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Consider the following problem.

I know that $\mathbb{Q}[x]/ (2x-3) $ is isomorphic to $\mathbb{Q}$. Use the evaluation map $x \to \frac{3}{2}$.

But why $\Bbb Z[x]/(2x-3)\cong\Bbb Z[1/2]$? Can I use the same evaluation map here and get $\Bbb Z[x]/(2x-3)\cong\Bbb Z$?

Please help me?

user26857
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    What evaluation map would you use over $\mathbb Z$? – Mathmo123 Oct 20 '21 at 13:57
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    $\Bbb Z[1/2]=\Bbb Z[3/2]$ – lhf Oct 20 '21 at 13:58
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    As long as you specify that your evaluation map goes from $\Bbb{Z}[x]$ to $\Bbb{Q}$ (or some other commutative ring containing $\Bbb{Z}$ and $3/2$) you will be fine. – Jyrki Lahtonen Oct 20 '21 at 13:58
  • @Mathmo123 The map $x \to \frac{3}{2}$. It works for the $\mathbb{Q}$. Would it works for this case? –  Oct 20 '21 at 13:59
  • Mind you, this looks a bit familiar. I think it likely that we have handled this question on the site already. – Jyrki Lahtonen Oct 20 '21 at 13:59
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    You can’t use the same evaluation map because $\frac32\notin \mathbb Z.$ – Thomas Andrews Oct 20 '21 at 14:01
  • Anyway, if $R$ is a commutative ring and $S$ is any ring such that its center contains a copy of $R$, then any $s\in S$ gives rise to an evaluation homomorphism $ev_s:R[x]\to S$, $ev_s(f)=f(s)$. It may be that $S$ is the field of fractions of $R$, like here, an extension field, or even a matrix algebra over $R$. – Jyrki Lahtonen Oct 20 '21 at 14:02
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    Here's an argument to tell you that the quotient and $\Bbb Z$ are not isomorphic: Consider the image $x$ in the quotient $\Bbb Z[x]/(2x + 3)$, let us call it $\bar x$. It has the following property: If you add $3$ to twice of $\bar x$, you get zero. (That is, $2 \bar x + 3 = 0$. Here, by $2$ and $3$, I mean $1 + 1$ and $1 + 1 + 1$.) But no element of $\Bbb Z$ has that property. – Aryaman Maithani Oct 20 '21 at 14:20
  • See here for a glimpse of what happens more generally. – Bill Dubuque Oct 20 '21 at 21:18

2 Answers2

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Well, how do you use the evaluation map in the rational case? You consider the map $$ e_{3/2}: \mathbb{Q}[x] \to \mathbb{Q} $$ given by $e_{3/2}(f) = f(3/2)$. Now, you notice that $e_{3/2}$ is surjective (why?), that $\ker e_{3/2} = (2x - 3)$ and you use the first isomorphism theorem to conclude that $$ \frac{\mathbb{Q}[x]}{(2x - 3)} \simeq \mathbb{Q}. $$

Now, consider the same map with domain $\mathbb{Z}[x]$. What can you say about its image ? Is it well defined as a function with co-domain $\mathbb{Z}$? Can you apply the same argument?

fbianchini
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Your ring is $\mathbb{Z}[x]/(2x -3) = \mathbb{Z}[x-1]/(2(x-1)-1)$. Now in general $A[x]/( a x -1) \simeq A[1/a]$.

orangeskid
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