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 Apr11 comment about hyperplanes in finite fields Hyperplanes are vector spaces and if $U ⊂ \mathbf F_q^k$ is a ($k-2$-dimensional) proper subspace and $x \notin U$, then $U ∪ \{x\}$ just won’t be a vector space (think of all the sums $x + u$ with $u ∈ U$ you are missing). Apr11 comment Definition of split of exact sequence @ringwith1 You are thinking of this: If $M$ is an abelian group, then for subgroups $A ⊂ M$ and $B ⊂ M$ it is equivalent that $M$ splits as an internal direct sum $M = A \oplus B$ if and only if for every $m ∈ M$ there are unique $a ∈ A$, $b ∈ B$ such that $m = a + b$. In your case, the latter part of the equivalence may be satisfied in some sense, but $ℤ/nℤ$ $(=B)$ just isn’t a subgroup of $ℤ$ $(=M)$ (and even not so if you regard $ℤ/nℤ$ as the subset $\{0,…,n-1\}$ of $ℤ$, because then the addition on $ℤ/nℤ$ is not inherited from the addition in $ℤ$). Apr11 comment Definition of split of exact sequence Why would you think that $ℤ \cong nℤ \oplus ℤ/nℤ$? Apr11 comment When is $G\cong\operatorname{End}(G)$? @Gaussler Yeah, but it doesn’t work for rings without one (like $2ℤ$) either. Apr11 comment When is $G\cong\operatorname{End}(G)$? @Gaussier For $R = ℤ$, $f = \mathrm{id}_ℤ + \mathrm{id}_ℤ$ is not multiplicative (nor does it preserve $1$), because $f(1·1) = 2 ≠ 4 = f(1)·f(1)$ (that’s right – $\mathrm{Ring}$ is not an additive category). Apr9 comment Find an automorphism Find instead two surjective homomorphisms $ℤ → G$ and use the first isomorphism theorem. Alternatively, directly find two different isomorphisms $ℤ/nℤ → G$ which send $1$ to …? Apr8 comment Why are functors exact if they preserve all short exact sequences? @lenticcatachresis I neither speak nor read Spanish but I might just be able to guess my way through this with a dictionary. With David’s answer, though, this will be unnecessary. Many thanks eitherway! Apr8 accepted Why are functors exact if they preserve all short exact sequences? Apr8 comment Why are functors exact if they preserve all short exact sequences? @Berci Thanks, so generally $\operatorname{ker}~if = \operatorname{ker}~f$ whenever $i$ is a monomorphism and $\operatorname{img}~fp = \operatorname{img}~f$ whenever $p$ is an epimorphism – which can be proven by checking the universal properties, right? Apr8 comment Why are functors exact if they preserve all short exact sequences? @Jim I know that, but that is hardly enlightening and not very satisfying unless you prove/understand Mitchell’s embedding theorem first which I haven’t and which seems to me too much of an effort for this problem. Apr8 comment Why are functors exact if they preserve all short exact sequences? Thanks! This seems to work for the category of modules, say, where there is a notion of injectivity and kernels/cokernels can be defined as submodules rather than some limits/colimits. Does this approach generalise to arbitrary abelian categories? (Sorry for being so lazy not to think about it in detail.) Apr8 accepted The pushout of an open/closed injective map is open/closed Apr8 asked Why are functors exact if they preserve all short exact sequences? Apr6 comment An upper bound of a complex number For real numbers $a$ and $t$, $e^{iat}$ is on the unit circle, hence $|e^{iat}| = 1$? Apr6 answered Relative homotopy and composition of maps Apr6 comment If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$. The Frobenius homomorphism is used in (or rather mentioned to justify) the step $X^p-α^p = (X-α)^p$. Divisors of polynomials don’t change by expanding the field of coefficients – if $g ∈ F[X]$ is monic and $g | f$ in $F[X]$, then trivially $g | f$ in $L[X]$, hence $g = (X-α)^n$ in $L[X]$ for some $n ∈ ℕ_0$. Then you can expand $g$ in $L$ and conclude like I’ve done it. Apr6 comment Why $dz\wedge d\bar{z} = d|z|\wedge d\phi$ for $z \in \mathbb{C}$? But what is the exterior derivative of a complex number? Apr6 comment Why $dz\wedge d\bar{z} = d|z|\wedge d\phi$ for $z \in \mathbb{C}$? What does “$dz$” mean if $z$ is an actual complex number? Apr6 revised the number of inversions in the permutation “reverse” added 6 characters in body Apr6 revised the number of inversions in the permutation “reverse” added 6 characters in body