Hello

room mode changed to Public: anyone may enter and talk

Okay, assuming you find your way here. I might as well get started.

Hey!

Howdy

I'm good. How are you?

Okay, so, i think you're a little bit confused about what a module, or at least a free module is.

A module is like a vector space, just over a ring instead of a field. I think I understand that well enough

A free module is just a module generated by a set of elements with no relations between the elements. Is that right?

Ya.

Right. So let us consider a free module over the ring $k[x_1,x_2,\dots,x_n]$ generated by an element of degree $d$. What does this statement mean?

Sure

So, forget about the grading/degree for a moment

ok

So, I think of a free module over a ring R as, like, a box that holds an element of R.

The box is the generator.

Okay

Hmm, sorry, having trouble expanding this analogy/picture. Let me try again.

Sure

Okay, but let's keep some simplification: forget polynomials in $n$ generators, let's just use one.

Okay

Okay, so polynomials have a degree and that gives a grading on the ring S.

You know all this, just establishing common ground.

Now, a module over a graded ring is not necessarily graded, itself.

For example?

Right, like, you could define a module with one generator $m$ and say $x\cdot m=0$.

Okay

So, having a grading on a module is special extra info. And the only requirements are the natural ones that make it coherent, so that $x^d$ times multiplied by a degree $a$ element of $M$ s degre $d+a$

So, you could say "$M$ is the free module generated by one element, $v$, and that element has degree $a$"

So then $v$ has degree $a$, $x\cdot a$ has degree $a+1$, etc.

oops!

I should have written $x\cdot v$ has degree $a+1$

Callus, I have to go to sleep now. It is quite late in the US. Would you be free any time tomorrow to discuss this?

It doesn't make sense to apply $x$ to $a$

Oh you're good. I figured that out

Maybe. I live in Singapore. At work, actually.

Oh! I've lived in Singapore for some time too

Loved my time there.

Try to think of $S(-a)$ not as a copy of $S$, but as a module with a generator $v$ in degree $-a$. Then, to get a degree $d$ element, you have to multiply $v$ by an element in $S$ of degree $d+a$.

Then, since there are no relations, the degree $d$ part of $S(-a)$ is just a copy of $S_{d+a}$

Oh okay. Sure I'll keep that in mind. Thanks

alright, best of luck.

Cool. Thanks. Have a good day