03
Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above.
– Dietrich Burde
Nov 6 '17 at 15:54
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Apparently, Bill Dubuque's argument is not really about polynomial rings. Here is a generalization.
Let $A$ be a commutative $mathbbN$-graded ring. Let $f in A$ be a zero divisor. Then there is some $0 neq a in A$ homogeneous such that $a f = 0$.
Proof. Choose some $0 neq g in A$ of minimal total degree with $fg=0$. Let $f=f_0+f_1+cdots$ and $g=g_0+g_1+cdots+g_d$ be the homogeneous decompositions with $g_d neq 0$. If $f g_d = 0$, we are done. Otherwise, we have $f_i g_d neq 0$ for some $i$, and hence $f_i g neq 0$. Choose $i$ maximal with $f_i g neq 0$. Then $0=fg=(f_0+cdots+f_i)g=(f_0+cdots+f_i)(g_0+cdots+g_d)$ implies $f_i g_d = 0$. Then $f_i g$ has smaller degree than $g$, but still satisfies $f(f_i g)=0$ and $f_i g neq 0$, a contradiction. $square$
This may be applied to $A=R[x]$ with the usual grading. Hence, any zero divisor in $R[x]$ is killed by some element of the form $r x^n$ ($r in R setminus 0$) and then also by $r$.
edited Aug 2 '16 at 15:06
user26857
38.7k123678
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
Apparently, Bill Dubuque's argument is not really about polynomial rings. Here is a generalization.
Let $A$ be a commutative $mathbbN$-graded ring. Let $f in A$ be a zero divisor. Then there is some $0 neq a in A$ homogeneous such that $a f = 0$.
Proof. Choose some $0 neq g in A$ of minimal total degree with $fg=0$. Let $f=f_0+f_1+cdots$ and $g=g_0+g_1+cdots+g_d$ be the homogeneous decompositions with $g_d neq 0$. If $f g_d = 0$, we are done. Otherwise, we have $f_i g_d neq 0$ for some $i$, and hence $f_i g neq 0$. Choose $i$ maximal with $f_i g neq 0$. Then $0=fg=(f_0+cdots+f_i)g=(f_0+cdots+f_i)(g_0+cdots+g_d)$ implies $f_i g_d = 0$. Then $f_i g$ has smaller degree than $g$, but still satisfies $f(f_i g)=0$ and $f_i g neq 0$, a contradiction. $square$
This may be applied to $A=R[x]$ with the usual grading. Hence, any zero divisor in $R[x]$ is killed by some element of the form $r x^n$ ($r in R setminus 0$) and then also by $r$.
edited Aug 2 '16 at 15:06
user26857
38.7k123678
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
edited Aug 2 '16 at 15:06
user26857
38.7k123678
edited Aug 2 '16 at 15:06
user26857
38.7k123678
edited Aug 2 '16 at 15:06
user26857
38.7k123678
38.7k123678
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
answered Sep 27 '14 at 13:57
Martin Brandenburg
105k13150316
105k13150316
3
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)?
– Pham Hung Quy
Dec 14 '14 at 14:11
1
I've asked about the correctness of this proof here
– Jay
Jul 31 '16 at 21:04
1
@Jay The above proof can be easily fixed if one considers $g$ as having the minimal number of non-zero homogeneous components among the polynomials with the property $fg=0$. (As you can notice, the answerer wanted to mimic the proof given for polynomials in the accepted answer, and this is an easy task.)
– user26857
Aug 2 '16 at 15:01
@user26857: Yeah that makes sense. That's a neat trick, looking at the number of homogeneous components instead of the degree. Good to know this proof can be fixed!
– Jay
Aug 2 '16 at 22:03
Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above.
– Dietrich Burde
Nov 6 '17 at 15:54
add a comment |Â
3
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)?
– Pham Hung Quy
Dec 14 '14 at 14:11
1
I've asked about the correctness of this proof here
– Jay
Jul 31 '16 at 21:04
1
@Jay The above proof can be easily fixed if one considers $g$ as having the minimal number of non-zero homogeneous components among the polynomials with the property $fg=0$. (As you can notice, the answerer wanted to mimic the proof given for polynomials in the accepted answer, and this is an easy task.)
– user26857
Aug 2 '16 at 15:01
@user26857: Yeah that makes sense. That's a neat trick, looking at the number of homogeneous components instead of the degree. Good to know this proof can be fixed!
– Jay
Aug 2 '16 at 22:03
Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above.
– Dietrich Burde
Nov 6 '17 at 15:54
3
3
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)?
– Pham Hung Quy
Dec 14 '14 at 14:11
why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)?
– Pham Hung Quy
Dec 14 '14 at 14:11
1
1
I've asked about the correctness of this proof here
– Jay
Jul 31 '16 at 21:04
I've asked about the correctness of this proof here
– Jay
Jul 31 '16 at 21:04
1
1
@Jay The above proof can be easily fixed if one considers $g$ as having the minimal number of non-zero homogeneous components among the polynomials with the property $fg=0$. (As you can notice, the answerer wanted to mimic the proof given for polynomials in the accepted answer, and this is an easy task.)
