5
$\begingroup$

I'm trying to understand Hopf Galois Theory, and I decided to try studying some example of a non commutative ring extension. The papers I've studied tell me that, for a strongly $G$-graded algebra $A$ over a fixed commutative ring $K$, the corresponding Hopf algebra $H$ (that will verify that $A$ is a right $H$-comodule algebra) is the group algebra $K[G]$.

So I tried to put the theory into an example where $G$ is the quaternion group $\{\pm1,\pm i,\pm j,\pm k\}$.

My question: It's about how are these $A$ and $H$. $A$ is said to be a strongly $G$-graded algebra over $K$ ($K$ being a field). That means $A=\bigoplus_{g\in G}A_g$, where each $A_g$ is a $K$-subspace of $A$; and that $A_gA_h\subseteq A_{gh}$, for all $g,h\in G$. I guess $A$ can be expressed like this: $$A = A_1\oplus A_i\oplus A_j \oplus A_k = K\oplus Ki\oplus Kj \oplus Kk \ \cong \ Ke_1\oplus Ke_i\oplus Ke_j\oplus Ke_k,$$ where $e_1=(1,0,0,0), e_i=(0,1,0,0), e_j=(0,0,1,0), e_k=(0,0,0,1)$ (I prefer these vector space notation). I think I need just $4$ subalgebras instead of $8$ because $-g=-1_K\cdot g$. On the other hand, I believe $H=K[G]$ can be represented the exact same way. Is this correct?

$\endgroup$

1 Answer 1

5
$\begingroup$

I think I need just $4$ subalgebras instead of $8$ because $-g=-1_K\cdot g$. On the other hand, I believe $H=K[G]$ can be represented the exact same way. Is this correct?

No: you have to be careful not to confuse the "-" in the quaternion group with the "-" in $K$.

Explicitly, $(1_K)(-g)$ and $(-1_K)(g)$ are distinct elements of $K[G]$, and furthermore $(1_K)(g)$ and $(-1_K)(-g)$ are distinct from each other and from the last two.

$K[G]$ is necessarily a free $K$-module of rank $|G|$, one basis element per group element.

The same would be said of $A=\bigoplus_{g\in G}A_g$: the fact that $g$ and $-g$ "look" like scalar multiples of each other is just a red herring. They are distinct, and produce separate coordinates in the sum. This will be of rank $|G|$ too. The "-" in front of $g$ is just a notation from the group $G$, and not from the additive group of the module you're making.

$\endgroup$
8
  • $\begingroup$ Ok, so that applies to $K[G]$. Is my $A$ correct? Thank you so much for answering $\endgroup$ Apr 13, 2022 at 15:16
  • $\begingroup$ @Cafeinicola Sorry, I missed that you wrote $H=K[G]$. I thought you meant $A=K[G]$. I do not know what you mean by $A$ in this case... $\endgroup$
    – rschwieb
    Apr 13, 2022 at 15:17
  • $\begingroup$ $A$ is a strongly $G$-graded algebra over $K$, not the same as the group algebra in general. $\endgroup$ Apr 13, 2022 at 15:20
  • $\begingroup$ I thought you chose $A$ and computed $H$. Without knowing anything about $A$, how do you know $H=K[G]$? $\endgroup$
    – rschwieb
    Apr 13, 2022 at 15:24
  • $\begingroup$ Because it's known that, given a strongly $G$-graded algebra over $K$, it always has right $K[G]$-comodule algebra structure (all the papers I've checked state the same) and I did in fact prove that $A$ has that structure (the wrong $H$ I used made no difference in the properties it had to verify). What happends is that I thought $A/K$ would be a right $H$-Galois extension, but considering $H$ is not as I thought then it's just a right $H$-extension. $\endgroup$ Apr 13, 2022 at 15:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .