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bio website people.bath.ac.uk/mdp33
location Bath, United Kingdom
age 25
visits member for 2 years, 11 months
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Postgraduate student at the University of Bath, UK, studying geometry and representation theory. Currently thinking about cluster algebras and related objects.


Oct
20
comment Prove that the linear map of the basis $V$ is a spanning set of the image of $f$
You don't need $f$ to be injective (at least on $V$) - it isn't in general. If $f(v)=0$ then $v\in\ker{V}$, so $v$ is in the span of $\{v_1,\dotsc,v_k\}$...
Oct
20
revised Question on Showing the dimensions of a Vector Space
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Oct
20
revised Question on Showing the dimensions of a Vector Space
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Oct
20
comment Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
@DanielRust Ah, of course - not quite sure what I was thinking when I wrote that!
Oct
20
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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Oct
20
comment Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
Small typo - in your first line, you mean $d(y)=x$.
Oct
20
answered Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
Oct
20
revised Showing that the image of a homomorphism $d$, with $d^2=0$, is contained in its kernel
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Oct
20
revised Algebraic Geometry: A question about radical ideal
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Oct
16
answered Can we say anything about the unit of a $k$-algebra $A$ in terms of the unit $1\in k$?
Oct
14
comment Why is a projective subspace itself a projective variety?
Pick a basis $w_1,\dotsc,w_n$ of $W$ and extend it to a basis $w_1,\dotsc,w_{n+1}$ of $K^{n+1}$. Let $\lambda_1,\dotsc,\lambda_{n+1}$ be the corresponding coordinate functions; then in these coordinates the equation of $W$ is $\lambda_{n+1}=0$. You can then change coordinates into something else, like the standard basis of $K^{n+1}$, if you like.
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
I'm not talking about injectivity at the moment - the important statement, which I sketched the proof of in my previous comment, is that for any homomorphism $\phi$ (not necessarily injective), $\phi(e_{G_1})=e_{G_2}$. You need to use this fact a lot in both part a) and b) of the exercise - you already have used it in your first two comments on this answer! For example, if $\phi$ is injective, and $\phi(g)=e_{G_2}$, then $\phi(g)=\phi(e_{G_1})$, so $g=e_{G_1}$ - this proves one direction of part b.
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
You should prove that $\phi(e_{G_1})=e_{G_2}$ - in the courses I took this would always be proved before setting the exercise you are doing, and in fact you have already used this fact. If you haven't seen a proof of it, then you can use that $\phi(e_{G_1})\phi(e_{G_1})=\phi(e_{G_1}^2)=\phi(e_{G_1})$, and now multiply by $\phi(e_{G_1})^{-1}$.
Oct
14
revised Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
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Oct
14
answered Why is a projective subspace itself a projective variety?
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
Sorry, maybe I got confused - to prove the kernel is non-empty (part of proving its a subgroup), it suffices to show $e_{G_1}$ is in it. To prove that the kernel is just $\{e_{G_1}\}$ if and only if $\phi$ is injective requires a different argument, sketched in point 3 of the answer.
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
The proof for the inverse is correct. Homomorphisms don't have to be injective, so that wasn't the right thing to say - the kernel is non-empty because it contains $e_{G_1}$; you even write this down in the proof that $\phi(x^{-1})=e_{G_2}$!
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
Yes, that's much better! You should also check that $\phi(x^{-1})=e_{G_2}$ if $\phi(x)=e_{G_2}$ (and that $\ker{\phi}$ isn't empty, but this should be easier!).
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
@Mainviel I mean that you have written (more or less) "if $x\in\ker{\phi}$ then $x=\phi(g_1)=e_{G_2}$"; this is not correct. Something in the kernel of $\phi\colon G_1\to G_2$ cannot be in the image, as previously discussed, and this would imply that the only element of the kernel is the identity of $G_2$, which is not true. What is true is that if $x\in\ker{\phi}$ then $\phi(x)=e_{G_2}$ (and vice versa; if $\phi(x)=e_{G_2}$ then $x\in\ker{\phi}$).
Oct
14
comment Prove that the kernel of a group homomorphism $\phi$ is a subgroup and that $\phi$ is injective
@Mainviel Don't do that! If $\phi$ isn't injective, which might be the case at that stage, then $\phi^{-1}(g_1)$ is a set with more than one element, which makes things even worse! At the moment you haven't written the correct definition of $x\in\ker{\phi}$; if you fix this, the problem of elements living in the wrong group should go away.