– user26857
Aug 2 '16 at 15:01
@Jay The above proof can be easily fixed if one considers $g$ as having the minimal number of non-zero homogeneous components among the polynomials with the property $fg=0$. (As you can notice, the answerer wanted to mimic the proof given for polynomials in the accepted answer, and this is an easy task.)
– user26857
Aug 2 '16 at 15:01
@user26857: Yeah that makes sense. That's a neat trick, looking at the number of homogeneous components instead of the degree. Good to know this proof can be fixed!
– Jay
Aug 2 '16 at 22:03
@user26857: Yeah that makes sense. That's a neat trick, looking at the number of homogeneous components instead of the degree. Good to know this proof can be fixed!
– Jay
Aug 2 '16 at 22:03
Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above.
– Dietrich Burde
Nov 6 '17 at 15:54
Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above.
– Dietrich Burde
Nov 6 '17 at 15:54
add a comment |Â
up vote
4
down vote
This is the case of Armendariz Rings, which I studied in last summer briefly. It is an interesting topic.
A ring $R$ is called Armendariz if whenever $f(x)=sum_i=0^ma_ix^i, g(x)=sum_j=0^nb_jx^j in R[x]$ such that $f(x)g(x)=0$, then $a_ib_j=0 forall i,j$.
In his paper "A NOTE ON EXTENSIONS OF BAER AND P. P. -RINGS" in 1973, Armendariz proved that Reduced rings are Armendariz which is a nice generalization of your result.
Proof- Let $fg=0$ and assuming $m=n$ is sufficient. We then have $$a_0b_0=0,\ a_1b_0+a_0b_1=0 ,\ vdots \a_nb_0+dots +a_0b_n=0$$
Since $R$ is reduced, $a_0b_0=0implies (b_0a_0)^2=b_0(a_0b_0)a_0=0 implies b_0a_0=0$.
Now left multiplying $a_1b_0+b_1a_0=0$ by $b_0$ we get $b_0a_1b_0=-b_0a_0b_1=0 implies (a_1b_0)^2=a_1(b_0a_1b_0)=0 implies a_1b_0=0$.
Similarly we get $a_ib_0=0 forall 1leq i leq n$.
Now original equations reduces to
$$a_0b_1=0\ a_1b_1+a_0b_2=0\ vdots\ a_n-1b_1+dots +a_0b_n=0$$ and then by same process first we will get that $a_0b_1=0$ and then multiplying on left of $a_1b_1+a_0b_2=0$ by $b_1$ we get $a_1b_1=0$, and so on we will get, $a_ib_1=0 forall 1le i le n$.
Similarly, Repeating it we get $a_ib_j=0 forall 1 leq i,j leq n$. $hspace5.5cmblacksquare$
Some other examples of Armendariz Rings are:
- $BbbZ/nBbbZ forall n$.
- All domains and fields (which are reduced).
- If A and B are Armendariz , then $A(+)B$ in which multiplication is defined by $(a,b)(a',b')=(aa',ab'+a'b)$ is Armendariz.
- Direct Product of Armendariz rings.
edited May 4 '17 at 19:28
Xam
4,45451445
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
add a comment |Â
up vote
4
down vote
This is the case of Armendariz Rings, which I studied in last summer briefly. It is an interesting topic.
A ring $R$ is called Armendariz if whenever $f(x)=sum_i=0^ma_ix^i, g(x)=sum_j=0^nb_jx^j in R[x]$ such that $f(x)g(x)=0$, then $a_ib_j=0 forall i,j$.
In his paper "A NOTE ON EXTENSIONS OF BAER AND P. P. -RINGS" in 1973, Armendariz proved that Reduced rings are Armendariz which is a nice generalization of your result.
Proof- Let $fg=0$ and assuming $m=n$ is sufficient. We then have $$a_0b_0=0,\ a_1b_0+a_0b_1=0 ,\ vdots \a_nb_0+dots +a_0b_n=0$$
Since $R$ is reduced, $a_0b_0=0implies (b_0a_0)^2=b_0(a_0b_0)a_0=0 implies b_0a_0=0$.
Now left multiplying $a_1b_0+b_1a_0=0$ by $b_0$ we get $b_0a_1b_0=-b_0a_0b_1=0 implies (a_1b_0)^2=a_1(b_0a_1b_0)=0 implies a_1b_0=0$.
Similarly we get $a_ib_0=0 forall 1leq i leq n$.
Now original equations reduces to
$$a_0b_1=0\ a_1b_1+a_0b_2=0\ vdots\ a_n-1b_1+dots +a_0b_n=0$$ and then by same process first we will get that $a_0b_1=0$ and then multiplying on left of $a_1b_1+a_0b_2=0$ by $b_1$ we get $a_1b_1=0$, and so on we will get, $a_ib_1=0 forall 1le i le n$.
Similarly, Repeating it we get $a_ib_j=0 forall 1 leq i,j leq n$. $hspace5.5cmblacksquare$
Some other examples of Armendariz Rings are:
- $BbbZ/nBbbZ forall n$.
