I stumbled across this isomorphism on the Wikipedia article for finite morphisms: $$k[x,t]/(x^n-t)\simeq k[t]\oplus k[t]\cdot x\oplus\ldots\oplus k[t]\cdot x^{n-1}$$ as $k[t]$-modules. I'm really struggling to understand what this quotient looks like and am wondering how one would construct this isomorphism.
$\begingroup$
$\endgroup$
6
-
4$\begingroup$ This looks more complicated than it is, but by generalizing just a little, we can make things clearer. This result is a special case of the following: let $R$ be a commutative ring, and let $f(X) \in R[X]$ be a polynomial of degree $n \geqslant 1$ whose leading coefficient is a unit of $R$. Then $R[X]/f(X)R[X]$ is a free $R$-module with basis $1, X, X^{2}, \ldots, X^{n-1}$. In our case, $R = k[T]$, and $f(X) = X^{n} - T$. If you want more details, let me know and I can post a full answer. $\endgroup$– Alex WertheimJun 10, 2022 at 6:02
-
$\begingroup$ @AlexWertheim some more details would be great, I think fully understanding this general result would be helpful. $\endgroup$– Thigh High CrocsJun 10, 2022 at 15:22
-
1$\begingroup$ @ThighHighCrocs the proof of the result explained by Alex is precisely the one in my answer. You just change the $p_i$ there for elements of $R$. If you have any questions about it, I would be glad to help. $\endgroup$– GabrielJun 11, 2022 at 11:27
-
$\begingroup$ @Gabriel could you just briefly clarify why we're just putting powers of $x$ into $p(x,t)$ as opposed to full polynomials in $x$? $\endgroup$– Thigh High CrocsJun 11, 2022 at 15:18
-
1$\begingroup$ Absolutely! An element of $k[x,t]$ is something like $3x^2y+xy^3+x^2y^2+1$. We can always factor everything "by powers of $x$". In this case, it would be $1+y^3x+(3y+y^2)x^2$. If you prefer to think abstractly, this is an instance of the isomorphism $k[x,t]\cong k[t][x]$. $\endgroup$– GabrielJun 11, 2022 at 18:46
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
An element of $k[x,t]/(x^n-t)$ is an equivalence class of a polynomial $p(x,t)\in k[x,t]$. We write $$p(x,t)=\sum_i p_i(t)x^i,$$ for some polynomials $p_i(t)\in k[t]$, and observe that $x^n=t$ in the quotient. This means that, in the quotient, we have that $$\begin{align*}p(x,t) &=p_0(t)+\cdots+p_{n-1}(t)x^{n-1}+p_n(t)x^n+p_{n+1}(t)x^{n+1}+\cdots \\ &=p_0(t)+\cdots+p_{n-1}(t)x^{n-1}+p_n(t)t+p_{n+1}(t)tx+\cdots\\ &=(p_0(t)+p_n(t)t+\cdots)+\cdots+(p_{n-1}(t)+p_{2n-1}(t)t+\cdots)x^{n-1}. \end{align*}$$
This makes it clear that the map $$\begin{align*}k[x,t]/(x^n-t)&\to k[t]\oplus k[t]\cdot x\oplus\ldots\oplus k[t]\cdot x^{n-1}\\ [p(x,t)]&\mapsto (p_0(t)+p_n(t)t+\cdots,\dotsc,(p_{n-1}(t)+p_{2n-1}(t)t+\cdots)x^{n-1})\end{align*}$$ is your desired isomorphism. (The map $(q_0(t),\dotsc,q_{n-1}(t)x^{n-1})\mapsto [q_0(t)+\cdots +q_{n-1}(t)x^{n-1}]$ is its inverse.)
