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revised References on completion and Tor/Ext
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revised How many commutative rings with exactly one non-zero zero divisor are there?
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comment A subgroup of a cyclic group is cyclic - Understanding Proof
You assume $m$ is the smallest positive integer such that $a^m\in H$, but then find that $a^r\in H$, and $r<m$. The only way this is possible is if $r=0$.
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comment Question about unity and composition of morphisms (Category theory)
@Kyson $\circ$ is a function $\operatorname{Hom}(A,B)\times\operatorname{Hom}(B,C)\to\operatorname{Hom}(A,C)$‌​, so by definition, $f\circ g$ is uniquely defined when $f\in\operatorname{Hom}(A,B)$ and $g\in\operatorname{Hom}(B,C)$. This is just part of the data of the category. (This is analagous to a requirement that groups are "closed" under the operation; which is not really a requirement at all once you have defined the operation to be a map $G\times G\to G$.)
Dec
19
revised Difference between Ring and Algebra?
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19
comment Difference between Ring and Algebra?
@RipanSaha Rings in general need not be commutative, but algebras should be over commutative rings. The bilinear map referred to on Wikipedia is the operation $\times$ - the fact that it is bilinear (over $R$) is one of the compatibility axioms between the multiplication in $A$ and the multiplication with elements of $R$.
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revised Question about homotopy equivalence
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comment Prove that $I$ is a maximal ideal
If this question is from a course or a book, then there will be theorems that you are expected to be able to just use without reproving. That $\mathbb{R}$ is a field is definitely one of them, but I don't know what others there might be without more context.
Nov
11
comment Prove that $I$ is a maximal ideal
It depends what theorems you know. There is a theorem that if $K$ is a field, then an ideal $I$ of $K[x]$ is maximal if and only if $I=\langle f\rangle$ for some irreducible $f$. The proof uses that $K[x]$ is a PID for any field $K$. If you can use this theorem, then it is enough to show $\mathbb{R}$ is a field (but you almost certainly already know this). If you can't, then you might want to prove it, at least in this case, which would require proving that $\mathbb{R}[x]$ is a principal ideal domain.
Nov
11
comment Prove that $I$ is a maximal ideal
Based on this, and the other comments, you seem to have misunderstood my point - which was that it is enough to prove that $f$ is irreducible or to prove that $K[x]/I$ is a field (and each will imply the other). You proved that $f$ is irreducible, so you're done. You could instead have proved that $K[x]/I$ is a field, but you don't have to do both. There are almost certainly other reasonable approaches as well.