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Let $A$ be an integral domain. Prove that if $(x+1)^2=x^2+1$ in $A[x]$, then $A$ must have characteristic $2$.

We have $x^2+2x+1=x^2+1$, so $2=0$, and hence the characteristic must be $1$ or $2$. Now, I don't see anything wrong with the characteristic being $1$. So, is the problem correct?

JJ Beck
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  • An integral domain is by definition a Ring where $0$ is a prime ideal. The $0$-ring by definition doesn't have any prime ideals. (in fact it's the only commutative ring with that property) – Louis Jan 15 '14 at 22:06
  • @Bill: I'm using the convention of the stacks project: http://stacks.math.columbia.edu/tag/00AQ

    In my opinion a prime ideal should be a proper ideal.

     – Louis Jan 15 '14 at 22:15
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    @Louis Indeed but one could argue otherwise in the zero ring. – Bill Dubuque Jan 15 '14 at 22:21
  • @Bill Can you explain from which point of view $0$ should be a prime ideal in the $0$ ring? The only thing I can imagine is the fact that we desire every ring to consider a prime ideal, this is "proven" by first showing maximal ideals exist, then that those are prime. However, this is not an axiom, but a fact that has to be proven and the proof fails in the case $A=0$. From any other point of view, I don't see a reasoning why we should regard the $0$ ideal as prime in this case. (It would also conflict the wish that we want $R/I$ to be an integral domain iff $I \subset R$ is prime) – Louis Jan 16 '14 at 00:26
  • I recognise you are the by far more experienced mathematician, so please don't see it as critique by any means, I only want to understand the other view point. – Louis Jan 16 '14 at 00:30
  • @Louis For example, if one deletes the non-equational axiom $,1\ne 0,$ from the domain axioms, then ${0}$ is a domain. Further, if one desires universality of the theorem that $R$ is a domain iff $0$ is a prime ideal of $R,,$ then one is forced to consider $,0,$ to be prime in ${0}$. Whether or not that is a natural thing to do depends on the context (as true for all conventions). – Bill Dubuque Jan 16 '14 at 00:35
  • @Louis The reason I deleted my initial comment was because extended discussions on the zero ring can often prove very confusing for students, what with modifying hardwired conventions, etc. For some extreme examples see the links in my Remark in this answer. – Bill Dubuque Jan 16 '14 at 01:22
  • @Bill With the definition of a domain I use, even without the $1 \neq 0$ axiom, $0$ is not an integral domain iff $0 \subset 0$ is not a prime ideal. Namely the 3rd (of the equivalent) definitions here: http://en.wikipedia.org/wiki/Integral_domain#Definitions

    So in this setting of definitions (which seems the more common one) everything works out quite well.

     – Louis Jan 16 '14 at 01:28

2 Answers2

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The characteristic of an integral domain is the smallest positive integer $n$ such that $n \cdot 1_A= 0_A$. So if $n=1$, then $1_A = 0_A$ in your integral domain $A$. But, we usually (i.e. pretty much always) assume that an integral domain must have unity different from the $0$. Note that a characteristic of an integral domain must always be prime.

Vladhagen
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What does characteristic equal to $1$ mean? It means that $1 x =x = 0$ for all $x.$ So, your $A$ is the zero ring, in particular it has no $1,$ in particular not an integral domain.

Igor Rivin
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