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3h
revised Extracting the Axis a Quaternion is rotating around from the Quaternion itself Directly
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4h
answered Extracting the Axis a Quaternion is rotating around from the Quaternion itself Directly
11h
comment $\Bbb R,\Bbb C,\Bbb H$ are the only complete normed division rings
@MarioCarneiro The type of norm involved here is this flavor, not norms like those for Banach algebras.
11h
comment $\Bbb R,\Bbb C,\Bbb H$ are the only complete normed division rings
The Frobenius Theorem doesn't make any mention of norms. I thought that "normed" and "finite dimensional" were simply two different conditions that were sufficient to make the theorem, but perhaps every proof of Frobenius Theorem depends on manufacturing a norm out of a finite dimensional algebra, somehow. Certainly we all know the norms that exist on those three algebras.
13h
revised When is an ideal also a ring, and could then be anything said about its relation to the original ring
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13h
comment When is an ideal also a ring, and could then be anything said about its relation to the original ring
@Stefan I went ahead and incorporated some of our discussion into the answer since I am at a real computer now (and not on a mobile device.)
13h
revised When is an ideal also a ring, and could then be anything said about its relation to the original ring
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15h
comment When is an ideal also a ring, and could then be anything said about its relation to the original ring
@ZelosMalum You don't believe in rings without identity? or just in the name "rng"?
17h
comment When is an ideal also a ring, and could then be anything said about its relation to the original ring
@ZelosMalum I think most people say "rng" as "rung," but my teacher joked it was pronounced "wrong." You might also find it amusing to know that semirings (rings where the underlying abelian group is relaxed to a commutative monoid, so there are 'no negatives') are sometimes called "rigs" (no n). Much easier to pronounce...
17h
comment When is an ideal also a ring, and could then be anything said about its relation to the original ring
@Stefan : Yes, that is correct, $eR$ is not necessarily an ideal. Here, we have it as an assumption. If $I$ is a two-sided ideal with an identity $e$, then $I=eI\subseteq eR\subseteq I$, so $I=eR$. Similarly $I=Re$. Both $Re$ and $eR$ are the ideal $I$ under these assumptions.
17h
comment When is an ideal also a ring, and could then be anything said about its relation to the original ring
@Stefan Oops, didn't realize I borked a dollar sign somewhere in my last comment. Fixing it here: "if $I$ is an ideal, and $x\in I$ and $S$ is any nonempty subset of and $R$ , then $Sx\subseteq I$ . Apply this to $I=eR$ and $S=(1-e)R$"
20h
comment When is an ideal also a ring, and could then be anything said about its relation to the original ring
@Stefan if $I$ is an ideal, and x\in I$ and S is any nonempty subset of $R$, then $Sx\subseteq I$. Apply this to $I=eR$ and $S=(1-e)R$
20h
comment When is an ideal also a ring, and could then be anything said about its relation to the original ring
@Stefan good, because that is a false statement that nobody claimed to be true. The conclusion is that $(1-e)Re\subseteq eR$
1d
answered Ideals and submodules are the same
1d
revised Definition of Inverse in Linear and Abstract Algebra
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1d
revised Definition of Inverse in Linear and Abstract Algebra
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1d
answered Definition of Inverse in Linear and Abstract Algebra
1d
comment Is it accurate to say that multiplication of two integers yields an integer?
Talking about factorization a is beside the point, considering that they are algebraically equivalent to their product, and their product is entirely algebra within the ring of integers.
1d
revised Is it accurate to say that multiplication of two integers yields an integer?
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1d
revised Is it accurate to say that multiplication of two integers yields an integer?
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