Proof about finite integral domains.
$begingroup$
Lemma: Let R be a finite integral domain. Assume the elements of R are {$a_1$,$a_2$,,,,$a_n$} for some n$in$ $mathbb{N}$. Then for c$in$R, A={c$a_1$,c$a_2$,,,,c$a_n$}=R.
My proof: Since R is a finite integral domain. By definition, R is a finite commutative ring with the property that $forall$ a,b,c $in$ R with c$neq0$ if $ca=cb$ then $a=b$. Let R have elements {$a_1$,$a_2$,,,,$a_n$}. Consider the set A={c$a_1$,c$a_2$,,,,c$a_n$}, for some c$in$R, then each element is distinct. However; since by the definition of a ring, the ring is closed under multiplication, it follows that A$subseteq$R. Now, we must show that R$subseteq$A. Since $1.x=x.1=x$, it follows that x$in$A; therefore, R$subseteq$A.
I am trying to avoid the theorem that every finite integral domain is a field, partly because i'm using the above lemma to prove that every finite integral domain is a field.
Is my proof correct? What can I do to make it better? I'm just worried about the 1.x =x.1 =x part because I assumed that it holds for an arbitrary c and now i'm assuming it holds for c=1.
abstract-algebra ring-theory
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
$endgroup$
add a comment |
$begingroup$
Lemma: Let R be a finite integral domain. Assume the elements of R are {$a_1$,$a_2$,,,,$a_n$} for some n$in$ $mathbb{N}$. Then for c$in$R, A={c$a_1$,c$a_2$,,,,c$a_n$}=R.
My proof: Since R is a finite integral domain. By definition, R is a finite commutative ring with the property that $forall$ a,b,c $in$ R with c$neq0$ if $ca=cb$ then $a=b$. Let R have elements {$a_1$,$a_2$,,,,$a_n$}. Consider the set A={c$a_1$,c$a_2$,,,,c$a_n$}, for some c$in$R, then each element is distinct. However; since by the definition of a ring, the ring is closed under multiplication, it follows that A$subseteq$R. Now, we must show that R$subseteq$A. Since $1.x=x.1=x$, it follows that x$in$A; therefore, R$subseteq$A.
I am trying to avoid the theorem that every finite integral domain is a field, partly because i'm using the above lemma to prove that every finite integral domain is a field.
Is my proof correct? What can I do to make it better? I'm just worried about the 1.x =x.1 =x part because I assumed that it holds for an arbitrary c and now i'm assuming it holds for c=1.
abstract-algebra ring-theory
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
$endgroup$
1
$begingroup$
Why "..., but clearly $1cdot x=xcdot 1=xin A$"? Is it obvious?
$endgroup$
– Song
Jan 9 at 12:13
$begingroup$
If $A$ is not all of $R$, then there would exist some element $a_k in R$ such that $ca_i = a_k=ca_j$
$endgroup$
– JavaMan
Jan 9 at 12:27
1
$begingroup$
I don’t think it works you say that 1x=x which is fine, but that doesn’t prove that R is a subset of A. That would only prove this if c were 1. Think about it, I give you an elememt x of R and you have to show me that it’s equal to cy for some (possibly different) y in R. The fact that 1x=x does not prove this, because c is fixed and differrent from 1 (if c=1 then this is trivial)
$endgroup$
– Ovi
Jan 9 at 12:28
$begingroup$
That's what I was worried about, what can I do to show that R is a subset of A?
$endgroup$
– mathsssislife
Jan 9 at 12:30
add a comment |
$begingroup$
Lemma: Let R be a finite integral domain. Assume the elements of R are {$a_1$,$a_2$,,,,$a_n$} for some n$in$ $mathbb{N}$. Then for c$in$R, A={c$a_1$,c$a_2$,,,,c$a_n$}=R.
My proof: Since R is a finite integral domain. By definition, R is a finite commutative ring with the property that $forall$ a,b,c $in$ R with c$neq0$ if $ca=cb$ then $a=b$. Let R have elements {$a_1$,$a_2$,,,,$a_n$}. Consider the set A={c$a_1$,c$a_2$,,,,c$a_n$}, for some c$in$R, then each element is distinct. However; since by the definition of a ring, the ring is closed under multiplication, it follows that A$subseteq$R. Now, we must show that R$subseteq$A. Since $1.x=x.1=x$, it follows that x$in$A; therefore, R$subseteq$A.
I am trying to avoid the theorem that every finite integral domain is a field, partly because i'm using the above lemma to prove that every finite integral domain is a field.
Is my proof correct? What can I do to make it better? I'm just worried about the 1.x =x.1 =x part because I assumed that it holds for an arbitrary c and now i'm assuming it holds for c=1.
abstract-algebra ring-theory
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
$endgroup$
Lemma: Let R be a finite integral domain. Assume the elements of R are {$a_1$,$a_2$,,,,$a_n$} for some n$in$ $mathbb{N}$. Then for c$in$R, A={c$a_1$,c$a_2$,,,,c$a_n$}=R.
