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!
Best Answer
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 $\mathbb Z_{14}$.