# Should the sum of zero divisors also a zero divisor?

In a general ring $A$ (commutative with $1$), should the sum of two zero divisors also a zero divisor? Could anyone give a proof or a countexample?

Moreover, consider the polynomial ring $A[x]$, where $A$ is a general ring, then could anyone ask the same question as above?

In fact, my question is: given two polynomials, both can be annihilated by some nonzero element in $A$, then is there exist some nonzero element in $A$ (or some nonzero element in $A[x]$ will also OK, in fact, the two cases are equivalent, see Introduction to Commutative Algebra, Atiyah and Macdonald, chapter 1, exercise 2) annihilate both the two polynomials? The essential is to annihilate the two ideals generated by these two elements simultaneously by an nonzero element.

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How about that we consider 0 as a zero divisor? – Lao-tzu Jan 6 '14 at 2:09

The elements $2, 3 \in \mathbb{Z}/6$ are zero divisors, but their sum, 5, is a unit.

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OK, thank you very much! – Lao-tzu Jan 6 '14 at 2:09

In response to your edit:

Making the ring bigger won't change the answer. If $A$ is a general ring, then $A[x]$ will have zero divisors iff $A$ has zero divisors.

In any case, you run into the same problems as before. Consider $\mathbb{Z}_6[x]$. As before, we have $2+3=5$. For a newer counterexample, we could take $2x + 3x = 5x$. Note that both $2x$ and $3x$ are zero divisors (annihilated by $3$ and $2$ or respectively), but $5x$ is not.

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How about the element $f_1+x^N f_2$ for $N$ sufficiently large? Is it a zero divisor when both $f_1, f_2$ are? – Lao-tzu Jan 6 '14 at 2:35
Let $f_1=2,f_2=3$. Then $f=2+3x^N$ for some $N\geq 0$. Now take $g\in \Bbb Z_6[x]$, say $g=g_0+g_1x+\dots+g_kx^k$. Then $fg=(2+3x^N)(g_0+g_1x+\dots+g_kx^k)=2g_0+(3g_0+2g_N)x^N+\dots$. If this is zero, then $g_0=3$ and $9+2g_N=0\implies 2g_N=3$, which is impossible. Therefore $\{2+3x^n\}$ are all not zero divisors. – Yong Hao Ng Jan 6 '14 at 3:03
But we can take $g_0=0$, then we can say nothing from your deduction. Note that $N$ is large means we can make "the two summands" separated as far as we hope. – Lao-tzu Jan 6 '14 at 10:17
@Lao-tzu If $g_0=0$, then $fg=(2+3x^N)(g_sx^s+\dots+g_kx^k)=(2+3x^N)(g_s+\dots+g_kx^{k-s})x^s$. If this is zero, then it must be the case that $(2+3x^N)(g_s+\dots+g_kx^{k-s})=0$, so we can replace $g$ by $g_s+\dots+g_kx^{k-s}$, which means we can assume $g_0\neq 0$. By the way you can reply with "@name" so that the person will receive a notification. You don't need to do so when replying to the answer though. – Yong Hao Ng Jan 6 '14 at 17:48
OK, I have known the answer (affirmative) from some other way (the proof is essentially the same as that of the hint of "Introduction to Commutative Algebra" by Atiyah and Macdonald, chapter 1, exercise 2), thank you anyway! – Lao-tzu Jan 7 '14 at 4:08