Here's an answer which I found yesterday while trying to identify the ring $\Bbb Z[x]/(x^2-3,2x+4)$. What interest me the most is that the calculation that one does in simplifying the given Ideal. Can anyone provide me the reason why we can do such calculation to make the Ideal simpler. The link is given below as well and I'm pasting the answer here to make it easier.
Hint $\rm\ \ \ mod\,\ (\color{#0a0}{3\!-\!x^2},\ \color{#c00}{2x\!+\!4})\!: $ $\quad 6\!\!\!\!\! \color{#0a0}{\ \overset{ \overset{\ \ \ \ \Large{3\ \ \equiv\ \ x^2}}{\phantom{M}} }\equiv}\! $ $ 2x^2\!\!\!\color{#c00}{\ \overset{ \overset{\ \Large{2x\ \ \equiv\ \ -4}}{\phantom{M}} }\equiv}\!\! $ $ -4x \!\!\!\!\! \color{#c00}{\ \overset{ \overset{\ \Large{-2x\ \ \equiv\ \ 4\ \ \ \ \ \ }}{\phantom{M}} }\equiv}\!\!\!\!\!\! 8 \,\ \Rightarrow\,\ 2\equiv 0$
So $\rm\ \ 2 \in I = (x^2\!-\!3,\, 2x\!+\!4)\ \Rightarrow\ I = (2,I) = (2,\, I\ mod\ 2) = (2, x^2\!+\!1) = (2,(x\!+\!1)^2)$
