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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 $ba_0=ba_1=\cdots=ba_n=0$?

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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.

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    $\begingroup$ 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)? $\endgroup$ – mathjacks Sep 7 '14 at 15:54
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    $\begingroup$ $F(fG)=0$ follows from $FG=0$. The discussion before was needed in order to ensure that $\deg(fG)<\deg(G)$. $\endgroup$ – Martin Brandenburg Sep 27 '14 at 13:51
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    $\begingroup$ Bill, what an elegant proof! Thanks for sharing this lovely argument. $\endgroup$ – Prism Jun 26 '15 at 20:43
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    $\begingroup$ @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. $\endgroup$ – Bill Dubuque Aug 23 '15 at 15:07
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    $\begingroup$ @user386627 e.g. $\,Fg = 0\ (\Rightarrow\, gF = 0),\,$ and $\,F(fG) = f(FG)= 0\ \ $ $\endgroup$ – Bill Dubuque Feb 28 '17 at 18:35
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Assume that $gf=0$ for some $g\in R[X]$ 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 ...

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Apparently, Bill Dubuque's argument is not really about polynomial rings. Here is a generalization.

Let $A$ be a commutative $\mathbb{N}$-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 $r$.

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    $\begingroup$ why is degree of $f_ig$ smaller than degree of $g$ (in graded ring)? $\endgroup$ – Pham Hung Quy Dec 14 '14 at 14:11
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    $\begingroup$ I've asked about the correctness of this proof here $\endgroup$ – Jay Jul 31 '16 at 21:04
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    $\begingroup$ @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.) $\endgroup$ – user26857 Aug 2 '16 at 15:01
  • $\begingroup$ @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! $\endgroup$ – Jay Aug 2 '16 at 22:03
  • $\begingroup$ Bill Dubuque is temporarily "Number $1$", in case someone does not know. See his answer above. $\endgroup$ – Dietrich Burde Nov 6 '17 at 15:54
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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 $f(x)=\sum_{i=0}^{m}a_ix^i, g(x)=\sum_{j=0}^{n}b_jx^j \in R[x]$ 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, $a_0b_0=0\implies (b_0a_0)^2=b_0(a_0b_0)a_0=0 \implies b_0a_0=0$.
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 $a_ib_0=0\ \forall\ 1\leq i \leq n$.

Now original equations reduces to
$$a_0b_1=0\\ a_1b_1+a_0b_2=0\\ \vdots\\ a_{n-1}b_1+\dots +a_0b_n=0$$ 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, $a_ib_1=0\ \forall\ 1\le i \le n$.

Similarly, Repeating it we get $a_ib_j=0\ \forall\ 1 \leq i,j \leq n$. $\hspace{5.5cm}\blacksquare$

Some other examples of Armendariz Rings are:

  • $\Bbb{Z}/n\Bbb{Z}\ \forall\ n$.
  • 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.
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