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I looked this up on here, but I couldn't find anything that explained it clearly enough for me. I'm doing problems in Artin, in particular 11.5.4, which asks:

Determine the structure of $R'$ obtained from adjoining $\alpha$ to $\mathbb{Z}$ satisfying the following relations: a) $2\alpha = 6, 6\alpha = 15$ b) $2\alpha = 6, 6\alpha = 10$ c) $\alpha^3+\alpha^2+1 = 0, \alpha^2 +\alpha = 0$

I'm working on a), and my professor said that the answer is that the ring is isomorphic to $\mathbb{Z}_3/(x-3)$. I still don't have much intuition for this, so it's hard for me to see why this is the answer.

I divided $6\alpha - 15$ by $2\alpha - 6$ and got a remainder of $3$, and $2\alpha-6$ is equivalent to $\alpha-3$, so we have $(x-3,3)$. Can we just say this is $\mathbb{Z}_3/(x-3)$ right away? Why?

In short, my question is: what is the process for questions like this?

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When you adjoin an element $\alpha$ to a ring $R$ subject to the relations $f_1(\alpha)=\cdots=f_n(\alpha)=0$, this means you are forming the ring $$\frac{R[x]}{(f_1(x),\ldots,f_n(x))}$$ so in a) you are forming $$\frac{\mathbb Z[x]}{(2x-6,6x-15)}\cong \frac{\mathbb Z[x]}{(x,3)}\cong \frac{\mathbb Z}{(3)}=\mathbb Z_3$$

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  • $\begingroup$ So, since $3=0$ is a relation, we have $x-3 = x$? $\endgroup$
    – Lost
    Dec 11 '13 at 3:46
  • $\begingroup$ @Lost Yes, I'm exploiting that. $\endgroup$ Dec 11 '13 at 3:47
  • $\begingroup$ Ah, thank you. So, basically, for this type of question, we use the division-remainder algorithm until everything is in reduced form, and then finding suitable n such that the structure fits that of $\mathbb{Z}_n$ (when working with $\mathbb{Z}$ of course)? $\endgroup$
    – Lost
    Dec 11 '13 at 3:50
  • $\begingroup$ @Lost That's not how I'd go about it exactly (if I understand what you're saying). I'd use a division-remainder algorithm to write the ideal in a nice form which I can take the quotient by easily. I'll edit my answer to make that a little more clear. $\endgroup$ Dec 11 '13 at 3:57
  • $\begingroup$ I may not be saying it very well, but that was what I was trying to get at - reducing the ideal adequately using the division-remainder theorem. Thanks for your help. $\endgroup$
    – Lost
    Dec 11 '13 at 4:24

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