# Finite integral domains

• Sep 3rd 2011, 06:21 AM
Bernhard
Finite integral domains
I am reading Dummit and Foote Chapter 10 - Introduction to Rings

On page 228 Corollary 3 reads:

Any finite integral domain is a field

Proof:

Let R be a finite integral domain and let a be a non-zero element of R.

By the cancellation law the map x $\displaystyle \mapsto$ ax is an injective function. Since R is finite, this map is surjective. In particular, there is some b $\displaystyle \in$ such that ab = 1, i.e. a is a unit in R. Since a was an arbitrary element, R is a field.

My problem is as follows: how do we explicitly demonstrate that since R is finite, the map x $\displaystyle \mapsto$ ax is injective?

Peter
• Sep 3rd 2011, 09:01 AM
FernandoRevilla
Re: Finite integral domains
Quote:

Originally Posted by Bernhard
My problem is as follows: how do we explicitly demonstrate that since R is finite, the map x $\displaystyle \mapsto$ ax is injective.

It is not necessary for $\displaystyle R$ to be finite. If $\displaystyle 0\neq a\in R$, the map $\displaystyle f:R\to R$ , $\displaystyle f(x)=ax$ is injective for any integral domain $\displaystyle R$ :

$\displaystyle f(x)=f(y)\Rightarrow ax=ay\Rightarrow a(x-y)=0\Rightarrow x-y=0\Rightarrow x=y$
• Sep 3rd 2011, 09:06 AM
ModusPonens
Re: Finite integral domains
You probably mean surjective in your question. Just make a venn diagram of an injective function from a 4 element set and another 4 element set. You'll get the principle by doing that.
• Sep 3rd 2011, 04:48 PM
Bernhard
Re: Finite integral domains
Sorry Fernando - my question was wrong!

I meant: How do we explicitly show that since R is finite, the map x $\displaystyle \mapsto$ ax is surjective.

Peter
• Sep 3rd 2011, 11:15 PM
FernandoRevilla
Re: Finite integral domains
Quote:

Originally Posted by Bernhard
I meant: How do we explicitly show that since R is finite, the map x $\displaystyle \mapsto$ ax is surjective.

One way: the map $\displaystyle f:R\to f(R)\subset R$ , $\displaystyle f(x)=ax$ is bijective i.e. $\displaystyle \textrm{card}(R)=\textrm{card}(f(R))$ . Suppose $\displaystyle f:R\to R$ is not surjective, then $\displaystyle f(R)\subset R$ and $\displaystyle f(R)\neq R$ . As $\displaystyle R$ is finite, $\displaystyle \textrm{card}(R)=\textrm{card}(f(R))<\textrm{card} (R)$ which is absurd
The text passage is a discussion from a mathematics forum about a proof in abstract algebra, specifically about finite integral domains. It includes a question about the proof and responses to that question.
no_question

the passage does not contain a mathematical question that requires a numerical or calculative answer. the question posed is about understanding a step in a proof, which is more conceptual than computational.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Finite integral domains

• Sep 3rd 2011, 06:21 AM
Bernhard
Finite integral domains
I am reading Dummit and Foote Chapter 10 - Introduction to Rings

On page 228 Corollary 3 reads:

Any finite integral domain is a field

Proof:

Let R be a finite integral domain and let a be a non-zero element of R.

By the cancellation law the map x $\displaystyle \mapsto$ ax is an injective function. Since R is finite, this map is surjective. In particular, there is some b $\displaystyle \in$ such that ab = 1, i.e. a is a unit in R. Since a was an arbitrary element, R is a field.

My problem is as follows: how do we explicitly demonstrate that since R is finite, the map x $\displaystyle \mapsto$ ax is injective?

Peter
• Sep 3rd 2011, 09:01 AM
FernandoRevilla
Re: Finite integral domains
Quote:

Originally Posted by Bernhard
My problem is as follows: how do we explicitly demonstrate that since R is finite, the map x $\displaystyle \mapsto$ ax is injective.

It is not necessary for $\displaystyle R$ to be finite. If $\displaystyle 0\neq a\in R$, the map $\displaystyle f:R\to R$ , $\displaystyle f(x)=ax$ is injective for any integral domain $\displaystyle R$ :

$\displaystyle f(x)=f(y)\Rightarrow ax=ay\Rightarrow a(x-y)=0\Rightarrow x-y=0\Rightarrow x=y$
• Sep 3rd 2011, 09:06 AM
ModusPonens
Re: Finite integral domains
You probably mean surjective in your question. Just make a venn diagram of an injective function from a 4 element set and another 4 element set. You'll get the principle by doing that.
• Sep 3rd 2011, 04:48 PM
Bernhard
Re: Finite integral domains
Sorry Fernando - my question was wrong!

I meant: How do we explicitly show that since R is finite, the map x $\displaystyle \mapsto$ ax is surjective.

Peter
• Sep 3rd 2011, 11:15 PM
FernandoRevilla
Re: Finite integral domains
Quote:

Originally Posted by Bernhard
I meant: How do we explicitly show that since R is finite, the map x $\displaystyle \mapsto$ ax is surjective.

One way: the map $\displaystyle f:R\to f(R)\subset R$ , $\displaystyle f(x)=ax$ is bijective i.e. $\displaystyle \textrm{card}(R)=\textrm{card}(f(R))$ . Suppose $\displaystyle f:R\to R$ is not surjective, then $\displaystyle f(R)\subset R$ and $\displaystyle f(R)\neq R$ . As $\displaystyle R$ is finite, $\displaystyle \textrm{card}(R)=\textrm{card}(f(R))<\textrm{card} (R)$ which is absurd
user
The text passage is a discussion from a mathematics forum about a proof in abstract algebra, specifically about finite integral domains. It includes a question about the proof and responses to that question.

TYPE: NO_QUESTION

The passage does not contain a mathematical question that requires a numerical or calculative answer. The question posed is about understanding a step in a proof, which is more conceptual than computational.