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The characteristic of an integral domain $R$ is $0$ (or prime).

My lecture has not yet covered infinite integral domain but I'll like to understand the proof.

Basic fact:

$R$ is an integral domain so $R$ is a commutative ring with unity (multiplicative inverse = $1$ exists) containing no zero-divisor.

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By definition of a commutative ring:

  • $\left ( R,+ \right )$ is an Abelian group.

  • "$\cdot$" is associative.

  • Distributive law holds.

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Definition (Zero-divisor):

Let $R$ be a ring. Then an element $a(\neq 0) \in R$ is called a zero-divisor if there exists $b(\neq 0) \in R$ s.t $a\cdot b=0$

By contraposition, since there exists no zero-divisor in the integral domain R, it is true that $a\neq 0$ and there exists no $b\neq 0$.

Can someone take me further? Thanks in advance.

Edit: finite integral domain in lecture is covered

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Mathematicing
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  • I don't understand exactly where you are. Specifically, left aside the well-known definitions, I don't understand if the fourth-to-last an third-to-last line are meant to contain a proof of anything or, if so, of what. –  May 10 '16 at 06:30

1 Answers1

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Suppose by absurd that the characterist of an integral domain is an integer $n$ not prime. Therefore $n=n_1n_2$ if $e$ is the multiplicative unity of the integral domain then $$ne=(n_1n_2)e= (n_1e)(n_2e)=0$$ but $R$ is an integral domain therefore or $n_1e=0$ or $n_2e=0$ but $n_2,n_1<n$ impossible

Domenico Vuono
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