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Every ring with identity and even order either has zero divisors or $1+1=0$ holds

Homomorphisms from a unital ring to a ring with no zero divisors preserve unity?The set of zero-divisors in a ring $R$ containing maximal idealsAre there any zero divisors in this ring?Is an ideal which is maximal with respect to the property that it consists of zero divisors necessarily prime?Show that a polynomial ring in one indeterminate over a field doesnt have zero divisorsIf $R$ is commutative ring with identity, prove that $R$ is and integral domain iff cancelation holds in $R$Is a finite commutative ring with no zero-divisors always equal to the ideal generated by any of its nonzero elementsNeed to prove that a given nonzero ring $R$ with no zero divisors has both left and right identityProperties of $Amapsto (Acap X, Ysetminus (Acap Y))$ given $X,Yneqemptyset$ and $Xcap Yneq emptyset$Algebraic set is irreducible $iff$ its algebra has no zero divisors

1

$begingroup$

I'm asking two things, whether this proof is correct, and if there is a more intuitive proof I haven't thought of.

We assume that $R$ is finite and that $1+1 neq 0$ and $R$ has no zero divisors. Let $x in R$ with $x = -x$. Then, $x+x = (1+1)x = 0$. Since $1+1 neq 0$ and $R$ has no zero divisors, this results in $x = 0$, which means that $0$ is the only self-inverse element of $R$ (regarding addition). Next, consider a set $A subset R$ such that $A cap A^-1 = emptyset$ (where $A^-1 = x^-1;$) and $Rsetminus0 = A cup A^-1$ (the existence of such a set can be proved with Zorn's lemma, or, considering $R$ is finite, via induction). Then, $|Rsetminus0| = |Acup A^-1|$, and since $A cap A^-1 = emptyset$, $|R| - 1 = |A| + |A^-1|$. Obviously, $|A| = |A^-1|$ (consider the bijection $x mapsto x^-1$) and thus, $|R| = 2|A|+1$, which proves that $|R|$ is odd.

abstract-algebra proof-verification ring-theory

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asked Mar 22 at 0:21

NicolasNicolas

1717

$endgroup$

add a comment | 

1

$begingroup$

I'm asking two things, whether this proof is correct, and if there is a more intuitive proof I haven't thought of.

We assume that $R$ is finite and that $1+1 neq 0$ and $R$ has no zero divisors. Let $x in R$ with $x = -x$. Then, $x+x = (1+1)x = 0$. Since $1+1 neq 0$ and $R$ has no zero divisors, this results in $x = 0$, which means that $0$ is the only self-inverse element of $R$ (regarding addition). Next, consider a set $A subset R$ such that $A cap A^-1 = emptyset$ (where $A^-1 = x^-1;$) and $Rsetminus0 = A cup A^-1$ (the existence of such a set can be proved with Zorn's lemma, or, considering $R$ is finite, via induction). Then, $|Rsetminus0| = |Acup A^-1|$, and since $A cap A^-1 = emptyset$, $|R| - 1 = |A| + |A^-1|$. Obviously, $|A| = |A^-1|$ (consider the bijection $x mapsto x^-1$) and thus, $|R| = 2|A|+1$, which proves that $|R|$ is odd.

abstract-algebra proof-verification ring-theory

share|cite|improve this question

asked Mar 22 at 0:21

NicolasNicolas

1717

$endgroup$

add a comment | 

$begingroup$

I'm asking two things, whether this proof is correct, and if there is a more intuitive proof I haven't thought of.

We assume that $R$ is finite and that $1+1 neq 0$ and $R$ has no zero divisors. Let $x in R$ with $x = -x$. Then, $x+x = (1+1)x = 0$. Since $1+1 neq 0$ and $R$ has no zero divisors, this results in $x = 0$, which means that $0$ is the only self-inverse element of $R$ (regarding addition). Next, consider a set $A subset R$ such that $A cap A^-1 = emptyset$ (where $A^-1 = x^-1;$) and $Rsetminus0 = A cup A^-1$ (the existence of such a set can be proved with Zorn's lemma, or, considering $R$ is finite, via induction). Then, $|Rsetminus0| = |Acup A^-1|$, and since $A cap A^-1 = emptyset$, $|R| - 1 = |A| + |A^-1|$. Obviously, $|A| = |A^-1|$ (consider the bijection $x mapsto x^-1$) and thus, $|R| = 2|A|+1$, which proves that $|R|$ is odd.

abstract-algebra proof-verification ring-theory

share|cite|improve this question

asked Mar 22 at 0:21

NicolasNicolas

1717

$endgroup$

share|cite|improve this question

asked Mar 22 at 0:21

NicolasNicolas

1717

share|cite|improve this question

asked Mar 22 at 0:21

NicolasNicolas

1717

asked Mar 22 at 0:21

NicolasNicolas

1717

asked Mar 22 at 0:21

NicolasNicolas

1717

1 Answer
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oldest

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2

$begingroup$

Here's a simpler proof.

