43,059 reputation
83599
bio website
location USA
age
visits member for 2 years, 3 months
seen 3 hours ago

PhD

Interests:

Ring and module theory

Clifford algebra/Geometric algebra

Mathematical physics

Applications of abstract algebra


Jul
9
revised Relation between von Neumann regular rings, Krull dimension 0, and rings with no nonzero nilpotents.
edited tags
Jul
9
comment Relation between von Neumann regular rings, Krull dimension 0, and rings with no nonzero nilpotents.
Here is another duplicate, if the other one doesn't suit you: math.stackexchange.com/q/238185/29335
Jul
9
comment Copy of C in H , trace is independent of the choice
What are "the choices" in your context? Or maybe this means "all choices"?
Jul
9
comment Jacobson radical in $A[x]$ where $A$ is a ring.
@Thisismuchhealthier. As I reread the rules for superclosure, I'm guessing this might have been caused by the "original tag" clause. I see this one hasn't had its tags changed, so I will be more confident in success :)
Jul
9
comment Jacobson radical in $A[x]$ where $A$ is a ring.
@Thisismuchhealthier. I know. I was just hedging in case I couldn't guarantee success, for some reason. I've observed in some cases posts with votes to close already and I had a gold badge in the main tag, and I was unable to close the question with superpower. I was worried the same might happen for reopening.
Jul
9
comment Jacobson radical in $A[x]$ where $A$ is a ring.
Agh, stupid superpower. Wanted to mark it as "possible duplicate" and let the votes decide. so.very.tired: if the dupe really isn't a dupe, just ping me and I'll do my best to reverse the closure. In the commutative case (the only one I know it works for sure) this is called Snapper's Theorem.
Jul
9
comment Jacobson radical in $A[x]$ where $A$ is a ring.
Commutative ring?
Jul
9
comment Help with a problem from Christian Peskine's book about Artinian rings
I briefly had hopes for an elementary proof: If there is a nonmaximal prime $P$, you have a nonunit, nonnilpotent $a\in R\setminus P$. The $b$ that you are given must then be in $P$, and it has its own nonmaximal prime $Q$ that it isn't in, but $a\in Q$. I didn't see the way to a contradiction, though :) This is an accessible proof though! +1
Jul
8
revised Mathematical background for TQFT
edited tags
Jul
8
comment Show that $A = \{(3x,y)~|~ x,y \in Z\}$ is a maximal ideal of $Z \oplus Z$. My Attempt Shown
Dear @VHP : Sure, I've added a section about it. It looks like maybe you've confused the coset and cross product concepts. The set of cosets $\{b+Z\mid b\in Z\}$ is completely different from $Z\oplus Z$. The first one denotes elements of a partition of $Z$ and the set for the second one consists of pairs of elements of $Z$.
Jul
8
revised Show that $A = \{(3x,y)~|~ x,y \in Z\}$ is a maximal ideal of $Z \oplus Z$. My Attempt Shown
added 287 characters in body
Jul
8
revised Show that $A = \{(3x,y)~|~ x,y \in Z\}$ is a maximal ideal of $Z \oplus Z$. My Attempt Shown
deleted 8 characters in body
Jul
8
answered Using the definition of adjoint to show that a linear transformation is self adjoint/normal
Jul
8
answered Show that $A = \{(3x,y)~|~ x,y \in Z\}$ is a maximal ideal of $Z \oplus Z$. My Attempt Shown
Jul
8
revised idempotents acting as local identities
useful title, minor text fixing
Jul
8
revised Isomorphism of quaternions with a matrix ring over real numbers
better title, English spelling of *morphism*
Jul
8
comment Getting coordinate vector in linear algebra
If this question is what it looks like (potentially it is homework for credit) then the first person you need to ask for hints is your teacher.
Jul
8
comment Showing that $\bigoplus_{i\in\mathbb N}\mathbb Z/2\mathbb Z$ is not a direct summand of $\prod_{i\in\mathbb N}\mathbb Z/2\mathbb Z$
@Nishant In other words, $\oplus k$ is an essential submodule of $\prod k$. The only essential direct summand of the regular module is the entire module.
Jul
8
comment Projective modules over $kG$ equivalent to injective.
We have to additionally observe that $kG$is Artinian to draw your first conclusion. What you wrote proves that free modules of finite rank are injective, but knowing the ring is Noetherian allows us to also say infinite direct sums of the regular module are injective too.
Jul
7
answered If an integral domain $R$ has a factorization basis, is it a UFD?