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Referring to this question, I want to explore the structure of $\mathbb{Z}[x]/(2x-6, 6x-15)$. Can I just automatically say that $\mathbb{Z}[x]/(2x-6, 6x-15)\cong \mathbb{Z}[x]/(x-3, 6x-15)$ by just dividng the first polynomial by $2$?

Then it seems we conclude $6x-15=6(x-3)+3$, and can substitute $6x-15$ by just 3, and have $\mathbb{Z}[x]/(x-3, 6x-15)\cong \mathbb{Z}[x]/(x-3, 3)\cong\mathbb{Z}_{3}[x]/(x-3) \cong \mathbb{Z}_{3}$.

I am confused on whether this is legal to do and it seems like not a strict proof.

user15277629
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    You certainly can't say it "automatically". The question, is: is $x-3$ in the ideal $(2x-6,6x-15)$? If yes, then you can replace $2x-6$ by $x-3$. Otherwise you can't. – Captain Lama Feb 11 '22 at 21:20

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Yes, we can cancel $2$. Call the ideal $I.\,$ $(2)+I = (2,2x-6,6x-15)= (2,15)=(1)\,$ so by Euclid's Lemma $\,I\supseteq 2(x\!-\!3)\Rightarrow I\supseteq (x\!-\!3),\,$ so $I = (x\!-\!3,I\bmod x\!-\!3)\!=\! (x\!-\!3,3)=(x,3)$

Simpler $\,(2x-6,6x-15) = (2x-6,3) = (x,3)\,$ using $\,(a,b) = (a,\, b\bmod a)$

Bill Dubuque
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