As a finite abelian group of even order, $R$ contains an element $a$ of order 2.

Thus $2a=0$. Either $2=0$, or $a$ is a zero-divisor.

Edit

Also, your proof is correct, though there's no need to use Zorn's lemma or induction explicitly. Instead it can be proved directly if you do the following:

Consider the equivalence relation $xsim y$ if $x=pm y$. This partitions $R$ into equivalence classes, all but one of which have two elements (as you observed, $0$ is the only element with $x=-x$). Thus $|R|=2n-1$, where $n$ is the number of equivalence classes.

share|cite|improve this answer

edited Mar 22 at 15:22

answered Mar 22 at 0:24

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, but let's note that the proof that any group of even order contains an element of order $2$ is usually given via a counting argument that's pretty much the same as what OP has given [but which can be written in a simpler way] — or else is seen as an implication of Cauchy's theorem (whose proofs are also quite involved).
  $endgroup$
  – M. Vinay
  Mar 22 at 12:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Vinay Good point, I hadn't considered that the OP might not have seen groups prior to seeing rings.
  $endgroup$
  – jgon
  Mar 22 at 15:18
  

  
  
  
  

add a comment | 

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2

$begingroup$

Here's a simpler proof.

As a finite abelian group of even order, $R$ contains an element $a$ of order 2.

Thus $2a=0$. Either $2=0$, or $a$ is a zero-divisor.

Edit

Also, your proof is correct, though there's no need to use Zorn's lemma or induction explicitly. Instead it can be proved directly if you do the following:

Consider the equivalence relation $xsim y$ if $x=pm y$. This partitions $R$ into equivalence classes, all but one of which have two elements (as you observed, $0$ is the only element with $x=-x$). Thus $|R|=2n-1$, where $n$ is the number of equivalence classes.

share|cite|improve this answer

edited Mar 22 at 15:22

answered Mar 22 at 0:24

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, but let's note that the proof that any group of even order contains an element of order $2$ is usually given via a counting argument that's pretty much the same as what OP has given [but which can be written in a simpler way] — or else is seen as an implication of Cauchy's theorem (whose proofs are also quite involved).
  $endgroup$
  – M. Vinay
  Mar 22 at 12:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Vinay Good point, I hadn't considered that the OP might not have seen groups prior to seeing rings.
  $endgroup$
  – jgon
  Mar 22 at 15:18
  

  
  
  
  

add a comment | 

2

$begingroup$

Here's a simpler proof.

As a finite abelian group of even order, $R$ contains an element $a$ of order 2.

Thus $2a=0$. Either $2=0$, or $a$ is a zero-divisor.

Edit

Also, your proof is correct, though there's no need to use Zorn's lemma or induction explicitly. Instead it can be proved directly if you do the following:

Consider the equivalence relation $xsim y$ if $x=pm y$. This partitions $R$ into equivalence classes, all but one of which have two elements (as you observed, $0$ is the only element with $x=-x$). Thus $|R|=2n-1$, where $n$ is the number of equivalence classes.

share|cite|improve this answer

edited Mar 22 at 15:22

answered Mar 22 at 0:24

jgonjgon

16.4k32143

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, but let's note that the proof that any group of even order contains an element of order $2$ is usually given via a counting argument that's pretty much the same as what OP has given [but which can be written in a simpler way] — or else is seen as an implication of Cauchy's theorem (whose proofs are also quite involved).
  $endgroup$
  – M. Vinay
  Mar 22 at 12:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @M.Vinay Good point, I hadn't considered that the OP might not have seen groups prior to seeing rings.
  $endgroup$
  – jgon
  Mar 22 at 15:18
  

  
  
  
  

add a comment | 

$begingroup$

Here's a simpler proof.

As a finite abelian group of even order, $R$ contains an element $a$ of order 2.

Thus $2a=0$. Either $2=0$, or $a$ is a zero-divisor.

Edit

Also, your proof is correct, though there's no need to use Zorn's lemma or induction explicitly. Instead it can be proved directly if you do the following:

Consider the equivalence relation $xsim y$ if $x=pm y$. This partitions $R$ into equivalence classes, all but one of which have two elements (as you observed, $0$ is the only element with $x=-x$). Thus $|R|=2n-1$, where $n$ is the number of equivalence classes.

share|cite|improve this answer

edited Mar 22 at 15:22

answered Mar 22 at 0:24

jgonjgon

16.4k32143

$endgroup$

share|cite|improve this answer

edited Mar 22 at 15:22

answered Mar 22 at 0:24

jgonjgon

16.4k32143

answered Mar 22 at 0:24

jgonjgon

16.4k32143

answered Mar 22 at 0:24

jgonjgon

16.4k32143