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Apr
17
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 $n$ …)
Apr
17
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?
Apr
17
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.
Apr
17
awarded  Investor
Apr
17
revised Values of the Herbrand quotient
deleted 1 character in body
Apr
17
comment Values of the Herbrand quotient
Why would anyone downvote this question?
Apr
17
revised Values of the Herbrand quotient
fixed latex, spelling of “Herbrand”
Apr
17
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?
Apr
11
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 $∀x ∈ R\colon x = x·1 = x·0 = 0$.
Apr
11
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.
Apr
11
answered about hyperplanes in finite fields
Apr
11
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).
Apr
11
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 $ℤ$).
Apr
11
comment Definition of split of exact sequence
Why would you think that $ℤ \cong nℤ \oplus ℤ/nℤ$?
Apr
11
comment When is $G\cong\operatorname{End}(G)$?
@Gaussler Yeah, but it doesn’t work for rings without one (like $2ℤ$) either.
Apr
11
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).
Apr
9
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 …?
Apr
8
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!
Apr
8
accepted Why are functors exact if they preserve all short exact sequences?
Apr
8
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?