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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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Jan
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comment Question about unity and composition of morphisms (Category theory)
@Kyson It's not a good idea to post new questions as comments on answers they don't relate to, and I don't know the proof that you are referring to. Just wait until you can post questions again and then ask it officially!
Jan
27
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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Dec
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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Nov
11
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.
Nov
10
answered Prove that $I$ is a maximal ideal
Nov
10
comment What does it mean for two polynomials to be the same in this fundamental field extension theorem?
It would be interesting to know what it is, and to check that the image of $\alpha$ does indeed have the same minimal polynomial as $\alpha$... This is an important observation - the theorem you are quoting gives necessary and sufficient conditions for a particular map $F(\alpha)\to F(\beta)$ to be an isomorphism, but the two fields may still be isomorphic even if this map is not an isomorphism.
Nov
10
comment What does it mean for two polynomials to be the same in this fundamental field extension theorem?
Does he say what the isomorphism is between $\mathbb{Q}(\alpha)$ and $\mathbb{Q}(\beta)$? Because if it doesn't send $\alpha$ to $\beta$ then the theorem doesn't apply.
Nov
10
comment Given $x$ and $y$ in $\mathbb{Z}[i]$, find $q$ and $r$ such that $x=qy+r$.
@Kieran That was an aside though - mixedmath's comment is correct, you are being asked to do division with remainder.