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| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 1 year, 10 months |
| seen | 1 hour ago | |
| stats | profile views | 218 |
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May 19 |
comment |
$[(|X| \ge \aleph_0) \;\wedge\; (A \subset X) \;\wedge\; (|A| = |X\backslash A|)] \Rightarrow |A| = |X|$ Thanks, but I think that what I'm trying to prove is essentially the fact $|A| + |A| = |A|$ that you invoke in your proof... (Establishing that $|A| \ge \omega$, as you did, does help though.) |
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May 19 |
revised |
$[(|X| \ge \aleph_0) \;\wedge\; (A \subset X) \;\wedge\; (|A| = |X\backslash A|)] \Rightarrow |A| = |X|$ edited title |
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May 19 |
revised |
$[(|X| \ge \aleph_0) \;\wedge\; (A \subset X) \;\wedge\; (|A| = |X\backslash A|)] \Rightarrow |A| = |X|$ deleted 54 characters in body; edited title |
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May 19 |
asked | $[(|X| \ge \aleph_0) \;\wedge\; (A \subset X) \;\wedge\; (|A| = |X\backslash A|)] \Rightarrow |A| = |X|$ |
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May 18 |
comment |
Can't argue with success? Looking for “bad math” that “gets away with it” @J.M.: Thanks, that's indeed quite the mother lode... |
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May 14 |
comment |
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? Thanks, I had not noticed (b). |
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May 14 |
comment |
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? Nice proof. Thanks. My only quibble with it (and it's tiny) is that its reference to "prime ideals" was unnecessary, since it already contains the proof of $xy \in \ker h \Rightarrow x \in \ker h$ or $y \in \ker h$, which is all the rest of the proof needs. |
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May 14 |
revised |
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? edited title |
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May 14 |
accepted | Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? |
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May 14 |
comment |
Do Boolean rings always have a unit element? Thanks for your answer. It flies so many miles above my head that I can't reasonably ask you to explain it further (my question requested elementary proofs for a reason). Needless to say, I can't even begin to evaluate it, but others may find it useful. |
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May 14 |
comment |
Do Boolean rings always have a unit element? @BorisNovikov: I don't think so. I happen to know some operations that work (namely, symmetric difference for $+$ and intersection for $\cdot$), and that's why I decided to accept your answer (reluctantly), but I consider them neither "evident" nor "natural". |
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May 14 |
comment |
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? @Stahl: thanks for the explanation. I did not realize that some definitions of a ring assume that the ring has a unit element. I have modified my question to explicitly deny this assumption. |
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May 14 |
revised |
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? added 48 characters in body; edited title |
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May 14 |
accepted | Do Boolean rings always have a unit element? |
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May 14 |
comment |
Do Boolean rings always have a unit element? @ZhenLin: What do you call something that is like $2\mathbb{Z}$, which is similar to what you call a "ring", except that it does not include a unit? I will gladly rephrase my question to use your terminology. |
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May 14 |
revised |
Do Boolean rings always have a unit element? deleted 3 characters in body |
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May 14 |
revised |
Do Boolean rings always have a unit element? added 121 characters in body |
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May 14 |
asked | Do Boolean rings always have a unit element? |
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May 14 |
comment |
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? Thanks again, but I'm embarrassed to admit that there are several steps in your proof I can't follow. For one, I don't know how you know that $B$ has a unit, $1_B$. Second, I don't see why $x - (x - 1_B)$ has to belong to $J$, even if I assume that $J$ is an ideal. |
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May 14 |
comment |
Is the kernel of a homomorphism from a Boolean ring to $\mathbb{Z}_2$ always a maximal ideal? Thanks, I'm looking for more elementary proofs... (I've clarified my question) |
