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Aug
21
revised In algebraic topology, for a function $f$ what does $f _\ast$ mean?
ce
Aug
21
comment Is there a binary operator (besides composition) closed under permutations or a notion of a metric space on permutations?
What do you mean "additive operation"?
Aug
21
answered Is there a binary operator (besides composition) closed under permutations or a notion of a metric space on permutations?
Aug
21
comment If $R_1,..$ are rings with identity and $I$ is an ideal in $R_1\times.. R_n$, then $I=A_1\times…\times A_m$,where each $A_i$ is an ideal in $R_i$.
@haiganghu might be worth answering your own question to explain what you realized then!
Aug
20
comment If $R_1,..$ are rings with identity and $I$ is an ideal in $R_1\times.. R_n$, then $I=A_1\times…\times A_m$,where each $A_i$ is an ideal in $R_i$.
Until you write how you "can get $I$ has the form $A_1\times...\times A_m$" we can't know for sure if you have done it without using identities. It is certainly easy if you are given identities.
Aug
19
answered Can we construct a homomorphism from a projective module into a free module?
Aug
19
comment Adjoining an identity to a ring
It's a neat little exercise. What algebra book was it, btw?
Aug
19
comment Adjoining an identity to a ring
@karparvar that also explains why $(0, R_0)$ does collapse to zero when localizing at any prime containing $R_0$ :)
Aug
19
comment Adjoining an identity to a ring
@karparvar Ah ok, e would definitely be zero in the localization, yes.
Aug
19
comment Adjoining an identity to a ring
@karparvar Under this multiplication, $(2,0)(0,x)=0$ for any $x$ in the $R_0$, and hence in any localization. Unless $(0,R_0)$ collapses to zero in the localization, it looks like (2,0) is a zero divisor.
Aug
19
comment Adjoining an identity to a ring
@karparvar Under this operation (which is not the only candidate, but I admit is probably the most likely candidate) $e^2=e$ for everything in $R_0$. Why would there be any expectation for $e=0$?
Aug
19
comment Omission in Jacobson's BAI regarding extension of isometries.
@dREaM basic algebra I (of two), a classic Jacobson text
Aug
14
comment How do modern algebraists think about diagonal matrices?
@goblin Of course. My comment above about the N-S theorem specified $M_n(\Bbb K)$ only.
Aug
14
comment How do modern algebraists think about diagonal matrices?
@goblin Every ring is an algebra over its center, and every simple ring is a central simple ring over its center. The catch is that you've got to be interested in the maps that are linear over this center. I thought you would be willing to restrict your attention to $\Bbb K$ linear automorphisms of $M_n(\Bbb K)$, but if you aren't you're right, the theorem doesn't directly apply.
Aug
14
comment How do modern algebraists think about diagonal matrices?
@goblin As for the two definitions you proffered being the same on $M_n(\Bbb K)$, yes (the Skolem-Noether theorem)
Aug
14
answered Question on prime ideals of ${\mathbb Z}[x]$
Aug
13
revised Is the standard definition of vector wrong?
added 206 characters in body
Aug
13
answered Is the standard definition of vector wrong?
Aug
12
comment In a ring: If every non-zero and non-unit element factors into prime elements, is factorization over irreducible elements unique?
It's interesting that "everything factors into (finitely many) primes" is more conclusive than "everything factors into (finitely many) irreducibles," since of course, there exist atomic domains which aren't unique factorization domains. +1
Aug
10
answered Algebra and Analysis