- All domains and fields (which are reduced).
- If A and B are Armendariz , then $A(+)B$ in which multiplication is defined by $(a,b)(a',b')=(aa',ab'+a'b)$ is Armendariz.
- Direct Product of Armendariz rings.
edited May 4 '17 at 19:28
Xam
4,45451445
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
add a comment |Â
up vote
4
down vote
up vote
4
down vote
This is the case of Armendariz Rings, which I studied in last summer briefly. It is an interesting topic.
A ring $R$ is called Armendariz if whenever $f(x)=sum_i=0^ma_ix^i, g(x)=sum_j=0^nb_jx^j in R[x]$ such that $f(x)g(x)=0$, then $a_ib_j=0 forall i,j$.
In his paper "A NOTE ON EXTENSIONS OF BAER AND P. P. -RINGS" in 1973, Armendariz proved that Reduced rings are Armendariz which is a nice generalization of your result.
Proof- Let $fg=0$ and assuming $m=n$ is sufficient. We then have $$a_0b_0=0,\ a_1b_0+a_0b_1=0 ,\ vdots \a_nb_0+dots +a_0b_n=0$$
Since $R$ is reduced, $a_0b_0=0implies (b_0a_0)^2=b_0(a_0b_0)a_0=0 implies b_0a_0=0$.
Now left multiplying $a_1b_0+b_1a_0=0$ by $b_0$ we get $b_0a_1b_0=-b_0a_0b_1=0 implies (a_1b_0)^2=a_1(b_0a_1b_0)=0 implies a_1b_0=0$.
Similarly we get $a_ib_0=0 forall 1leq i leq n$.
Now original equations reduces to
$$a_0b_1=0\ a_1b_1+a_0b_2=0\ vdots\ a_n-1b_1+dots +a_0b_n=0$$ and then by same process first we will get that $a_0b_1=0$ and then multiplying on left of $a_1b_1+a_0b_2=0$ by $b_1$ we get $a_1b_1=0$, and so on we will get, $a_ib_1=0 forall 1le i le n$.
Similarly, Repeating it we get $a_ib_j=0 forall 1 leq i,j leq n$. $hspace5.5cmblacksquare$
Some other examples of Armendariz Rings are:
- $BbbZ/nBbbZ forall n$.
- All domains and fields (which are reduced).
- If A and B are Armendariz , then $A(+)B$ in which multiplication is defined by $(a,b)(a',b')=(aa',ab'+a'b)$ is Armendariz.
- Direct Product of Armendariz rings.
edited May 4 '17 at 19:28
Xam
4,45451445
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
This is the case of Armendariz Rings, which I studied in last summer briefly. It is an interesting topic.
A ring $R$ is called Armendariz if whenever $f(x)=sum_i=0^ma_ix^i, g(x)=sum_j=0^nb_jx^j in R[x]$ such that $f(x)g(x)=0$, then $a_ib_j=0 forall i,j$.
In his paper "A NOTE ON EXTENSIONS OF BAER AND P. P. -RINGS" in 1973, Armendariz proved that Reduced rings are Armendariz which is a nice generalization of your result.
Proof- Let $fg=0$ and assuming $m=n$ is sufficient. We then have $$a_0b_0=0,\ a_1b_0+a_0b_1=0 ,\ vdots \a_nb_0+dots +a_0b_n=0$$
Since $R$ is reduced, $a_0b_0=0implies (b_0a_0)^2=b_0(a_0b_0)a_0=0 implies b_0a_0=0$.
Now left multiplying $a_1b_0+b_1a_0=0$ by $b_0$ we get $b_0a_1b_0=-b_0a_0b_1=0 implies (a_1b_0)^2=a_1(b_0a_1b_0)=0 implies a_1b_0=0$.
Similarly we get $a_ib_0=0 forall 1leq i leq n$.
Now original equations reduces to
$$a_0b_1=0\ a_1b_1+a_0b_2=0\ vdots\ a_n-1b_1+dots +a_0b_n=0$$ and then by same process first we will get that $a_0b_1=0$ and then multiplying on left of $a_1b_1+a_0b_2=0$ by $b_1$ we get $a_1b_1=0$, and so on we will get, $a_ib_1=0 forall 1le i le n$.
Similarly, Repeating it we get $a_ib_j=0 forall 1 leq i,j leq n$. $hspace5.5cmblacksquare$
Some other examples of Armendariz Rings are:
- $BbbZ/nBbbZ forall n$.
- All domains and fields (which are reduced).
- If A and B are Armendariz , then $A(+)B$ in which multiplication is defined by $(a,b)(a',b')=(aa',ab'+a'b)$ is Armendariz.
- Direct Product of Armendariz rings.
edited May 4 '17 at 19:28
Xam
4,45451445
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
edited May 4 '17 at 19:28
Xam
4,45451445
edited May 4 '17 at 19:28
Xam
4,45451445
edited May 4 '17 at 19:28
Xam
4,45451445
4,45451445
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
answered Jun 7 '15 at 8:46
Bhaskar Vashishth
7,43411950
7,43411950
add a comment |Â
add a comment |Â
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