Ryker

@MartinSleziak, awesome, thanks for the insight.

@MartinSleziak, the reason why I was asking was that I wanted to know whether you assign homework, and if so, how much time one would spend on it weekly and how much it would account for when calculating the final grade. Do you have any further input on that? I realize it differs from country to country, but how is it at your university?

Is there anyone here that's studying or teaching at a European university?

Or something similar, it might be reversed.

Hey guys, what's that inequality we learn in first year calculus again? I forgot both the name and the inequality, but it's something of the form $(1 + b)^a \leq c$.

@seaturtles, like I said, I'm going from that definition, there are no restraints.

@seaturtles This is the definition I'm using - en.wikipedia.org/wiki/Tensor#As_multilinear_maps.

@seaturtles, I don't know, I'm asking in general for any two tensors. A friend is telling me that's true, but I don't see why this would be the case.

I was wondering whether for two antisymmetric tensors, $T_1$ and $T_2$, $T_1 \bigotimes T_2 = - T_2 \bigotimes T_1$.

Is anyone here familiar with tensor products and antisymmetric tensors?

No, it's the direct sum.

Two sets.

Well, anything.

I'm sure it's true in finite cases, but would infinite dimension mess things up?

It seems to me it is, but I just wanted to check.

Hi guys, quick question. Is the direct sum unique? That is, is it true that $H = A \bigoplus B = A \bigoplus C \implies A = C$?

Thanks for the help.

That's what I was saying.

If $\phi$ is a homomorphism, then if $\phi(1)=12$, but $\phi(1)\phi(1)=0$, a contradiction.

No worries, I have it figured out.

$\phi(1\times1)=\phi(1)\phi(1)$.

I meant $12^2 = 0 \mod 24$.

No, as I said, we're looking at rings.

Yes, but $12^2 \mod 0$.

So we have only two homomorphisms, including the trivial one.

@PedroTamaroff, I believe you mean $16$, not 12.

Am I right in thinking there are three?

Well, ultimately I want to determine all homomorphisms from $\mathbb{Z}/\mathbb{Z}_{30}$ to $\mathbb{Z}/\mathbb{Z}_{24}$.

$\phi(1) = \{0, 1, 9, 16\}$

Exactly.

Never mind, I have it figured out, I think.

Except for 1.

Yes, but you can't send $\phi$ to those.

Yes, I found 4 possibilities.

No, I need to use that.

Never mind, it actually is.

Yet, $\mathbb{Z}_{30}$ is not in the kernel of $\phi$ if $\phi(1) = 16$.

But for that statement to be true, we need $I_2$ to be in the kernel.

The statement from above.

No, I need to do it via the factorization theorem.

It seems to me that $\phi(1) = 16$ is one such homomorphism.

For example, I want to get the homomorphisms from $\mathbb{Z}_{30}$ to $\mathbb{Z}_{24}$.

@PedroTamaroff, no other ones?

But could we have a homomorphism that doesn't arise from one from the parent ring?

Then if we have a homomorphism from the parent ring $R$ to a quotient ring $R/ I_1$, this gives rise to another homomorphism from any $R/I_2$ s.t. $I_2$ is a subset of the kernel of the original homomorphism.

Say we want to find homomorphisms between two quotient rings.

To me, it doesn't seem to.

I have a quick algebra question. Namely, does the factorization theorem for rings give all homomorphisms?

@KarlKronenfeld It's actually on the homework that's due tomorrow, but we've never introduced this notation formally, and it's too late to e-mail the professor.

@PedroTamaroff, thanks.