Isomorphism between cyclotomic rings

Question

This question is similar to the one posted here.

Let $m=m_1\cdots m_k$ be a positive integer decomposed into coprime prime powers $m_i$. For an integer $n\geq 1$, denote the $n$-th cyclotomic polynomial by $\Phi_n(X)$. Introduce the rings

$$\mathcal A:=\frac{\mathbb Z[X]}{(\Phi_m(X))},\qquad \mathcal B:=\frac{\mathbb Z[Y_1,\dots,Y_k]}{(\Phi_{m_i}(Y_i))_{i=1,\dots,k}}.$$

In this document listed also in the other post, it is claimed that $\mathcal A$ and $\mathcal B$ are isomorphic. In fact, if we denote the image of $X$ in $\mathcal A$ by $x$, and the image of the $Y_i$ in $\mathcal B$ by $y_i$, it is claimed that sending each $y_i\in\mathcal B$ to $x_i:=x^{m/m_i}$ yields the desired isomorphism.

I have difficulties to show this. As always, I think the starting point is to consider a ring isomorphism

$$\psi:\mathbb Z[Y_1,\dots,Y_k]\to\mathcal A,\qquad f\mapsto f(x_1,\dots,x_k).$$

The goal would be to prove that:

1. $\psi$ is surjective (for which it suffices to show that $x$ lies in the image of $\psi$);
2. $\ker\psi=(\Phi_{m_i}(Y_i))_{i=1,\dots,k}$ (likely by proving both inclusions).

One could then conclude using the First Isomorphism Theorem. Unfortunately, for both points, I am fully lost.

Answer 1

In order to prove this result, it is enough to check it for $m=pq$, $p,q$ two coprime numbers, then make an induction. We have $A=Z[X]/ \Phi_m [X], B= Z[Y,Z]/\Phi _p[Y], \Phi_q[Z]$

Choose two Bézout numbers $u,v$ so that $u.p+v.q=1$

Let us check successively three points.

1 There exists an homomorphism $\psi :B\to A$ such that $\psi(y)=x^q, \psi(z)=x^p$. Indeed, if $x$ is a $m$-primitive roots of 1, $x^q$ is a $m$-primitive roots of $1$, so that $\phi _m(X)=0 \implies \phi _p(X^q)=0$, and a similar argument for $q,Z$.

2. There exists an homomorphism $\phi : A \to B$ such that $\phi (X)=y^v.z^u=y'z'$. Indeed if $y$ is a $p$-primitive roots of 1, $y^v=y'$ is also a $p$-primitive roots of 1, and similarly, $z'$ is a $q$ primitive roots of 1.

3. We then have $\psi (\phi (x))=\psi (y^v.z^u)=x^{vq}x^{u q}=x$.

And also $\phi (\psi (y)=\phi (x^q)=y^{vq}z^{uq}=y.(y^{-up}z^{uq})$. However in $B$ $y^p=1,z^q=1$, indeed the polynomials $\phi _p, \Phi _q$ divides $X^p-1, X^q-1$, so that $\phi \circ \psi (y)=y)$, and similarly for $z$.