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Generalizations of the Notion of Primes J
Exercise 0.5 crazy
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Thayaden
1361 posts
Thayaden
#1 h  Sep 24, 2024, 11:08 PM
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Let $S$ be a finite integral domain that is not a field. For some $t\in S \not\ \{ 0\}$ there is no $t^{-1}\in S$ for some many $t$ as $R$ is not a field. Consider the powers of $t$,
$$t,t^2,t^3,t^4, ...$$let,
$$\underbrace{t,t^2,t^3,...,t^{|S|}}_{|S|},t^{|S|+1}$$Consider the first $|S|$ powers as unique, implying that $t^{|S|+1}$ is not unique or consider the first $|S|$ as not unique. In any given case for some $m>n$,
$$t^m=t^n$$$$t^m-t^n=0$$$$t^n \cdot (t^{m-n}-1)=0$$If $t^n=0$ that implies that $t=0$ (this can be shown inductively) and we know that $t \neq 0$ so that implies that,
$$t^{m-n}-1=0$$$$t^{m-n}=1$$$$t^{m-n-1}\cdot t=1$$Although that implies that $t$ has an inverse, that begins $t^{m-n-1}$ the only numbers that could not have an inverse indeed do have an inverse thus all numbers in $S$ have an inverse thus $S$ is a field!
This post has been edited 1 time. Last edited by Thayaden, Nov 7, 2024, 7:43 PM
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felixgotti
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felixgotti
#2 h  Oct 31, 2024, 11:38 PM
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Good job! Your solution is correct. However, there are many typos. Could you please fix them? Thanks!
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Exercise 0.5 crazy
Thayaden   1
N Oct 31, 2024 by felixgotti
Let $S$ be a finite integral domain that is not a field. For some $t\in S \not\ \{ 0\}$ there is no $t^{-1}\in S$ for some many $t$ as $R$ is not a field. Consider the powers of $t$,
$$t,t^2,t^3,t^4, ...$$let,
$$\underbrace{t,t^2,t^3,...,t^{|S|}}_{|S|},t^{|S|+1}$$Consider the first $|S|$ powers as unique, implying that $t^{|S|+1}$ is not unique or consider the first $|S|$ as not unique. In any given case for some $m>n$,
$$t^m=t^n$$$$t^m-t^n=0$$$$t^n \cdot (t^{m-n}-1)=0$$If $t^n=0$ that implies that $t=0$ (this can be shown inductively) and we know that $t \neq 0$ so that implies that,
$$t^{m-n}-1=0$$$$t^{m-n}=1$$$$t^{m-n-1}\cdot t=1$$Although that implies that $t$ has an inverse, that begins $t^{m-n-1}$ the only numbers that could not have an inverse indeed do have an inverse thus all numbers in $S$ have an inverse thus $S$ is a field!
1 reply
Thayaden
Sep 24, 2024
felixgotti
Oct 31, 2024
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Exercise 0.5
elee310   3
N Aug 4, 2024 by elee310
What I've tried so far:
$\text{As written in Resource 0, all rings R we use are commutative.}$
$\text{Thus, we only have to prove that all elements have inverses to prove that } (R \backslash \{0\}, *) \text{is an abelian group.}$
$\text{As } (R \backslash \{0\}, *) \text{ is cancellative, we have to prove that cancellation in multiplicative monoids of finite rings imply existence of inverse.}$
$\text{Once that is proved, all elements, which are cancellative, have inverses, making } (R \backslash \{0\}, *) \text{an abelian group.}$

Where I'm stuck:

How should I prove that cancellation imply inverse in finite monoids?
My guess is that "cancellation" in finite rings must happen by multiplying another element, which makes that element an inverse?
3 replies
elee310
Jul 29, 2024
elee310
Aug 4, 2024
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Exercise 0.5
hkoo   13
N Jun 18, 2024 by hkoo
What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>

Where I'm stuck:

<Describe what's confusing you, or what your question is here!>

I can prove finite R is cancellative. but can't prove commutativity for multiply operation. i can find wedderburn's theorem in the google, which is used to prove commutativity. i want to prove wedderburn's theorem by myself. Can i get a hint or motivation of this theorem?
13 replies
hkoo
May 30, 2024
hkoo
Jun 18, 2024
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Posible solution of Exercise 5.
mitosaurio47   7
N May 24, 2024 by hkoo
Well I have this idea for Exercise 5 :

Let $R = \{0, 1 = a_0, a_1, a_2, \cdots, a_n \}$ a finite integral domain. If $R$ is not a field then we have a not invertible $a_k$. So in the set of products $S_k= \{ a_k \cdot a_i: i \in \mathbb{N}_0, 0 \leq i \leq n\}$ they can only take the values $R$ \ $\{0, 1\}$ (because if it's 1 then $a_k$ would be invertible and if it's 0 then $R$ it's not an integral domain). By Pigeonhole Principle there are $a_{j_1} \neq a_{j_2}$ such that:
$$a_k \cdot a_{j_1} = a_j = a_k \cdot a_{j_2}$$But the multiplicative monoid of an integral domain is cancellative, then $a_{j_1} = a_{j_2}$ wich is a contradiction. So $a_i$ is invertible for all $i$, then $R$ is a field. $\square$

7 replies
mitosaurio47
Jan 21, 2024
hkoo
May 24, 2024
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No more topics!
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Exercise 0.5 crazy
Thayaden   1
N Oct 31, 2024 by felixgotti
Let $S$ be a finite integral domain that is not a field. For some $t\in S \not\ \{ 0\}$ there is no $t^{-1}\in S$ for some many $t$ as $R$ is not a field. Consider the powers of $t$,
$$t,t^2,t^3,t^4, ...$$let,
$$\underbrace{t,t^2,t^3,...,t^{|S|}}_{|S|},t^{|S|+1}$$Consider the first $|S|$ powers as unique, implying that $t^{|S|+1}$ is not unique or consider the first $|S|$ as not unique. In any given case for some $m>n$,
$$t^m=t^n$$$$t^m-t^n=0$$$$t^n \cdot (t^{m-n}-1)=0$$If $t^n=0$ that implies that $t=0$ (this can be shown inductively) and we know that $t \neq 0$ so that implies that,
$$t^{m-n}-1=0$$$$t^{m-n}=1$$$$t^{m-n-1}\cdot t=1$$Although that implies that $t$ has an inverse, that begins $t^{m-n-1}$ the only numbers that could not have an inverse indeed do have an inverse thus all numbers in $S$ have an inverse thus $S$ is a field!
1 reply
Thayaden
Sep 24, 2024
felixgotti
Oct 31, 2024
J