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$R$ is a commutative ring. Let $(x)$ denote the ideal generated by $x \in R$. If $(a) = (b)$, is it true that there exists a unit $u$ such that $a =ub$?

If $R$ is integral domain, then it's simple. Simply choose $a = ub$ and $va =b$ so that $b(uv - 1) = 0$. But what if $R$ is not integral domain? I wrote a short code to run all $\mathbb Z/n\mathbb Z$ for $n < 10000$ and can't find any contradiction. I also tried construction like $\mathbb Z/n\mathbb Z \times \mathbb Z/n\mathbb Z$ and let $b = (0, x)$ and $uv = (y, 1)$ so that $b(uv-1) = 0$. None of them works. Any suggestions?

user26857
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The One
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  • I would consider a free construction $\Bbb Z[x,y,u,v]/(uy-x,, vx-y)$. If you can prove here that there can be found (perhaps different) $u',v'$ units for the job, then it must be also the case in every such situation. Otherwise, you got a counterexample. – Berci Jan 14 '21 at 01:22
  • You did not find any contradiction since in $\mathbb Z_n$ the property holds. Try to prove it! – user26857 Jan 14 '21 at 15:05

1 Answers1

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Let $R$ be the ring of all polynomials in $x$ vanishing at zero (and having integer coefficients, or rational coefficients, or real coefficients, doesn't really matter). Then $x$ and $-x$ generate the same ideal, but there's no unit $u$ such that $-x=ux$. Indeed, there's no identity element, so there are no units at all.

Gerry Myerson
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    This is cheating! Rings have identities! – Qiaochu Yuan Jan 14 '21 at 02:04
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    @QiaochuYuan, tell it to Noether, Emmy (1921). "Idealtheorie in Ringbereichen". Math. Annalen. 83 (1–2): 24–66. doi:10.1007/bf01464225. S2CID 121594471, page29. Or Hungerford, Thomas W. (1997). Abstract Algebra: an Introduction, Second Edition. Brooks/Cole. ISBN 9780030105593, page 42. – Gerry Myerson Jan 14 '21 at 03:50
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    But I will grant that the problem is more interesting when restricted to rings with identity. – Gerry Myerson Jan 14 '21 at 03:57
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    The OP clearly uses the term "unit"; what does "unit" even mean in a ring without identity? – Qiaochu Yuan Jan 14 '21 at 04:33
  • @QiaochuYuan OP asks whether it's true that there exists a unit such that .... I give an example where (as I point out) there are no units at all, because there's no identity. It's not like I'm trying to hide anything. All my cards are on the table. There's nothing wrong with notifying OP that not everyone insists that rings have multiplicative identities. – Gerry Myerson Jan 14 '21 at 06:46
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    I don't think it's accurate to say that there are no units. A unit is, by definition, an element $a$ such that there exists $b$ such that $ab = ba = e$, where $e$ is the identity. This definition doesn't even parse if there isn't an identity; it's not that it isn't satisfied. – Qiaochu Yuan Jan 14 '21 at 08:31
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    Or, without reference to identity: $u$ is a unit iff $uR=R$. – lhf Jan 14 '21 at 10:57