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Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Check the comments above, under the other Travis' answer. When you choose $j^2=-1, ij=k, ji=k$ and get a commutative set, it's isomorphic to $\mathbb C^2$.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Yep, that checks out. i $=(i,i)$, j $=(i,-i)$, and k $=(-1,1)$.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Perhaps, but any ring forms a left-algebra over itself though, right? So if they had other division rings, wouldn't they already have other noncommutative algebras? I'd assumed that he avoided commutativity because he was writing this after having figured them out, and by using one less assumption he was able to simultaneously prove that no field over a basis of 3 elements existed, and none existed even if he gave up commutativity, an objection someone would likely raise as he was giving up commutativity while jumping from 2 to 4 elements.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
From reading the text above, it seems to me that he was looking for a field larger than $\mathbb C$ and was proving that one didn't exist. Since the quaternions were the first noncommutative division algebra discovered, it seems to me that they would have helped lead to these now-well-understood properties.
Nov
16
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
Like this. I figured I'd try getting Mathematica to symbolically solve for a set of 2x2 matrices that might commute like this, and right away it reduced the equations to several possibilities. I started plugging in values and ... oh, right, I have to tell it none of them equal each other. I told it to run again and it didn't finish, so I don't have high hopes
Nov
15
comment Why is $(-1) \cdot j = j \cdot (-1)$ for quaternions?
@JamesS.Cook It's wrong to say $Q_8$ (or $\mathbb H$) comes from the cross-product. That might be the perspective of people whose experiences are based in application. Hamilton's perspective is the modern one: In order to define cross-product of vectors (a nonassociative algebra), you need a vector space, which requires a field, which requires a ring, which requires a group. $Q_8$ comes from trying to develop a group that can be turned into a field larger than $\mathbb C$. Hamilton proved this cannot be done, but the quaternions form a noncommutative division ring, the next best thing.
Nov
15
comment Why should I consider the components $j^2$ and $k^2$ to be $=-1$ in the search for quaternions?
I have long noticed that one could define an associative (and commutative) product with $i^2=j^2 = -1$ and $k^2 = 1$, but wondered why I never heard about it. I appreciate @Micah 's comment above about getting zero divisors when trying to make a field out of a multiplicative group that contains a second element that squares to a positive real. Would this product I've specified also lead to the split quaternions, or something else? I'm guessing not since, if the split quaternions are isomorphic to the 2 by 2 matrix ring, they're also noncommutative.
Nov
13
comment What is the difference between ring homomorphism and module homomorphism?
I'm saying it wasn't a bad analogy, and examples help people who are trying to learn new concepts.
Nov
13
comment What is the difference between ring homomorphism and module homomorphism?
Clarifying your driving/flying analogy: Person $A$ and $B$ are both pilots (isomorphic under the structure: "having passed the flying exam"). Person $A$ and $C$ are both drivers (isomorphic under the structure: "having passed the driving exam"). But if person $B$ is not a driver, and person $C$ is not a pilot, these two are not isomorphic with respect to either structure.
Nov
13
comment Shorter proof of $R/I$ is a field if and only if $I$ is maximal
Your example gives a field, but I saw how $I + (x) = R$ meant that after sending $I$ to $0$, if I sent $x$ to $0$, everything collapsed, which meant $x$ was a unit in $R/I$. Further, $I$ needn't be maximal, i.e. this should work when $R/I$ is not a field. I think a more illuminating example is $I = 4\mathbb Z = (4)$. The lattice involves $(1), (2), (4)$, and $(3), (6), (12)$ "below" them. Sending $(4)$ to $(0)$, also sends $(12)$ there, meaning $(6)$ collapses to $(2)$ and $(3)$ collapses to $(1)$. $(2)$ is the only remaining maximal ideal and $3 + I$ isn't in it, so it's a unit.
Nov
12
comment What is the relationship between prime ideals and their generators?
When writing my question, I edited for simplicity, and removed everything about UFDs, but then tried to imply my question included UFDs when I said "what additional criteria are needed". Thanks for helping me understand generators better. I think over the weekend I'll type up a similar question just for UFDs and reference this one.
Nov
12
comment What is the relationship between prime ideals and their generators?
Something's not right about the 'visa-versa' part, but between Wikipedia, you, and then me, I'm not sure where the miscommunication / misunderstanding lies. In IDs (where irreducible is defined), every prime is irreducible. In UFDs, every irreducible is prime so they are equivalent (since every UFD is an ID). I originally removed this from my 'Known' section, but I've added it back.
Nov
12
comment What is the relationship between prime ideals and their generators?
In my mind "$\mathbb{Z}[x]$ and the like" can be very general. The way I understand things, $\mathbb{Z}$ is the free ring on $\{0,1\}$ (and thus the smallest free ring with unity). Then by adjoining indeterminates (but not necessarily allowing them to commute with existing elements) you can get any larger free ring, and then by imposing relations (by modding out ideals in that free ring), one can get any ring that exists.
Nov
12
revised What is the relationship between prime ideals and their generators?
added 105 characters in body
Nov
12
accepted What is the relationship between prime ideals and their generators?
Nov
12
comment What is the relationship between prime ideals and their generators?
Clearly my questions (about multiple generators) only apply outside a PID. I know that this example $\mathbb Z [\sqrt{-5}]$ is an ID, but not a UFD. I'm now wondering if this relies on the multiple ways to factor ... if this still holds in a UFD where prime and irreducible coincide.
Nov
12
comment What is the relationship between prime ideals and their generators?
Thanks, I've worked out everything except why $R/(p)$ must be finite for any $p \in I$. Why is that?
Nov
11
asked What is the relationship between prime ideals and their generators?
Nov
10
comment What's an example of an ideal in $\mathbb{Z}[\sqrt{-n}]$ that is not principal?
@goblin The most widely used convention is that integral domains are commutative, and domains are not. On the other hand, the most widely used convention for 'prime' is only defined in a commutative ring, and 'irreducible' is only defined in an integral domain.
Nov
8
revised How do I show a mapping is a homomorphism?
typo