1001: For Integral Domain and Nonzero Element, Multiplication Map by Element Is Injection

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description/proof of that for integral domain and nonzero element, multiplication map by element is injection

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Topics

About: ring

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

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Starting Context

• The reader knows a definition of integral domain.
• The reader knows a definition of injection.

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Target Context

• The reader will have a description and a proof of the proposition that for any integral domain and its any nonzero element, the multiplication map by the element is an injection.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$R$: $\in \{\text{ the integral domains }\}$
$r$: $\in R\setminus \{0\}$
$f_r$: $:R\to R,r^{\prime }\mapsto r{r}^{\prime }$
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Statements:
$f_r\in \{\text{ the injections }\}$
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2: Note

Replacing $r{r}^{\prime }$ with $r^{\prime }r$ does not make any difference, because any integral domain is commutative.

$f_r$ is not any ring homomorphism unless $r=1$: $f_r(1)=r1=r\ne 1$.

When $r=0$, $f_r(r^{\prime })=0r^{\prime }=0$, which is of course non-injective.

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3: Proof

Whole Strategy: Step 1: let $r^{\prime },r^{″}\in R$ such that $r^{\prime }\ne r^{″}$, suppose that $f_r(r^{\prime })=f_r(r^{″})$, and find a contradiction.

Step 1:

Let $r^{\prime },r^{″}\in R$ be any such that $r^{\prime }\ne r^{″}$.

Let us suppose that $f_r(r^{\prime })=f_r(r^{″})$.

$r{r}^{\prime }=r{r}^{″}$ and $r(r^{\prime }-r^{″})=0$. As $R$ is an integral domain, $r=0$ or $r^{\prime }-r^{″}=0$, but $r\ne 0$, so, $r^{\prime }-r^{″}=0$, so, $r^{\prime }=r^{″}$, a contradiction.

So, $f_r(r^{\prime })\ne f_r(r^{″})$.

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References

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