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Consider the ring $\mathbb Z_4[x]$. Clearly the elements of the form $2f(x)$ are zero divisors.

1. Is it true that they are all the zero divisors? I mean is it true that if $p(x)$ is a zero divisor then it is of the form $$ 2f(x) $$ for some $f(x) \in \mathbb Z_4[x]$? In other words, is the set of zero-divisors exactly the ideal $(2)$?

I believe it is true, but I do not know how to prove it.

Secondly, the elements $1+g(x)$, with $g(x)$ zero divisors, are clearly units: $(1+g(x))^2=1$.

2. Is it true that they are all the units? I mean is it true that if $p(x)\in \mathbb Z_4[x]$ is a unit then it is of the form $$ 1+g(x) $$ for some zero divisor $g(x)$?

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Can you translate your questions about $\mathbb{Z}_4[x]$ into questions about $\mathbb{Z}[x]$? – Hurkyl Sep 13 '12 at 11:16
@Hurkyl What do you mean? Sorry, but I don't understand your question. – Romeo Sep 13 '12 at 12:03
@Romeo Why should $(1+g(x))^2=1$? If $g(x)^2=0$ that would be true, but that doesn't always happen for zero divisors. Do you actually mean to ask about nilpotent elements? – rschwieb Sep 13 '12 at 12:17
@rschwieb If the point 1 is true (and I think so) then zero divisors are all of the form $2f(x)$, which means they are nilpotent. Do you agree? Indeed, if $g(x)=2p(x)$ then $(1+g(x))^2=1+g(x)^2+2g(x)=1$. – Romeo Sep 13 '12 at 12:24
@Romeo Ah, so you're banking on all zero divisors being nilpotent. I see then :) – rschwieb Sep 13 '12 at 12:31
up vote 3 down vote accepted

You're right on both accounts.

1. First notice that if $p(x)$ has zero constant coefficient, say $p(x)=x^mq(x)$, then $p(x)$ is a zero divisor if and only if $q(x)$ is.

So suppose $p(x) = \sum_0^n a_ix^i \in \mathbb{Z}_4[x]$ has an odd coefficient and a nonzero constant coefficient, and let $k \le n$ be least such that $a_k$ is odd. Then all the $a_j$ for $j < k$ are even. If $k=0$ then clearly $p$ is not a zero divisor (why?) so we must have $a_0=2$.

If $q(x) = \sum_0^m b_ix^i \in \mathbb{Z}_4[x]$ is another polynomial (with nonzero constant coefficient, for the same reason as above), then the coefficient of $x^k$ in $p(x)q(x)$ is $$\sum_{i+j=k} a_ib_j$$ If $p(x)q(x)=0$ then we must have $b_0=2$. But then $$\sum_{i+j=k} a_ib_j = 2a_k + \cdots \equiv 2 + \cdots \pmod 4$$ and so $b_{\ell}$ is odd for some $\ell<k$; otherwise the other terms would all disappear and we'd be left with $2$.

But $q$ is a zero divisor with nonzero constant coefficient, and so interchanging the roles of $p$ and $q$ in the above, we see that $a_i$ is odd for some $i<\ell<k$, contradicting minimality of $k$.

2. Certainly if $p(x)q(x)=1$ then their constant coefficients are either both $1$ or both $3$. So $p(x)=1+g(x)$ and $q(x)=1+h(x)$ for some $g,h$ with equal even constant coefficients; and then $g(x)+h(x)+g(x)h(x)=0$.

Let $g(x)=\sum_0^n c_ix^i$ and $h(x)=\sum_0^m d_ix^i$, and let $k$ be least such that at least one of $c_k$ or $d_k$ is odd. [We must have $k \ge 1$.] Then $$c_k + d_k + \sum_{i+j=k} c_id_j = 0$$ We must therefore have $c_k$ and $d_k$ both odd. But $c_0=d_0$ is even, and so $$c_k + d_k + \sum_{i+j=k} c_id_j = c_k+d_k+c_0d_k+c_kd_0 = (1+c_0)(c_k+d_k) \equiv 2 \pmod 4$$ since the rest of the terms in the sum are zero.

This is a contradiction, so $g$ and $h$ must both be divisible by $2$, and so indeed $p(x)=1+g(x)$ where $g$ is a zero divisor.

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Thanks for your answer. For point 1, I did almost the same things you did, but I wasn't able to finish. I have 2 questions: you say "If $p(x)$ as odd constant term, then it cannot be a zero divisor". Why? Suppose $a_0=1$ and multiply it by a polynomial $q$ with zero constant term... Secondly, I don't understand what you mean with "a_k=2".. Wasn't $k$ the least integer such that $a_k$ is odd? Thanks again. – Romeo Sep 13 '12 at 12:33
If $p$ has odd constant term, $q(x) = x^{\ell}r(x)$ ($\ell \ge 0$) and $r$ has nonzero constant term, then the coefficient of $x^{\ell}$ in $p(x)q(x)$ is necessarily nonzero, so $p(x)q(x) \ne 0$. As for $a_k=2$ I meant $a_0=2$, I'll correct that now :) – Clive Newstead Sep 13 '12 at 12:36
Thanks for your kind explanations: I got it. Now I'm going on reading your proof. Thanks again. – Romeo Sep 13 '12 at 12:40
Point 1 is great! Very very nice proof, it seems to me correct. Thanks! Now I go on with point 2. – Romeo Sep 13 '12 at 12:43
Also point 2 seems correct to me. I thank you for your help. – Romeo Sep 13 '12 at 13:01

Your first conjecture can be easily deduced by translating the problem to that of $\mathbb{Z}[x]$.

The ring $\mathbb{Z}_4[x]$ is (isomorphic to) the quotient ring $\mathbb{Z}[x] / (4)$, so we can write elements of $\mathbb{Z}_4[x]$ as (equivalence classes of) integer polynomials. The condition to be a zero divisor of $\mathbb{Z}_4[x]$ is

u is a zero divisor if and only if there exists nonzero v such that $uv = 0$

(I'm adopting the convention that 0 is a zero divisor)

Lifting this condition to $\mathbb{Z}[x]$ gives

The equivalence class of u is a zero divisor if and only if there exists $v \not\equiv 0 \pmod 4$ such that $uv \equiv 0 \pmod 4$.

Rephrasing in terms of divisibility, this means the prime factorization of $v$ may not have $2$, but the prime factorization of $uv$ must have at least 2 copies of two. Therefore, the prime factorization of $u$ must have a copy of $2$.

Therefore, if the equivalence class of $u$ is a zero divisor, then $u = 2 g$ for some integer polynomial $g$.

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Both results follow from general characterizations of zero-divisors and units in polynomial rings. It's more insightful (and just as simple) to give the general proofs, then specialize them as below.

$\rm(1)\ $ McCoy's theorem states: $ $ if $\rm\:f\in R[x]\:$ is a zero-divisor then $\rm\:r\,f = 0\:$ for some nonzero $\rm\:r\in R.\:$ Thus $\rm\,f\,$ zero-divisor in $\rm\:\Bbb Z_4[x]\:\Rightarrow\:c\,f = 0\:$ for $\rm\:0\ne c\in \Bbb Z_4,\,$ so in $\rm\:\Bbb Z[x],\,\ 4\:|\:cf,\ 4\nmid c,f\:\Rightarrow\:2\:|\:c,f.\ $

$\rm(2)\ $ It is very easy to prove that $\rm\:f(x)\in R[x]\:$ is a unit iff $\rm\:f(0)\:$ is a unit and all other nonzero coefficients are nilpotent. But the only nilpotent in $\rm\,\Bbb Z_4\,$ is $\,2$.

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