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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?

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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 at 22:06
@Bill: I'm using the convention of the stacks project: stacks.math.columbia.edu/tag/00AQ In my opinion a prime ideal should be a proper ideal. –  Louis Jan 15 at 22:15
@Louis Indeed but one could argue otherwise in the zero ring. –  Bill Dubuque Jan 15 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 at 0:26
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 at 0:30

2 Answers 2

up vote 1 down vote accepted

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.

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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.

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The zero ring certainly does have a $1$, namely $0.\ \ $ –  Bill Dubuque Jan 15 at 22:08

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