MIT PRIMES/Art of Problem Solving
CROWDMATH 2024: Generalizations of the Notion of Primes
Generalizations of the Notion of Primes
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Exercise 0.4
Thayaden 3
N
Nov 9, 2024
by felixgotti
In general, is a ring. A ring is and integral domain iff is the only zero-divisor. Another way of putting this is for some prime number show that,
where where . Consider for some where is still some prime number. This implies although by definition ONLY for some this implies that either or although in modulus we have thus or must be a zero-divisor thus we can conclude that for any given prime is an integral domain. Consider if we have that implies that , in this case, or where is equal to the other that implies that is a zero divisor and if is non-prime that implies that for some non- divisor, is a zero divisor.
where where . Consider for some where is still some prime number. This implies although by definition ONLY for some this implies that either or although in modulus we have thus or must be a zero-divisor thus we can conclude that for any given prime is an integral domain. Consider if we have that implies that , in this case, or where is equal to the other that implies that is a zero divisor and if is non-prime that implies that for some non- divisor, is a zero divisor.
3 replies
Exercise 0.3
Thayaden 2
N
Nov 7, 2024
by Thayaden
Crazy Idea but I think the condtions work
and
yet
and
yet
2 replies
Exercise 0.8
Thayaden 1
N
Oct 31, 2024
by felixgotti
Consider and such that we clearly see that,
Recall thus we can clearly see that as they are each other inverse that thus let,
Thus we see,
In other words and are elements in the group of units of although since it is also the group of units of therefor
Recall thus we can clearly see that as they are each other inverse that thus let,
Thus we see,
In other words and are elements in the group of units of although since it is also the group of units of therefor
1 reply
units...
Thayaden 1
N
Oct 31, 2024
by felixgotti
Part 1:
Notice thus for we might enharite this too. This might be easily proven by letting the coefficient of the unreal part be . Thus we might know that at least is a unit. Taking that idea once again letting the real part be we have,
Thus as we are communitive is in the set this makes sense from an algebraic perspective on the complex plane when multiplying a unit we have a change of direction likewise this is also true for real numbers thus
Notice thus for we might enharite this too. This might be easily proven by letting the coefficient of the unreal part be . Thus we might know that at least is a unit. Taking that idea once again letting the real part be we have,
Thus as we are communitive is in the set this makes sense from an algebraic perspective on the complex plane when multiplying a unit we have a change of direction likewise this is also true for real numbers thus
1 reply
Exercise 0.5 crazy
Thayaden 1
N
Oct 31, 2024
by felixgotti
Let be a finite integral domain that is not a field. For some there is no for some many as is not a field. Consider the powers of ,
let,
Consider the first powers as unique, implying that is not unique or consider the first as not unique. In any given case for some ,
If that implies that (this can be shown inductively) and we know that so that implies that,
Although that implies that has an inverse, that begins the only numbers that could not have an inverse indeed do have an inverse thus all numbers in have an inverse thus is a field!
let,
Consider the first powers as unique, implying that is not unique or consider the first as not unique. In any given case for some ,
If that implies that (this can be shown inductively) and we know that so that implies that,
Although that implies that has an inverse, that begins the only numbers that could not have an inverse indeed do have an inverse thus all numbers in have an inverse thus is a field!
1 reply
Exercise 0.7
Thayaden 0
Oct 2, 2024
If is an intergral domain then is too,
Let such that they are non zero, denote
Taking let be the leading coefficient of . Notice that thus so in short for the leading coefficient of cannot be thus must be an integral domain.
Going the other way say is an integral domain,
For the sake of contradiction let such that and , let taking there product,
This contradicts our previous statement thus is an integral domain!
Finally iff is an integral domain then must also be an integral domain
Let such that they are non zero, denote
Taking let be the leading coefficient of . Notice that thus so in short for the leading coefficient of cannot be thus must be an integral domain.
Going the other way say is an integral domain,
For the sake of contradiction let such that and , let taking there product,
This contradicts our previous statement thus is an integral domain!
Finally iff is an integral domain then must also be an integral domain
0 replies
Exercise 0.2
Thayaden 0
Sep 16, 2024
Let be a submonoid of , where runs over is the index set. Let be the intersection of all possible submonoids of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a submonoid. Now we just need to prove that it is a monoid with respect to the operation of . Let by intersection rules and and notice is a submonoid in itself and thus the operation for every thus by intersection rules again. Thus is a submonoid in itself!
