Challenge Math Challenge - January 2019

Math_QED

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Part 1 of (9)

##\phi: R \to S## is a ring epimorphism. Define ##I:= \phi^{-1}(J)##. It is well known that the inverse image of an ideal is an ideal, thus ##I## is an ideal.

Define ##\psi: R/I \to S/J: [r] \mapsto [\phi(r)]##

This is well defined: If ##r \in I##, then ##\phi(r) \in J##.

Clearly, this is also a ring morphism.

For injectivity, assume ##[\phi(r)] = 0##, then ##\phi(r) \in J##, and ##r \in \phi^{-1}(J) = I##, thus ##[r] = 0##. The kernel is trivial and the map is injective.

Surjectivity follows immediately by surjectivity of ##\phi##.

It follows that ##\psi## is an isomorphism, and thus ##R/I \cong S/J##.
 
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fresh_42

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Part 1 of (9)

##\phi: R \to S## is a ring epimorphism. Define ##I:= \phi^{-1}(J)##. It is well known that the inverse image of an ideal is an ideal, thus ##I## is an ideal.

Define ##\psi: R/I \to S/J: [r] \mapsto [\phi(r)]##

This is well defined: If ##r \in I##, then ##\phi(r) \in J##.

Clearly, this is also a ring morphism.

For injectivity, assume ##[\phi(r)] = 0##, then ##\phi(r) \in J##, and ##r \in \phi^{-1}(J) = I##, thus ##[r] = 0##. The kernel is trivial and the map is injective.

Surjectivity follows immediately by surjectivity of ##\phi##.

It follows that ##\psi## is an isomorphism, and thus ##R/I \cong S/J##.
Correct.

You could have saved some lines by setting ##\psi\,' := \pi \circ \phi ## with the canonical projection ##\pi \, : \,S \twoheadrightarrow S/J## and ##I=\operatorname{ker}(\pi \phi)## plus the isomorphism theorem.

I'll wait for the second part before I mark it as solved.
 

Math_QED

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Correct.

You could have saved some lines by setting ##\psi\,' := \pi \circ \phi ## with the canonical projection ##\pi \, : \,S \twoheadrightarrow S/J## and ##I=\operatorname{ker}(\pi \phi)## plus the isomorphism theorem.

I'll wait for the second part before I mark it as solved.
I'm studying exams, so I don't have much time to try now. It is sufficient to prove that ##Z(S) = \phi(Z(R))## and the inclusion ##\supseteq## is immediate by writing out the definitions. My first guess would be that the other inclusion can fail. A counterexample (if the statement isn't true) should be in non-commutative ring theory. Will come back to the problem once I got more time.

Maybe the first part of the question is relevant to this part?
 

fresh_42

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Maybe the first part of the question is relevant to this part?
No.
... should be in non-commutative ring theory ...
Nobody said the rings were commutative!

I know that some distinguish between rings = commutative and pseudo-rings = not necessarily commutative, but a) I haven't learnt it this way, and b) do not see any advantage in it. A ring is an associative and distributive structure on an Abelian additive group. That's it. No mention of commutativity, nor unity.
 

scottdave

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For future reference, I think that #2 (the pirates and coconuts) should be slightly reworded for clarity. It states that each pirate takes a third of the available coconuts, but then there is an odd number left, so he gives one away.

If we start with a quantity of coconuts, which is divisible by 3, let's say that is 3N coconuts, where N is a whole number. The first pirate takes a third of that (N coconuts), leaving 2N remaining, which is an even number. Perhaps the storyteller meant to say that it is not evenly divisible by 3, or something to that effect?
 

fresh_42

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For future reference, I think that #2 (the pirates and coconuts) should be slightly reworded for clarity. It states that each pirate takes a third of the available coconuts, but then there is an odd number left, so he gives one away.

If we start with a quantity of coconuts, which is divisible by 3, let's say that is 3N coconuts, where N is a whole number. The first pirate takes a third of that (N coconuts), leaving 2N remaining, which is an even number. Perhaps the storyteller meant to say that it is not evenly divisible by 3, or something to that effect?
If he leaves 2N coconuts, then there is no need to give away one for the monkeys!
 

scottdave

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If he leaves 2N coconuts, then there is no need to give away one for the monkeys!
That's what I was thinking. But it appears that the problem intends something like this:
I am a pirate and wake up to steal some coconuts for myself and hide them. I'll take my "third", but I want to make sure the remaining pile is evenly divisible by 3. If the remaining pile is not, I will give away 1 or 2 (apparently only 1 was necessary) to make the remainder divisible by 3.
I expect to go back to sleep and wake up in the morning and we all get our "even share"

The next pirate wakes up with the same thinking. And the last pirate wakes up and does the same thing.
 

fresh_42

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That's what I was thinking. But it appears that the problem intends something like this:
I am a pirate and wake up to steal some coconuts for myself and hide them. I'll take my "third", but I want to make sure the remaining pile is evenly divisible by 3. If the remaining pile is not, I will give away 1 or 2 (apparently only 1 was necessary) to make the remainder divisible by 3.
I expect to go back to sleep and wake up in the morning and we all get our "even share"

The next pirate wakes up with the same thinking. And the last pirate wakes up and does the same thing.
Well, the given solution is the same as I intended and which was given where I took the riddle from. So instead to actively cheat the other pirates, the assumption seems to be passive, that every pirate wants to save his share before the others cheat on him.
 

PeroK

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That's what I was thinking. But it appears that the problem intends something like this:
I am a pirate and wake up to steal some coconuts for myself and hide them. I'll take my "third", but I want to make sure the remaining pile is evenly divisible by 3. If the remaining pile is not, I will give away 1 or 2 (apparently only 1 was necessary) to make the remainder divisible by 3.
I expect to go back to sleep and wake up in the morning and we all get our "even share"

The next pirate wakes up with the same thinking. And the last pirate wakes up and does the same thing.
I think you're right. The problem says "an odd number", but what is meant is "an extra coconut upon division by three".
 

fresh_42

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... but what is meant is "an extra coconut upon division by three"
No, because the pirate who took away his share, expects the rest to be divided by two, not three. No cheating intended here.
 

PeroK

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No, because the pirate who took away his share, expects the rest to be divided by two, not three. No cheating intended here.
Okay, honest pirates. Except, they all end up with different numbers of coconuts and they will all expect the other two to share the remaining coconuts. The whole funny business will be discovered and I suspect a knife fight would ensue in any case!
 

scottdave

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Okay, honest pirates. Except, they all end up with different numbers of coconuts and they will all expect the other two to share the remaining coconuts. The whole funny business will be discovered and I suspect a knife fight would ensue in any case!
So if I wake up and take my third of the pile. If the pile is evenly divisible by 3 before I take any, then I can take away a whole number, and the remaining must be an even number, regardless of my intentions.

Let's say we do start with a multiple of 3. Let's represent that by 3n, where n is a natural number. Now when I take away n coconuts (my third), I leave 2n, which is an even number, not odd.
 

fresh_42

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So if I wake up and take my third of the pile. If the pile is evenly divisible by 3 before I take any, then I can take away a whole number, and the remaining must be an even number, regardless of my intentions.

Let's say we do start with a multiple of 3. Let's represent that by 3n, where n is a natural number. Now when I take away n coconuts (my third), I leave 2n, which is an even number, not odd.
You're right. Had to be even ##(52,34,22)## and the next pirate to give away one for the monkey with 1 left for the monkeys at the end: ##78-26=52\; , \;52-1-17=34;34-1-11=22=3 \cdot 7 +1## makes ##(26+7,17+7,11+7,3)=(33,24,18,3)##.
 

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