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 Apr17 comment Proving every element of $\mathbb{F}[x]/(p(x))$ can be expressed uniquely @DH. Well, maybe you can say at least something about the degree of $a - a_1$, even if you can’t know the degree of $a + b$ for polynomials $a$ and $b$ for sure if you only know their degrees. (You can’t know it because, after all, if $a = X^n + X^{n-1} + … + 1$ and $b = - (X^n + X^{n-1} + … + X^{k+1})$ for some $k ∈ \{0,…,n\}$, then $a + b = X^k + X^{k-1} + … + 1$ has degree $k$ which could be anything. Well, of course anything between $0$ and $n$ …) Apr17 comment Proving every element of $\mathbb{F}[x]/(p(x))$ can be expressed uniquely What’s the degree of $a(x) - a_1(x)$, what’s the degree of $q(x)·p(x)$? Put differently: Can you bound the former from above, the latter from below? Apr17 comment Let $X$ be a normal space then there exists a continuous map $f : X → [0, 1]$ such that $f^{−1} (0) = A$ You probably need to say a thing or two about why $f$ still is continuous. Apr17 awarded Investor Apr17 revised Values of the Herbrand quotient deleted 1 character in body Apr17 comment Values of the Herbrand quotient Why would anyone downvote this question? Apr17 revised Values of the Herbrand quotient fixed latex, spelling of “Herbrand” Apr17 comment Let $X$ be a normal space then there exists a continuous map $f : X → [0, 1]$ such that $f^{−1} (0) = A$ @5xum Is it? The property doesn’t say that the closed sets in $X$ are exactly countable intersections of open sets, so there might be countable intersections of open sets (such as trivial ones) that aren’t closed, right? Apr11 comment I have a question about the multiplicative inverse in any field. It should. If you tell us the field axioms you are using, I’m sure we can point out which one assures $0 ≠ 1$. Anyway, the only ring $R$ with one in which $0 = 1$ is $R = 0$, since $∀x ∈ R\colon x = x·1 = x·0 = 0$. Apr11 comment I have a question about the multiplicative inverse in any field. If $α$ was a multiplicative inverse of $0$, then $α·0 = 1$, but also $α·0 = 0$ (which is true for all elements $a$ of the field since $a·0 = a·(0+0) = a·0 + a·0$ from which you can additively cancel $a·0$ (by adding $-a·0$ to both sides)). Hence $0 = 1$ which is not allowed by the very definition of a field. Apr11 answered about hyperplanes in finite fields 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?