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awarded  Nice Answer
May
23
comment Writing the ideal $m=\langle X, Y \rangle$ in $R=k[X, Y]$ as a countable union of prime ideals
@MartinBrandenburg: Wait, the "linear algebra fact" I quoted above is false! See Pete L Clark's answer here. So how do we show that if $k$ is uncountable, then $\mathfrak{m}$ cannot be written as a countable union of strictly smaller prime ideals?
May
23
comment Writing the ideal $m=\langle X, Y \rangle$ in $R=k[X, Y]$ as a countable union of prime ideals
In that case, $R$ is countable and so $\mathfrak{m}$ is countable. And I can just proceed in the same way as "slick solution" in the beginning of the post. Thanks so much Martin! Feel free to post to answer box below, so I can accept your answer.
May
23
comment Writing the ideal $m=\langle X, Y \rangle$ in $R=k[X, Y]$ as a countable union of prime ideals
Yeah so if $k$ is uncountable, I guess the answer will be no. It just occurred to me that it might follow from linear algebra fact: a $k$-vector space $V$ cannot be written as a union of $n$ proper subspaces if $n$ is strictly less than the cardinality of $k$. So that should deal with the case when $k$ is uncountable.
May
23
asked Writing the ideal $m=\langle X, Y \rangle$ in $R=k[X, Y]$ as a countable union of prime ideals
Mar
27
awarded  Popular Question
Mar
13
comment Prove the fractional field $Q(\mathbb{R})$ of the integral domain $\Bbb R$ is $\mathbb{R}$.
The fraction field of an integral domain $R$ is the smallest field containing $R$ (in an appropriate sense). But if $R$ is already a field . . .
Mar
9
answered Group Theory Lemma Proof
Mar
9
comment Does a Banach space always contain an element of arbitrarily large norm?
@N.S.: Touché :)
Mar
9
answered Does a Banach space always contain an element of arbitrarily large norm?
Mar
8
comment If $\phi(g)=g^3$ is a homomorphism and $3 \nmid |G|$, $G$ is abelian.
I am pretty sure this problem also appears in Herstein's classic "Topics in Algebra".
Feb
23
comment Is every real $\in(0,1)$ between the reciprocals of two consecutive integers?
@user7530: Good point. Since $\alpha<1$ and $m\in S$, we know that $m>m\alpha\geq 1$, so $m>1$.
Feb
23
revised Is every real $\in(0,1)$ between the reciprocals of two consecutive integers?
added 184 characters in body
Feb
23
answered Is every real $\in(0,1)$ between the reciprocals of two consecutive integers?
Feb
12
comment Does interior of closure of open set equal the set?
Of course, the punctured disk. Thanks a lot, Brian. You are too awesome :)
Feb
12
comment Does interior of closure of open set equal the set?
Does there exist such counterexample with $A$ connected? It seems no, to me.
Feb
12
comment Convergence in measure implies convergence almost everywhere (on a countable set!)
Explained here (just adding this so Davide sees it, in case he hasn't).
Feb
12
comment Convergence in measure implies convergence almost everywhere (on a countable set!)
@Aubrey: No problem! Thanks for linking the new question.
Jan
30
comment How does $n(n-1)(n-2)\cdots(n-m+1) \cdot \frac{(n-m)(n-m-1)\cdots1}{(n-m)(n-m-1)\cdots1} = \frac{n(n-1)(n-2)\cdots1}{(n-m)(n-m-1)\cdots1} $
@user1766555: Right click on the formula --> "Show Math as" --> "TeX Commands".
Jan
27
awarded  Nice Question