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I'm taking an abstract algebra class and we are introducing quotient rings specifically polynomial quotient rings and I'm trying to work out some example problems, but I cannot figure out a general way to approach problems where we're asked to identify something like $\Bbb Z[x]/I$.

The example I have is that we can identify $\Bbb Z[i]/(i-2)$ to $\Bbb F_5$ first by killing $g=x-2$ and then killing $f=x^2+1$.

So with this approach if I have for example $$\Bbb Z[x]/ (6, 2x-1)$$ then if I first kill $2x-1$ and then $6$ I should be able to arrive to some isomorphic ring to this?

Killing $2x-1$ is apparrently done by defining a map $\Bbb Z[x] \to \Bbb Z[1/2]$ where $x \mapsto 1/2$ and since the kernel of this map is generated by $(2x-1)$ I'll get an isomorphism $$\Bbb Z[x]/(2x-1) \cong \Bbb Z[1/2].$$

Now following this I should kill $6$ in $\Bbb Z[1/2]$ to get some kind of isomorphism from $\Bbb Z[1/2]$ to some other ring and I suppose then I can conclude that $\Bbb Z[x]/ (6, 2x-1)$ is isomorphic to the obtained ring?

How do I go about killing $6$ in $\Bbb Z[1/2]$? I don't think I understand how to construct something like this and what should the target ring even be in this case?

user26857
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Jacob Sanders
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1 Answers1

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Solution 1. As you were told, $\mathbb Z[x]/(6,2x−1)\simeq\mathbb Z_6[x]/(2x-1)$. But $$\mathbb Z_6[x]/(2x-1)\simeq \mathbb Z_2[x]/(2x-1)\times\mathbb Z_3[x]/(2x-1)\simeq\mathbb Z_3.$$

Solution 2. You can use the rings of fractions: $$\mathbb Z_6[x]/(2x-1)\simeq S^{-1}\mathbb Z_6,$$ where $S=\{\bar 1,\bar 2,\bar 4\}$, and then notice that $S^{-1}\mathbb Z_6\simeq\mathbb Z_3$.

Solution 3. As you noticed, $$\mathbb Z[x]/(2x-1)\simeq\mathbb Z[\frac12].$$ Then $$\mathbb Z[x]/(6,2x-1)\simeq\mathbb Z[\frac12]/6\mathbb Z[\frac12].$$ But $6\mathbb Z[\frac12]=3\mathbb Z[\frac12]$, and you arrived at $\mathbb Z[\frac12]/3\mathbb Z[\frac12]$. This is isomorphic to $S^{-1}\mathbb Z_3$, where $S=\{\bar 1,\bar 2\}$. Since $\bar 2$ is invertible in $\mathbb Z_3$, you actually got $\mathbb Z_3$.

user26857
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  • Thanks for the multiple approaches! Solution 3 makes most sense to me at this point. May I ask how do you get the isomorphism $\mathbb Z[x]/(6,2x-1)\simeq\mathbb Z[\frac12]/6\mathbb Z[\frac12]$ from $\mathbb Z[x]/(2x-1)\simeq\mathbb Z[\frac12]$? – Jacob Sanders Mar 28 '22 at 17:25
  • Also btw is it true in general that $\Bbb Z[x]/(n, p(x)) \cong \Bbb Z_n[x]/(p(x))$ for $p(x) \in \Bbb Z[x]$ and $n \in \Bbb Z$? I think that you're using something like this in Solution 1? – Jacob Sanders Mar 28 '22 at 17:28
  • If you have an isomorphism of rings $f:R\to S$, and $I$ is an ideal of $R$, then we get an isomorphism $\bar f:R/I\to S/f(I)$. This is what I used in solution 3. – user26857 Mar 28 '22 at 17:34
  • What you say in the second comment is true. Here one uses the following isomorphism theorem: $(R/I)/(J/I)\simeq R/J$. – user26857 Mar 28 '22 at 17:36
  • Solution 4: Consider the ideal $I = \langle 6, 2x - 1 \rangle$. Then $x(6) - 3(2x-1) = 3$ is also an element of $I$. So also, $x(3) - (2x-1) = x+1$ is in $I$. From here, it should be straightforward to show $I = \langle 3, x+1 \rangle$. Once you have that representation of $I$, it should be easy enough to either mod out by 3 and then by $x+1$, or vice versa. – Daniel Schepler Mar 28 '22 at 21:54
  • @DanielSchepler Why don't post it as an answer? Comments are for referring to an existing answer. – user26857 Mar 28 '22 at 21:57