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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 10 months
seen Dec 11 at 18:13

Postgraduate student at the University of Bath, UK, studying geometry and representation theory. Currently thinking about cluster algebras and related objects.


Dec
9
awarded  Caucus
Nov
13
revised Question about homotopy equivalence
added 1 character in body
Nov
13
revised Why is this kernel isomorphic to $\Omega^{n-2}(E-2D)$?
deleted 4 characters in body; edited title
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.
Nov
10
revised Given $x$ and $y$ in $\mathbb{Z}[i]$, find $q$ and $r$ such that $x=qy+r$.
added 70 characters in body; edited title
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 Two questions about triangle that blocked at rectangle…
By "a triangle blocked by a rectangle", I assume you mean the rectangle two of whose edges are the short edges of the triangle, and whose diagonal is the hypotenuse of the triangle? If yes, then the answer to the 1st question is yes, because the rectangle is built from two copies of the triangle, and the answer to the second question is no, because the center of the rectangle is on the hypotenuse. I'm not sure if there is a nice way to describe the centroid of the triangle in terms of the rectangle.
Nov
10
comment Two questions about triangle that blocked at rectangle…
Thanks. (To prove it wasn't a stupid question, here is a table including 16 other definitions of the center of a triangle: en.wikipedia.org/wiki/…).
Nov
10
comment Two questions about triangle that blocked at rectangle…
What do you mean by the "center" of the triangle here? There are many different definitions.
Nov
10
comment Smallest Subring of R that contains $S \cup \{a\}$
You seem to be thinking along the right lines. Can you show that any subring $S'$ containing $S\cup\{a\}$ in fact contains the set in question? This is enough to show that your set is contained in the intersection of all subrings containing $S\cup\{a\}$, and the other direction is easy since you have already shown that your set is such a subring.
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?