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Determine the structure of the ring $R_0$ obtained from $\mathbb{Z}$ by adjoining an element $\alpha$ satisfying each set of relations: $$ \begin{cases} 2x-6 &= 0\\ x-10 &=0 \end{cases} $$

I know that the solution is just $\mathbb{Z}/14$. I am struggling to reach that conclusion.

I understand that finding a linear combination and setting this equal to zero, we find that $$2x-6-2(x-10)=14=0,$$ so we much be in $\mathbb{Z}/14$.

What I am confused about now is how $2x-6$ and $x-10$ simply just go away? Any help on this would be great. Thanks

gt6989b
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Jacob Rodgers
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    Do not ask the same question again. – Dietrich Burde May 12 '16 at 21:32
  • I saw the other post and I see this one now: how can an infinite ring to which is been added an element and relations become a finite ring? Besides the doubt that I am missing something, could be we're simply taking a quotient (homomorphic image) of the ring $;\Bbb Z;$ by adding some relations to some of its generators? – DonAntonio May 12 '16 at 21:41

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