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First, I am sorry to ask this question because I cannot find the detail of the proof of the theorem.

Consider two quotient rings $ \mathbb Z [x]/(x^2+ax+b) $ and $ \mathbb Z [x]/(x^2+cx+d) $ We know that two rings are isomorphic if they have the same discriminant. What I am not sure about is how to construct this isomorphism. If these two polynomials do not have roots in $\mathbb Z$, I believe the isomorphism will be the one by sending one root of one polynomial because we can write these two rings as $\mathbb Z[x]$ adjoining by root.

I would appreciate it if someone can give me a reference of the proof of the theorem and tell me whether the isomorphism I am thinking about is correct.

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  • $\begingroup$ It's not correct. Sending a root of $x^2-2$ to a root of $x^2-8$ is not going to be a ring isomorphism. $\endgroup$ – Gerry Myerson Oct 2 '15 at 9:40
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Try looking for a map $\Bbb Z[x]/(x^2+ax+b) \to \Bbb Z[y]/(y^2+cy+d)$ determined by something like $f(x) = (y+e)$ for some integer $e$.
$f(x)^2+af(x)+b = y^2+(2e+a)y + (e^2+ae+b)$, so we will need $2e = c-a$.

Since $a^2-4b = c^2-4d$ we have $4 \mid a^2-c^2$ and so $a \equiv c \pmod 2$, so we can pick an integer $e$ such that $2e=c-a$.

Then $e^2+ae+b = \frac 14 (c^2-2ac+a^2+2ac-2a^2+4b) = \frac 14 ((c^2-4d)+(-a^2+4b))+d = \frac 14 (\Delta - \Delta) +d = d$, so $f(x)^2+af(x)+b = 0$ and $f$ is a well-defined map between the quotient spaces.

$f$ has an obvious inverse so it is an isomorphism.

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define $\phi:Z[x]\to Z[x]/(x^2+cx+d)$ by $\phi(a)=c/(x^2+cx+d),\phi(b)=d/(x^2+cx+d)$ and canonical projection otherwise.

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  • $\begingroup$ This can't work since $\phi$ must be $\mathbb Z$-linear so $\phi(a)=a$ and $\phi(b)=b$ necessarily. $\endgroup$ – Ferra Oct 2 '15 at 9:46
  • $\begingroup$ Can you explain more about your answer? I do not quite get the why you send $a$ to $c$ and $b$ to $d$. Can we know that the kernel of the map is $x^2+ax+b$? If so (this should be true), I think we haven't use the condtion that two polynomials share the same discriminant. Can you say somthing more? Thanks a lot. $\endgroup$ – user194201 Oct 2 '15 at 9:46
  • $\begingroup$ @ferra why $\phi$ must be $\mathbb{Z}$ linear as it's a map from $\mathbb{Z}[X]$ to a quotient ring of it? $\endgroup$ – user300 Oct 2 '15 at 9:51
  • $\begingroup$ @RSBST i thought of having the kernel as ideal $(x^2+ax+b)$ may be i'm wrong. $\endgroup$ – user300 Oct 2 '15 at 9:53
  • $\begingroup$ every homomorphism of rings is $\mathbb Z$-linear! Therefore you cannot choose the image of the elements of $\mathbb Z$, if you want a homomorphism of course. $\endgroup$ – Ferra Oct 2 '15 at 9:55

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