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Jul
16
comment Calculating a Factorial Base Representation
While it is hard to be more explicit in your statement of the algorithm, a proof of it would be nice (I have proven it myself--I am just suggesting an improvement for your answer)
Jun
27
comment Definition of relatively prime in UFD´s
BTW, it's great that you are comparing the definitions from two sources.
Jun
27
comment Definition of relatively prime in UFD´s
If it was not a Unique Factorization Domain, you would be 100% correct.
Jun
27
comment why is the answer 21,845 and not 218,450?
This is a great example indicating the value in estimating an answer prior to computing it. $5.47/6.26$ is larger than $5/7$ which is $0.7142857...>7/10$. So the true answer is greater than $7*25,000=175,000$.
Jun
26
comment Definition of irreducible element.
@Sodan I believe it is a mistake on the part of Ireland & Rosen. Cf. Corollary 1.3.2 to see an instance where they implicitly assume irreducible implies nonunit.
Jun
25
comment If $\mathbb{Z}$ satisfies an identity $\eta$, then every **commutative** ring satisfies $\eta$? And related questions.
Is it even true in $\mathbb Z$? :)
Jun
21
comment $ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
I see an approach to the second implication involving the fact that every divisor of $x^2+y^2$ is a sum of two squares if $x$ and $y$ are relatively prime. Presumably it is considered too high-powered, but on the other hand it makes the generalization to primes other than $7$ obvious.
Jun
8
answered Is this example right (ideals of $\mathbb{Z}[x]$)?
Jun
4
answered Why cant the proof for Image of Union applicable to Image of Intersection?
May
28
revised Alg Geom - Generators of a Ring
deleted 1 character in body
May
26
comment Prove $G$ is a group under $\ast$
Abelian groups must be commutative, while arbitrary groups may not be commutative. With this in mind, have you verified the defining properties of a group hold for $G$?
May
23
comment For a ring if we have: $\forall a\in R$ with $za=az=z$ does that mean $z=0$
In fact $z+z=0$ even when $R$ is a ring without unity. For if $z\ne0$, then $z^2=z$. So $z+z=z^2+z^2=z(z+z)$ implies either $z+z=0$ or it is nonzero and $z+z=z(z+z)=z$, the latter being a contradiction. In particular, you do not have to split between two cases like you have above.
May
18
revised Schur's first lemma for finitely generated continuous groups of $SU(d)$
Corrected spelling of commute
May
18
comment What does the topology on $\operatorname{Spec}(R)$ tells us about $R$?
I looked up the reference to number 2 (because surely commutative is misspelled) and I only found a section entitled "Zero-dimensional rings" without "commutative". Could you confirm this?
May
18
revised 4 dimensional numbers
Corrected spelling of commutativity
May
18
revised Center of a Group
Corrected spelling of commute
May
18
answered Show that a map is not an automorphism in an infinite field
May
18
revised How can I express such function as known functions or power series?
Corrected a couple misspelled words
May
18
revised Quaternionic veronese Embedding
Corrected spelling of commutativity
May
18
revised Monomials not in an ideal
Fixed some spelling