My proof: Since R is a finite integral domain. By definition, R is a finite commutative ring with the property that $forall$ a,b,c $in$ R with c$neq0$ if $ca=cb$ then $a=b$. Let R have elements {$a_1$,$a_2$,,,,$a_n$}. Consider the set A={c$a_1$,c$a_2$,,,,c$a_n$}, for some c$in$R, then each element is distinct. However; since by the definition of a ring, the ring is closed under multiplication, it follows that A$subseteq$R. Now, we must show that R$subseteq$A. Since $1.x=x.1=x$, it follows that x$in$A; therefore, R$subseteq$A.
I am trying to avoid the theorem that every finite integral domain is a field, partly because i'm using the above lemma to prove that every finite integral domain is a field.
Is my proof correct? What can I do to make it better? I'm just worried about the 1.x =x.1 =x part because I assumed that it holds for an arbitrary c and now i'm assuming it holds for c=1.
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
edited Jan 9 at 12:24
mathsssislife
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
asked Jan 9 at 12:08
mathsssislifemathsssislife
408
408
1
$begingroup$
Why "..., but clearly $1cdot x=xcdot 1=xin A$"? Is it obvious?
$endgroup$
– Song
Jan 9 at 12:13
$begingroup$
If $A$ is not all of $R$, then there would exist some element $a_k in R$ such that $ca_i = a_k=ca_j$
$endgroup$
– JavaMan
Jan 9 at 12:27
1
$begingroup$
I don’t think it works you say that 1x=x which is fine, but that doesn’t prove that R is a subset of A. That would only prove this if c were 1. Think about it, I give you an elememt x of R and you have to show me that it’s equal to cy for some (possibly different) y in R. The fact that 1x=x does not prove this, because c is fixed and differrent from 1 (if c=1 then this is trivial)
$endgroup$
– Ovi
Jan 9 at 12:28
$begingroup$
That's what I was worried about, what can I do to show that R is a subset of A?
$endgroup$
– mathsssislife
Jan 9 at 12:30
add a comment |
1
$begingroup$
Why "..., but clearly $1cdot x=xcdot 1=xin A$"? Is it obvious?
$endgroup$
– Song
Jan 9 at 12:13
$begingroup$
If $A$ is not all of $R$, then there would exist some element $a_k in R$ such that $ca_i = a_k=ca_j$
$endgroup$
– JavaMan
Jan 9 at 12:27
1
$begingroup$
I don’t think it works you say that 1x=x which is fine, but that doesn’t prove that R is a subset of A. That would only prove this if c were 1. Think about it, I give you an elememt x of R and you have to show me that it’s equal to cy for some (possibly different) y in R. The fact that 1x=x does not prove this, because c is fixed and differrent from 1 (if c=1 then this is trivial)
$endgroup$
– Ovi
Jan 9 at 12:28
$begingroup$
That's what I was worried about, what can I do to show that R is a subset of A?
$endgroup$
– mathsssislife
Jan 9 at 12:30
1
1
$begingroup$
Why "..., but clearly $1cdot x=xcdot 1=xin A$"? Is it obvious?
$endgroup$
– Song
Jan 9 at 12:13
$begingroup$
Why "..., but clearly $1cdot x=xcdot 1=xin A$"? Is it obvious?
$endgroup$
– Song
Jan 9 at 12:13
$begingroup$
If $A$ is not all of $R$, then there would exist some element $a_k in R$ such that $ca_i = a_k=ca_j$
$endgroup$
– JavaMan
Jan 9 at 12:27
$begingroup$
If $A$ is not all of $R$, then there would exist some element $a_k in R$ such that $ca_i = a_k=ca_j$
$endgroup$
– JavaMan
Jan 9 at 12:27
1
1
$begingroup$
I don’t think it works you say that 1x=x which is fine, but that doesn’t prove that R is a subset of A. That would only prove this if c were 1. Think about it, I give you an elememt x of R and you have to show me that it’s equal to cy for some (possibly different) y in R. The fact that 1x=x does not prove this, because c is fixed and differrent from 1 (if c=1 then this is trivial)
$endgroup$
– Ovi
Jan 9 at 12:28
$begingroup$
I don’t think it works you say that 1x=x which is fine, but that doesn’t prove that R is a subset of A. That would only prove this if c were 1. Think about it, I give you an elememt x of R and you have to show me that it’s equal to cy for some (possibly different) y in R. The fact that 1x=x does not prove this, because c is fixed and differrent from 1 (if c=1 then this is trivial)
$endgroup$
– Ovi
Jan 9 at 12:28
$begingroup$
That's what I was worried about, what can I do to show that R is a subset of A?
$endgroup$
– mathsssislife
Jan 9 at 12:30
$begingroup$
That's what I was worried about, what can I do to show that R is a subset of A?
$endgroup$
– mathsssislife
Jan 9 at 12:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Of course this only works if $cneq 0$. You must show that the map $x mapsto cx$ is injective and then, by finiteness of the set, you are done.
