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Apr
20
revised Show that $\tilde{X} \rightarrow X$ is a covering map.
added 50 characters in body
Apr
20
answered Show that $\tilde{X} \rightarrow X$ is a covering map.
Apr
18
comment Can one conjugate any element in $S_3$ to any other element?
yes the problem is to show that there are no automorphisms of $S_3$ other than conjugation. I really did not understand the spoiler 2, in particular the definition of the set of three involutions..
Apr
18
revised To show a set is open
added 18 characters in body
Apr
18
answered To show a set is open
Apr
16
answered Question about degrees of maps from $S^1 \rightarrow S^1$
Apr
16
comment $p(x)=x^4-2x^2-4$ is irreductible over $\mathbb Q.$
yes you don't have to because even if the product of three of these terms give a degree three polynomial in $\mathbb Q[X]$, then still the fourth remaining term is not in $\mathbb Q[X]$.
Apr
16
comment $p(x)=x^4-2x^2-4$ is irreductible over $\mathbb Q.$
This factorisation in $\mathbb R[X]$ is unique a[nd none of these four terms is in $\mathbb Q[X]$ so all we have to do is to check that none of the products of two of these four terms give a quadratic polynomial in $\mathbb Q[X]$ which is the case, hence the polynomial is not irreducible in $\mathbb Q[X]$.
Apr
13
comment Commutativity of two endomorphims
@MattN. If we write $E=\ker g \oplus H$ where $H$ is a complementary subspace of $\ker g$ in $E$. Then $x=x_1+x_2$ hence since $\ker g$ is stabilised by $f$ then $g\circ f(x)=gof(x_2)$ and $f\circ g(x)=f\circ g(x_2)$. Hence it suffice to show that $f$ and $g$ commute on $H$. But i don't see how to go further.
Apr
13
comment Showing an endomorphism is not surjective
@Najib Idrissi Thank you i see what you mean now.
Apr
13
comment Showing an endomorphism is not surjective
@Najib Idrissi I'm not sure I understand. What does $det(f_A(B))=0$ for each $B$ imply?
Apr
13
comment Showing an endomorphism is not surjective
Actually any matrix $$\begin{pmatrix}x&y\\x&y\end{pmatrix}$$ is in the kernel of $f_A$.
Apr
13
asked Showing an endomorphism is not surjective
Apr
11
reviewed Approve suggested edit on Is a cartesian product a group?
Apr
9
revised A basis for a nilpotent endomorphism
edited body
Apr
8
comment A basis for a nilpotent endomorphism
It looks fine !!! $b_2$ exists since $f$ is not the zero endomorphism. $b_1=f(b_2)\in \ker f$. The only problem is $b_3$ why does it exist and is there a canonical way how to choose one that is independent from $b_1$?
Apr
8
comment A basis for a nilpotent endomorphism
Thank you !! Is there any good reference where I can find this proof to apply it to my problem?
Apr
8
comment A basis for a nilpotent endomorphism
Dear DonAntonio, yes this is basically the same problem but put in another context, I couldn't change it in that post since comments are made there that will be irrelevant if I change the question in this way. I really want to solve this problem but without mentionning the notion of Jordan normal form.
Apr
8
asked A basis for a nilpotent endomorphism
Apr
8
reviewed Approve suggested edit on Question regarding ratio