1d
revised Are $e : 1 \rightarrow X$ and $\mathrm{id}_X : X \rightarrow X$ the only operations of $\mathsf{Grp}$ that are homomorphisms?
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1d
comment Are $e : 1 \rightarrow X$ and $\mathrm{id}_X : X \rightarrow X$ the only operations of $\mathsf{Grp}$ that are homomorphisms?
@goblin: Well, you do have some extras, because of $e : 1 \to X$: in a theory without constants, $\hom(1, X) = \varnothing$.
1d
answered Are $e : 1 \rightarrow X$ and $\mathrm{id}_X : X \rightarrow X$ the only operations of $\mathsf{Grp}$ that are homomorphisms?
1d
comment Reducing primes in $\mathbb Z[i]$
@J.A: Yes: it's generated by $\gcd(p, i-b)$. It's the same idea as the fact, in the integers, that every number of the set $\{ am + bn | a,b \in \mathbf{Z}\}$ is a multiple of $\gcd(m,n)$.
1d
comment For every natural number $n$, $\gcd(an,bn)=n\gcd(a,b).$
It might be easier to use the greatest common divisor version of $\gcd$, rather than the least linear combination version.
1d
comment Reducing primes in $\mathbb Z[i]$
@J.A: But I guess what I should have said is that saying "$x$ is a prime element of $R$" effectively means the same thing as saying "the set $\{ xr | r \in R \}$ is a prime ideal of $R$". In a "principal ideal domain" (of which $\mathbf{Z}[i]$ is an example), every ideal is of the form $\{xr | r \in R\}$. But there is a more general definition, and in general rings the notion of prime ideal is better behaved than the notion of prime element. e.g. here my argument quickly shows that $\{ ap + b(i-m) | a,b \in \mathbf{Z}[i]\}$ is a prime ideal, but it takes more effort to show that it's principal
1d
revised Reducing primes in $\mathbb Z[i]$
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2d
comment Reducing primes in $\mathbb Z[i]$
@J.A: Ah! I'm not sure how to talk about ring theory without knowing about ideals! Maybe think in terms of modular arithmetic then; if $p$ were prime, then when doing modular arithmetic modulo $p$, you could invert any nonzero number (e.g. in this case we could use the extended Euclidean algorithm to find the inverse). However, $(m-i)(m+i) = m^2 + 1 \equiv 0 \bmod p$, but $m-i \not\equiv 0 \bmod p$ and $m+i \not\equiv 0 \bmod p$, because the only Gaussian integers divisible by $p$ are of the form $pa + pbi$. Therefore we have a contradiction, and so $p$ is not prime.
2d
answered Reducing primes in $\mathbb Z[i]$
2d
answered Showing local ring isomorphisms
2d
answered An argument with my friend over $\bigcup [a+\frac{1}{n},b]$
2d
comment Ship of Theseus applied to GPL - Can I relicense my program if I replace all of the derivative parts?
I'm not sure I like the choice of the analogy: the original question sounds more like they think they were originally bundling their stuff with Harry Potter, but now want to bundle it with their own fantasy epic novel with magic and villains that takes place primarily in a school in a castle.
2d
comment Lucas' Theorem for $p$-adic integers?
@DAS: Nope: $p^{n-1}$ isn't a multiple of $p^n$ in $\mathbf{Z}_p$, so it can't be zero in $\mathbf{Z}_p / p^n$.
Oct
18
revised Gelfand triple: what happens if we don't identify the pivot Hilbert space with its dual?
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Oct
18
answered Gelfand triple: what happens if we don't identify the pivot Hilbert space with its dual?
Oct
18
revised The smallest girl in the world
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Oct
18
answered The smallest girl in the world
Oct
18
comment Does $1.0000000000\cdots 1$ with an infinite number of $0$ in it exist?
It might be more to the point to note that $1.0\dots 01$ doesn't even have an interpretation as a string comprised of digits and a decimal point; the problem comes even before you can start wondering about numbers!
Oct
17
revised how to define good language for the theory of vector space?
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Oct
17
answered how to define good language for the theory of vector space?
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