Proof about finite integral domains.












3












$begingroup$


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.










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$endgroup$








  • 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


















3












$begingroup$


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.










share|cite|improve this question











$endgroup$








  • 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
















3












3








3


1



$begingroup$


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.










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 9 at 12:24







mathsssislife

















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
















  • 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












1 Answer
1






active

oldest

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3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

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active

oldest

votes









3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















3












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














3












3








3





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 13:58

























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


















  • $begingroup$
    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


















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