7,961 reputation
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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 24
visits member for 2 years, 4 months
seen 18 mins ago

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


Jan
31
comment Maximal Ideals in the Integers
This is not true for arbitrary rings - you need your ring to be a principal ideal domain. (What you have stated is essentially the defining property of a PID.)
Jan
31
comment What are the elements of $k[X,Y]/(X^2-Y^3)$ like?
But you already have infinitely many representations of every element of $k[X]/(x^2+2x+3)$; you can take one and add on any multiple of $x^2+2x+3$. OK, in this case you can pick the one of smallest degree, but there's nothing particularly canonical about that. In the two variable case it's just harder to decide which is the best representation.
Jan
31
answered The elements of topological space can be not open?
Jan
30
answered Maximal Ideals in the Integers
Jan
30
comment Maximal Ideals in the Integers
One direction, that $(a,b)\subseteq(\gcd(a,b))$, is fairly clear. The other is a consequence of Bézout's identity, which follows from Euclidean algorithm. Wikipedia sketches a proof: en.wikipedia.org/wiki/…
Jan
30
comment Maximal Ideals in the Integers
So, in particular, if you do stipulate that $p_2$ is prime (and different from $p$), then $(p,p_2)=\mathbb{Z}$ is not proper, therefore not maximal.
Jan
30
comment Understanding Isomorphism on Binary Operation
Yes, that's the right definition here. I'm being a bit careful because you've just said "binary structure" rather than group, and that definition is for isomorphism of groups. In your example, the binary structures are groups, but maybe you have other examples where they are not.
Jan
30
comment Understanding Isomorphism on Binary Operation
Do you understand what a homomorphism is? You will need to check that $\phi$ is a homomorphism and that it has an inverse which is also a homomorphism. (In this case checking that $\phi$ is bijective will be enough to give you the second part.)
Jan
30
revised Understanding Isomorphism on Binary Operation
added 24 characters in body
Jan
28
comment “Easy” (maybe not) question about dual spaces (Lineal Algebra).
I would argue it is better to say "is" (if you can) than "is isomorphic to". This is quite a subtle issue; when dealing with vector spaces, you don't often see genuine equality! Certainly $V$ is not equal to $(V^*)^*$, but the two are more than just isomorphic, because there is a unique "best" way of identifying them. A lot of people (maybe unfortunately) will say that one vector space "is" another one (without the word equal) when they are naturally isomorphic, but not when they are only unnaturally isomorphic.
Jan
28
comment “Easy” (maybe not) question about dual spaces (Lineal Algebra).
A little bit about the last part; a finite dimensional vector space $V$ is isomorphic to the dual of any vector space $W$ with the same dimension as $V$, but this should not be thought of as an identification because there are lots of choices of isomorphism. However, $V$ is isomorphic to $(V^*)^*$ in a natural way, that involves no choices, and so it is more reasonable to treat this as an identification and say that $V$ is a dual space (it's the dual of its dual).
Jan
28
comment Understanding of a formula with matrix summation
$T$ has to be the length of the vector $e$, and the number of vectors $x_i$. As tabstop says, it is not a matrix. If your problem is with dividing a matrix by $T$; you do this by dividing each entry by $T$.
Jan
28
revised spectrum of a ring
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Jan
28
comment spectrum of a ring
Do you mean the part i) $\implies$ iii)?
Jan
28
reviewed Leave Closed Let V be a complex vector space. If $\langle T(v),v\rangle\in \mathbb R$ for every $v \in V$, then is T self adjoint?
Jan
27
comment Solve the following equation with combinatorics…
@user123499 In relation to the comment you asked for an explanation for - if $r>n$, there are $0$ ways of choosing $r$ things from a set of $n$. If you're question is related to a polynomial expansion, it's unlikely you care about this, and instead you want $n$ such that the two numbers are equal and non-zero. It might be worth editing your question to include the exercise you are trying to solve.
Jan
27
comment Solve the following equation with combinatorics…
@SpamIAm So do I - using the C-notation, the original post (before my edit) said that the 3 and the 12 should be at the top - it's possible this means one should read $\mathrm{C}^3_n$ as $\binom{3}{n}$; I don't remember. Either way, my edit to the actual question reflects what the OP described.
Jan
27
revised Solve the following equation with combinatorics…
added 13 characters in body
Jan
27
comment Solve the following equation with combinatorics…
Your question is a little unclear; do you want to find $n$ such that $\binom{3}{n}=\binom{12}{n}$? (Or perhaps you prefer the notation $\mathrm{C}^3_n=\mathrm{C}^{12}_n$.)
Jan
27
answered Can prime quadruplets be adjacent