Let be a subgroup of , where runs over is the index set. Let be the intersection of all possible subgroups of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of . Let by intersection rules and and notice is a subgroup in itself and thus the operation for every thus by intersection rules again. Finally, we need to show that every element has an inverse. Let and let represent the inverse of . Notice if then it follows that although notice that all are subgroups of and thus and since for any then by intersection thus is a subgroup of
Let be a subgroup of , where runs over is the index set. Let be the intersection of all possible subgroups of . Notice now we just need to prove . Let be the identity of notice that and in turn therfor has the identity of the first characteristic of a subgroup. Now we need to prove that it is a subgroup with respect to the operation of . Let by intersection rules and and notice is a subgroup in itself and thus the operation for every thus by intersection rules again. Finally, we need to show that every element has an inverse. Let and let represent the inverse of . Notice if then it follows that although notice that all are subgroups of and thus and since for any then by intersection thus is a subgroup of
0 replies
Exercise 0.1(part 1)
Thayaden 1
N
Sep 11, 2024
by Thayaden
let by definition,
this is true for all thus let this might lead us to (from ),
Now let (using ),
By the transitive property,
as the first condition is true, we might then say that given all identity elements of denoted as we might have,
and inductively,
thus,
. Or in plain text, we might say that as one identity is equal to another then if we have some identity it is the same as the identity we already have.
this is true for all thus let this might lead us to (from ),
Now let (using ),
By the transitive property,
as the first condition is true, we might then say that given all identity elements of denoted as we might have,
and inductively,
thus,
. Or in plain text, we might say that as one identity is equal to another then if we have some identity it is the same as the identity we already have.
1 reply
Exercise 3.4
FoxProdigy 1
N
Aug 19, 2024
by felixgotti
Problem Statement: Prove that a monoid is a GL-monoid if and only if it is prime-like.
Proof:
Prime-like GL:
Let be a prime-like monoid. Suppose such that . Suppose further towards a contradiction that ; that is, that there exists some divisor with . Then because is prime-like, has some divisor with either or . Either way, we also must have , which means that either or . This is a contradiction. Thus, , which means is GL.
GL prime-like:
Let be a GL monoid, and suppose with . Assume towards a contradiction that for all , and (in other words, shares no divisors with or ). This implies that . Since is GL, this means that . This contradicts the assumption that . Thus, there must exist some with or , which means is prime-like.
Proof:
Prime-like GL:
Let be a prime-like monoid. Suppose such that . Suppose further towards a contradiction that ; that is, that there exists some divisor with . Then because is prime-like, has some divisor with either or . Either way, we also must have , which means that either or . This is a contradiction. Thus, , which means is GL.
GL prime-like:
Let be a GL monoid, and suppose with . Assume towards a contradiction that for all , and (in other words, shares no divisors with or ). This implies that . Since is GL, this means that . This contradicts the assumption that . Thus, there must exist some with or , which means is prime-like.
1 reply
Exercise 0.1
palindrome868 0
Aug 5, 2024
What I've tried so far:
(1) If are two distinct identity elements of , then , a contradiction.
(2) Let be the identity element of . If are two distinct inverses of , then
a contradiction.
(3) Claim 1: Associativity holds in
.
Claim 2: Identity holds in
The identity element of is the inverse of itself, so it is also in . Since the operations in and are the same, it is the identity of as well.
Claim 3: Closure holds in
Let the identity element be .
.
, so is invertible (namely, its inverse is ), which implies .
Claim 4: Every element in has an inverse
This is true by the definition of .
By definition, these four claims are sufficient to determine is a group.
Where I'm stuck:
Please check :)
(1) If are two distinct identity elements of , then , a contradiction.
(2) Let be the identity element of . If are two distinct inverses of , then
a contradiction.
(3) Claim 1: Associativity holds in
.
Claim 2: Identity holds in
The identity element of is the inverse of itself, so it is also in . Since the operations in and are the same, it is the identity of as well.
Claim 3: Closure holds in
Let the identity element be .
.
, so is invertible (namely, its inverse is ), which implies .
Claim 4: Every element in has an inverse
This is true by the definition of .
By definition, these four claims are sufficient to determine is a group.
Where I'm stuck:
Please check :)
0 replies
Exercise 0.1(part 1)
Thayaden 1
N
Sep 11, 2024
by Thayaden
let by definition,
this is true for all thus let this might lead us to (from ),
Now let (using ),
By the transitive property,
as the first condition is true, we might then say that given all identity elements of denoted as we might have,
and inductively,
thus,
. Or in plain text, we might say that as one identity is equal to another then if we have some identity it is the same as the identity we already have.
this is true for all thus let this might lead us to (from ),
Now let (using ),
By the transitive property,
as the first condition is true, we might then say that given all identity elements of denoted as we might have,
and inductively,
thus,
. Or in plain text, we might say that as one identity is equal to another then if we have some identity it is the same as the identity we already have.
1 reply