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This is an exercise in Artin,

Describe the ring obtained from $\mathbb{Z}$ by adjoining an element $\alpha$ satisfying the two relations $2\alpha - 6 = 0$ and $\alpha - 10 = 0$.

I have seen examples in the text where one relation is imposed on a ring, but not two. I would very much like to see how this is done. Thanks!

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The idea it's simple indeed, what you're doing is considering the quotient ring obtained by $\mathbb Z[x]$ and its ideal generated by the polynomials $2x - 6$ and $x - 10$ (i.e. $(2x -6 , x - 10)$).

Through some calculations it's easy to see that $$(2x - 6,x - 10) = (x - 10, 14) = (x + 4, 14)$$ an so you can also easily prove that the ring you obtain is $$\mathbb Z_{14}[x]/(x + [4]_{14})$$ and as YACP pointed out this quotient is isomorphic to $\mathbb Z_{14}$.

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  • $\begingroup$ Thanks! Could you specify what you mean by "some calculations". How do you go from $(2x-6,x-10)$ to $(x+4,14)$? $\endgroup$
    – Carolus
    Nov 21, 2012 at 8:56
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    $\begingroup$ Simple: $2x - 6 = 2(x - 10) + 14$ and clearly $14 = (2x - 6) - 2(x - 10)$, so we have that $2x - 6, x - 10 \in (x - 10, 14)$ and, on the other end $14, x - 10 \in (x - 10, 2x - 6)$ so these two ideals are one contained in the other one, and so they're equal. $\endgroup$
    – Giorgio Mossa
    Nov 21, 2012 at 9:22
  • $\begingroup$ Great, thanks again! $\endgroup$
    – Carolus
    Nov 21, 2012 at 9:28
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    $\begingroup$ @ineff Actually you didn't finish the job: $\mathbb{Z}[X]/(X+4,14)\cong\mathbb{Z}_{14}$. $\endgroup$
    – user26857
    Nov 21, 2012 at 10:15
  • $\begingroup$ Right @YACP. Now I complete :) thanks. $\endgroup$
    – Giorgio Mossa
    Nov 22, 2012 at 16:44

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