1,625 reputation
1728
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location United Kingdom
age 23
visits member for 2 years, 9 months
seen 15 hours ago

Currently studying for an Msc in Mathematics


May
19
revised Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
added 73 characters in body
May
19
comment Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
No I am using the interior of the unit disc in $\mathbb{R}^2$ sorry I should have mentioned that I will edit my question
May
19
asked Showing geodesics in $\mathbb{H}^2$ have unique common perpendicular or common endpoint
May
18
accepted $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$
May
18
accepted Question on Frobenious Reciprocity
May
17
accepted Baumslag–Solitar $B(1,2)$ is not hyperbolic
May
16
comment Baumslag–Solitar $B(1,2)$ is not hyperbolic
@N.Owad No, could you give me reference/ explain to me?
May
16
asked Baumslag–Solitar $B(1,2)$ is not hyperbolic
May
9
awarded  Popular Question
May
8
revised $\mathbb{1}\uparrow_H^{G}$ is the permutation representation on $G/H$
added 23 characters in body
May
8
asked $\mathbb{1}\uparrow_H^{G}$ is the permutation representation on $G/H$
May
7
revised Showing that $g$ and $g^{-1}$ are conjugate iff $\chi(g)$ is real
deleted 2 characters in body
May
7
asked Showing that $g$ and $g^{-1}$ are conjugate iff $\chi(g)$ is real
May
5
asked Question on Frobenious Reciprocity
May
5
comment Proving that the tensor product is generated by $a\otimes b$
@Magdiragdag yes sorry that is what I meant, the stuff generated by the tensors $a\otimes b$. I think there is a problem with this approach though. Say we let $K$ be the module generated by stuff of the form $a\otimes b$. Then take $x,y\in A\otimes B-K$. Then I see no reason as to why we couldn't have $x+y\in K$ and $x+y\neq 0$ which would be a problem for the map.
Apr
30
accepted Proving The Diamond Lemma
Apr
30
comment $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$
Is there a slight subtly here in that we can write a map $f:\mathbb{C}\otimes \mathbb{C}\rightarrow \mathbb{C}$ such $f(a\otimes b)=ab$ and then show that this map is injectvive/surjective. However do we also need to check that this map is well defined, which is a problem as we don't know what $a\otimes b$ is and so we have to define $g:\mathbb{C}\times \mathbb{C}\rightarrow \mathbb{C}$ such that $g(a,b)=a\otimes b$ and then use the universal property to show that the map from before actually exists and is well defined?
Apr
30
accepted Example of a functor on products
Apr
30
accepted Proving that $V(R^*)=V(R)-1$
Apr
30
comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple
@JackSchmidt Are you saying that $\{0,g\}$ is an ideal? I am now confused, don't we have $\{0,g\}g=\{0,e\}$ hence this is not an ideal?