Let $f\in\mathbb{Z}[X]$ be a monic irreducible polynomial, $\alpha$ a root of $f$ and $k\in \mathbb{Z}$. Show that the map $$\varphi : \mathbb{Z}/f(k)\mathbb{Z} \to \mathbb{Z}[\alpha]/(k - \alpha)\mathbb{Z}[\alpha], \quad z + f(k)\mathbb{Z} \mapsto z + (k - \alpha)\mathbb{Z}[\alpha]$$ is a ring-isomorphism.

I'm looking for a direct proof, using the theory taught in standard classes in abstract algebra.

I was able to show that $\varphi$ is well-defined. The homomorphism property is clear. But now I have problems showing that $\varphi$ is injective, which I want to do by showing that the kernel of $\varphi$ is trivial.

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    $\begingroup$ You should write $\mathbb{Z}[\alpha]/(k-\alpha)\mathbb{Z}[\alpha]$ for the codomain of $\varphi$. $\endgroup$ – Michael Joyce Feb 20 '13 at 17:03
  • $\begingroup$ Maybe I'm missing something, but I don't see why this is supposed to be well defined. First as Joyce points out if you want this to be a map of rings then I presume you meant to factor out the ideal generated by $(k - \alpha)$. But then doesn't $\alpha \mapsto k$ give an iso $\frac{\mathbb Z[\alpha]}{(k - \alpha)\mathbb Z[\alpha]} \simeq \mathbb Z$ and so $\varphi\colon\mathbb Z/(f(k)) \to \mathbb Z$ is given by $\overline{1} \mapsto 1$, which is not a well defined map. $\endgroup$ – Jim Feb 20 '13 at 17:18
  • $\begingroup$ @Jim: You can't send $\alpha \mapsto k$ because $f(k)$ is not necessarily $0$. $\endgroup$ – Michael Joyce Feb 20 '13 at 17:50
  • $\begingroup$ @MichaelJoyce: Ahh, I see, I was thinking of $\alpha$ as a variable. $\endgroup$ – Jim Feb 20 '13 at 18:33
  • $\begingroup$ @MichaelJoyce Thank you. I edited my question accordingly. $\endgroup$ – azimut Feb 20 '13 at 21:56

I know you want a direct proof, but I can't help but point out what I think is the clearest way to think of this isomorphism. Note that $\mathbb{Z}[\alpha] \cong \mathbb{Z}[x] / (f(x))$. Thus $$ \mathbb{Z}[\alpha]/(k-\alpha) \cong \mathbb{Z}[x]/(f(x), k - x) \cong \mathbb{Z} / (f(k)). $$ The first isomorphism comes from reducing $\mathbb{Z}[x]$ by $(f(x))$ first, then quotienting out by $(k - \alpha)$; the second isomorphism comes from reducing $\mathbb{Z}[x]$ by $(k - x)$ first, then quotienting out by $(f(k))$.

As for the direct proof, it is equivalent to show that the (principal) ideal $$I = \{ n \in \mathbb{Z} : n = c \cdot (k - \alpha) \text{ for some } c \in \mathbb{Z}[\alpha] \}$$ is generated by $f(k)$. Suppose $n = c \cdot (k - \alpha)$, where $c = p(\alpha)$ for some $p(x) \in \mathbb{Z}[x]$. Consider the polynomial $g(x) = n - p(x)(k-x)$; since $g(\alpha) = 0$, $f(x)$ divides $g(x)$. So $n = p(x)(k-x) + b(x)f(x)$ for some $b(x) \in \mathbb{Z}[x]$. Evaluating at $x = k$ gives $n = b(k)f(k)$, or $f(k)$ divides $n$, so $I = (f(k))$. If you work through this argument, you'll see it is simply making the isomorphisms mentioned above explicit.

  • $\begingroup$ To make sure that I got everything: The polynomial $b$ is from $\mathbb{Q}[x]$ at first. But applying the fact that the content of polynomials is multiplicative to the equation $g = fb$ ($f$ is primitive) we get $b\in\mathbb{Z}[X]$, right? $\endgroup$ – azimut Feb 20 '13 at 23:46
  • $\begingroup$ Yeah, I glossed over that point, but you use contents to conclude that $b(x) \in \mathbb{Z}[x]$ and so $b(k) \in \mathbb{Z}$. $\endgroup$ – Michael Joyce Feb 21 '13 at 1:56

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