Note that $cx=cy$ implies $cx-cy=0$ which implies $c(x-y)=0$. Then use the definition of integral domain to conclude that $x=y$.
Note that the even integers $2mathbb{Z}$ are a proper subset of the integers $mathbb{Z}$ so you definately have to use the finiteness of $R$ in the proof.
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
$endgroup$
$begingroup$
Is there a way to do it without the need to show that the map is injective?
$endgroup$
– mathsssislife
Jan 9 at 15:55
$begingroup$
@mathsssislife not that I know
$endgroup$
– Anguepa
Jan 10 at 1:15
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Of course this only works if $cneq 0$. You must show that the map $x mapsto cx$ is injective and then, by finiteness of the set, you are done.
Note that $cx=cy$ implies $cx-cy=0$ which implies $c(x-y)=0$. Then use the definition of integral domain to conclude that $x=y$.
Note that the even integers $2mathbb{Z}$ are a proper subset of the integers $mathbb{Z}$ so you definately have to use the finiteness of $R$ in the proof.
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
$endgroup$
$begingroup$
Is there a way to do it without the need to show that the map is injective?
$endgroup$
– mathsssislife
Jan 9 at 15:55
$begingroup$
@mathsssislife not that I know
$endgroup$
– Anguepa
Jan 10 at 1:15
add a comment |
$begingroup$
Of course this only works if $cneq 0$. You must show that the map $x mapsto cx$ is injective and then, by finiteness of the set, you are done.
Note that $cx=cy$ implies $cx-cy=0$ which implies $c(x-y)=0$. Then use the definition of integral domain to conclude that $x=y$.
Note that the even integers $2mathbb{Z}$ are a proper subset of the integers $mathbb{Z}$ so you definately have to use the finiteness of $R$ in the proof.
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
$endgroup$
$begingroup$
Is there a way to do it without the need to show that the map is injective?
$endgroup$
– mathsssislife
Jan 9 at 15:55
$begingroup$
@mathsssislife not that I know
$endgroup$
– Anguepa
Jan 10 at 1:15
add a comment |
$begingroup$
Of course this only works if $cneq 0$. You must show that the map $x mapsto cx$ is injective and then, by finiteness of the set, you are done.
Note that $cx=cy$ implies $cx-cy=0$ which implies $c(x-y)=0$. Then use the definition of integral domain to conclude that $x=y$.
Note that the even integers $2mathbb{Z}$ are a proper subset of the integers $mathbb{Z}$ so you definately have to use the finiteness of $R$ in the proof.
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
$endgroup$
Of course this only works if $cneq 0$. You must show that the map $x mapsto cx$ is injective and then, by finiteness of the set, you are done.
Note that $cx=cy$ implies $cx-cy=0$ which implies $c(x-y)=0$. Then use the definition of integral domain to conclude that $x=y$.
Note that the even integers $2mathbb{Z}$ are a proper subset of the integers $mathbb{Z}$ so you definately have to use the finiteness of $R$ in the proof.
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
edited Jan 9 at 13:58
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
answered Jan 9 at 12:26
AnguepaAnguepa
1,401819
1,401819
$begingroup$
Is there a way to do it without the need to show that the map is injective?
$endgroup$
– mathsssislife
Jan 9 at 15:55
$begingroup$
@mathsssislife not that I know
$endgroup$
– Anguepa
Jan 10 at 1:15
add a comment |
$begingroup$
Is there a way to do it without the need to show that the map is injective?
$endgroup$
– mathsssislife
Jan 9 at 15:55
$begingroup$
@mathsssislife not that I know
$endgroup$
– Anguepa
Jan 10 at 1:15
$begingroup$
Is there a way to do it without the need to show that the map is injective?
$endgroup$
– mathsssislife
Jan 9 at 15:55
$begingroup$
Is there a way to do it without the need to show that the map is injective?
$endgroup$
– mathsssislife
Jan 9 at 15:55
$begingroup$
@mathsssislife not that I know
$endgroup$
– Anguepa
Jan 10 at 1:15
$begingroup$
@mathsssislife not that I know
$endgroup$
– Anguepa
Jan 10 at 1:15
add a comment |
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1
$begingroup$
Why "..., but clearly $1cdot x=xcdot 1=xin A$"? Is it obvious?
$endgroup$
– Song
Jan 9 at 12:13
$begingroup$
If $A$ is not all of $R$, then there would exist some element $a_k in R$ such that $ca_i = a_k=ca_j$
$endgroup$
– JavaMan
Jan 9 at 12:27
1
$begingroup$
I don’t think it works you say that 1x=x which is fine, but that doesn’t prove that R is a subset of A. That would only prove this if c were 1. Think about it, I give you an elememt x of R and you have to show me that it’s equal to cy for some (possibly different) y in R. The fact that 1x=x does not prove this, because c is fixed and differrent from 1 (if c=1 then this is trivial)
$endgroup$
– Ovi
Jan 9 at 12:28
$begingroup$
That's what I was worried about, what can I do to show that R is a subset of A?
$endgroup$
– mathsssislife
Jan 9 at 12:30