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May
7
comment Universal property of natural number semi-ring
@user134070 "universal properties" are a basic concept of category theory, so that's why I thought you are familiar with at least the basics.
May
7
revised Prove there generally is no isomorphism between $R[x]/(x^2-a)$ and $R^2$
added 22 characters in body
May
7
revised Prove there generally is no isomorphism between $R[x]/(x^2-a)$ and $R^2$
added 2 characters in body
May
7
comment Prove there generally is no isomorphism between $R[x]/(x^2-a)$ and $R^2$
The question seems clear and error-free to me, although I could be mistaken...
May
7
answered Prove there generally is no isomorphism between $R[x]/(x^2-a)$ and $R^2$
May
7
comment Universal property of natural number semi-ring
@robarthan I can't deny that, but my take on the situation is that in the context of category theory and universal properties, one would most likely stick to the categories I describe. It's a judgement call, of course, unless the OP decisively says differently.
May
6
answered Universal property of natural number semi-ring
May
6
comment For fractional ideal $I$ why is $I\cap R \supsetneq \{0\}$?
Oh, I see from your other post that you're using a definition that precludes the zero ideal. Never mind.
May
6
comment Question about working in modulo?
@user121615 do you understand that the order of an element in a finite group divides the size of the group?
May
6
comment Question about working in modulo?
@user121515 the order of an element of a group divides the size of the group, that's why.
May
6
answered Hints on how to approach a problem concerning rings/field in Abstract Algebra
May
6
answered Question about working in modulo?
May
5
revised Definition of angle between non-differentiable curves
spelling, tag
May
5
awarded  Revival
May
5
comment In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all element non zero in $A$ is invertible.
Why did you accept an answer that doesn't even address the hard half of the question?
May
5
revised In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all element non zero in $A$ is invertible.
added 82 characters in body
May
5
answered In a ring $(A,+, \cdot)$ if $aba = a$ then $bab = b$ and all element non zero in $A$ is invertible.
May
4
revised Is there any geometrical interpretation as to why matrix product is not commutative?
added 262 characters in body
May
4
answered Is there any geometrical interpretation as to why matrix product is not commutative?
May
4
comment Does $\text{End}_K(K^n) \cong \text{Mat}(n\times n, K)$?
@PavelČoupek Please put solutions in the solutions section. Consider transferring it out of the comments. Thanks!