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A very naturel question come in my mind but I do not know how to answer, I need a help.

For this ring $(\mathbb{Z}_2\times \mathbb{Z}_3)[X]/ (X^2)$, I was wondreing if thiere elements are like $(0,1)\overline{X}+(1,2)$ but I never see this notation!! If it is true where can I read about it?

Another question, I know that this is finite ring, so we can express all its elemnt. I think that is like $\mathbb{Z}_p[X]$ so I can use euclidian division, but here $\mathbb{Z}_2\times \mathbb{Z}_3$ is not a ring, is there a method to do that?

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    $\begingroup$ $Z_2xZ_3 \cong Z_6$ $\endgroup$ Commented May 1, 2019 at 21:37
  • $\begingroup$ thanks but this do not answer my question, I want to know this ring not what it is isomorphic to $\endgroup$
    – Mary Maths
    Commented May 1, 2019 at 21:40

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You claim that $(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/3\Bbb{Z})$ is not a ring, but it is in fact canonically a ring. In general, the product of two rings is again a ring. In this particular case we even have the very nice canonical isomorphism $$(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/3\Bbb{Z})\cong\Bbb{Z}/6\Bbb{Z},$$ so in stead of representing the coefficients of polynomials by pairs of integers, we can represent them by integers from the set $\{0,1,2,3,4,5\}$. And indeed, all elements of the quotient $$(\Bbb{Z}/6\Bbb{Z})[X]/(X^2),$$ are of the form $a\overline{X}+b$ with $a,b\in\{0,1,2,3,4,5\}$. A popular shorthand is to write $\varepsilon:=\overline{X},$ and $$(\Bbb{Z}/6\Bbb{Z})[\varepsilon]:=(\Bbb{Z}/6\Bbb{Z})[X]/(X^2),$$ in analogy with the infinitesimal $\varepsilon$ from calculus, as it satisfies $\varepsilon^2=0$.

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