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I am a graduate student at UCLA.


Dec
11
comment Prove or disprove: if $f$ is one-to-one and $g \circ f=h \circ f$ then $g=h$
Hint: If $f$ is not onto, consider the values of $g$ and $h$ on $x \notin \operatorname{im}(f)$.
Dec
8
awarded  Caucus
Dec
8
comment How to prove this polynomial has an imaginary root?
Note that the term "imaginary number" typically refers to a pure imaginary number, i.e., $ai$ where $a \in \mathbb R$ and $i^2 = -1$. For that interpretation of this problem, it is false; take, for instance, $a_i = 1$ for all $i$, and $n$ to be odd.
Dec
6
comment Prime ideals in $A$ and prime ideals in $S^{-1}A$
The canonical association of ideals forward, for $I$ an ideal of $A$, is $I \mapsto S^{-1} I$, and the canonical association of ideals in the reverse, for $J$ an ideal of $S^{-1} A$, is $J \mapsto f^{-1}(J)$, presuming $f$ is the name of the canonical map of rings that you mention.
Nov
29
comment Every nonabelian group of order divisible by 6 contains a subgroup of order 6
@BillyThorton $A_4$ has no subgroup of order $6$, even though its order $12$ is a multiple of $6$.
Nov
28
comment Prove that $A \subset B$ if and only if $A \setminus B = \emptyset$
The symbol $\subset$ must mean regular inclusion, same as in your linked answer, because otherwise it would be false. So the only difference between your question and your linked question is that yours is an equivalence, so you need to prove the converse as well.
Nov
27
comment Which is the identity element of $S_3 \times S_3$?
@coffeemath That doesn't mean it's worth a downvote. Just not be worth an upvote. I should think a downvote should be reserved for answers that are incorrect, or poorly formatted, or in some other way harmful.
Nov
26
comment Which is the identity element of $S_3 \times S_3$?
uh, why the downvote on this answer... ?
Nov
19
comment Why is indefinite integral called so?
The connection between them is given by the fundamental theorem of calculus.
Nov
18
answered Proving Product rule with Abstract Algebra Methods
Nov
16
comment Definition of orientation preserving linear map: is this welldefined?
It's only well-defined for a linearly independent pair of vectors in the plane.
Nov
15
comment How many $\mathbb R$s must a Mathematician walk down?
Also, it's pretty easy to get "far" from the origin; If your meter currently reads $d$ and you want to get at least distance $N$ away from the origin, you just travel a distance of $N+2d$ in any direction.
Nov
15
comment How many $\mathbb R$s must a Mathematician walk down?
@tomasz In addition to what you've said, he is blind to where the real axis is; the data he has access to only helps him to find, roughly, the origin and concentric circles at integer distances from the origin.
Nov
15
comment principal ideals explanation question
For the particular case of $\sqrt 2$, $\mathbf Z[\sqrt 2] = \{a+b\sqrt 2\mid a, b \in \mathbf Z\}$. For more general numbers $\alpha$ in place of $\sqrt 2$, higher powers of $\alpha$ may be necessary.
Nov
15
comment principal ideals explanation question
It's the set of polynomials in $\sqrt 2$ with coefficients in $\mathbf Z$.
Nov
15
comment principal ideals explanation question
well, yes to $n\mathbb Z$, and $2\mathbb Q$ is indeed a principal ideal in $\mathbb Q$, but... it's equal to all of $\mathbb Q$, because $2$ is a unit in $\mathbb Q$.
Nov
15
comment principal ideals explanation question
an ideal that is generated by a single element.
Nov
12
comment Addition in linear vector spaces
Actually, it's a subtle fact that commutativity of addition follows from the remaining axioms of a vector space ($2(a+b) = 2a+2b= a+a+b+b$ on the one hand and $2(a+b) = a+b+a+b$ on the other hand).
Nov
11
reviewed Approve How to prove that $abd = abcd + abc'd$ for all general occassions
Nov
11
comment How to prove that $abd = abcd + abc'd$ for all general occassions
$abcd+abc'd = abd(c+c') = abd(1) = abd$.