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Apr
3
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 $r=0$.
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
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 $R$.
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
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.
Nov
10
comment Given $x$ and $y$ in $\mathbb{Z}[i]$, find $q$ and $r$ such that $x=qy+r$.
You are correct that there are very few possible $r\in\mathbb{Z}[i]$ with $|r|<2$; there are two more than you have written though.
Nov
10
comment What does it mean for two polynomials to be the same in this fundamental field extension theorem?
I think you meant to say that $\eta$ is a root of $x^2+x-1$...or I messed up my calculation.
Nov
10
comment Basis of image and kernel of a linear transformation
I am guessing (and it really is just a guess) that the "dimension theorem" that isn't supposed to be used is another name for the rank-nullity theorem. But on the other hand, it isn't so hard to show that this set is independent directly.
Nov
10
comment Basis of image and kernel of a linear transformation
Look at what kinds of polynomials appear in the image, in particular at their coefficients. Can you see how, for a polynomial in the image, the coefficient of $x^2$ is determined by the other two coefficients?
Nov
10
comment Prove that $(x^3-2)$ is a maximal ideal of $\Bbb Q[x]$
To get better answers, you should add some context to your question - what have you already tried? Do you know some theorems that might help? For example, would it help you to know that $x^3-2$ is irreducible? Can you prove that it is?
Nov
7
comment Proving that $f$ is a bijection.
@egreg I'm not sure I'm following - the OP has a definition ($g(y)=x\iff f(x)=y$) in the question.
Nov
6
comment Proving that $f$ is a bijection.
It doesn't appear to be the definition the OP is using, given the phrasing of both the problem they are trying to solve and the clause beginning "I basically need to show...".
Nov
5
comment Give a counterexample to show that $(AB)^{-1} \neq A^{-1}B^{-1}$
To make the same point in another way - it is not true that $AB\ne BA$ for all $A,B$, but there do exist some $A,B$ for which $AB\ne BA$ - so find a pair (that are also invertible), and you have an explicit counterexample.
Nov
5
comment Shape of the visible part of the Moon
Pas de problème. ;) I have just discovered that both of these things are "ellipse" in French, which makes your mistake even more forgivable than it already was!
Nov
5
comment Shape of the visible part of the Moon
Since you asked for language corrections - I'm pretty certain that you meant "ellipse" in "other cases". This is the shape (singular of ellipses), whereas ellipsis refers to the three dots immediately preceding the word!
Nov
5
comment re-writing a mathematical expression
@EmmaTebbs I agree this is a slightly difficult question to tag - myself and another user have now retagged it. Physics and matlab are both relevant to the context, and I also added algebra-precalc because it's essentially about manipulating expressions (although the expressions are a little more involved than those that would normally occur under this tag!).