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What is the factor ring $R:=\mathbb{Z}_{6}[x]/\langle x^{2}\rangle$?

Since $\mathbb{Z}_{6}[x]/\langle x^{2}\rangle=\{a+bx+\langle x^{2}\rangle\,:\,a,b\in\mathbb{Z}_{6}\}$, i guess $R\cong\mathbb{Z}_{6}\times\mathbb{Z}_{6}$, but i'm not sure about it.

Q1) Is it possible to compute the following?: $$R\cong\mathbb{Z}_{2}[x]/\langle x^{2}\rangle\times\mathbb{Z}_{3}[x]\langle x^{2}\rangle\cong\left(\mathbb{Z}_{2}\times\mathbb{Z}_{2}\right)\times\left(\mathbb{Z}_{3}\times\mathbb{Z}_{3}\right)\cong\mathbb{Z}_{6}\times\mathbb{Z}_{6}.$$

Q2) If the above computation is true, I wonder what kind of theorem can guarantee that such computations are valid in general situation for example `on $\mathbb{Z}_{n}'$ where $n$ is composite.

Give some advice or comments. Thank you!

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    $\begingroup$ It's an example of a dual number ring. Such rings prove handy for algebraically modelling tangent spaces and higher-order jets spaces, and for transfering properties of homs to derivations. Follow the link for citations. $\endgroup$ – Gone Nov 12 '19 at 16:19
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It is not true that $R \stackrel{\sim}{\to} \mathbb{Z}_6 \times \mathbb{Z}_6$. For example, in $R$, there is an element whose square is zero (namely $x$), but there are no elements in $\mathbb{Z}_6 \times \mathbb{Z}_6$ with this property.

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Consider the homomorphism $f:R\to \Bbb Z_6\times \Bbb Z_6$ defined by $$f:a+bx+\langle x^2\rangle\mapsto (a,b).$$ This preserves addition, but not, multiplication.

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  • $\begingroup$ I don't think this is a ring homomorphism. We have $(a+bx)(a'+b'x) = aa' + (b+b')x$, so $f((a+bx)(a'+b'x)) = (aa', b+b') \neq (a, b)*(a', b')$. $\endgroup$ – hunter Nov 12 '19 at 4:39
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If I recall correctly, when we write

$R \cong \Bbb Z_6 \times \Bbb Z_6 \tag 1$

as a ring isomorphism, we usually mean the multiplication in $\Bbb Z_6 \times \Bbb Z_6$ to take place component-wise, that is,

$(a, b) \cdot (c, d) = (ac, bd); \tag 2$

however, in $\Bbb Z_6[x]/\langle x^2 \rangle$,

$x^2 \in \langle x^2 \rangle \Longrightarrow x^2 \equiv 0 \mod \langle x^2 \rangle; \tag 3$

therefore, in $\Bbb Z_6[x]/\langle x^2 \rangle$

$((a+ bx) + \langle x^2 \rangle) \cdot ((c + dx) + \langle x^2 \rangle) = (a + bx) \cdot (c + dx) + \langle x^2 \rangle = (ac + (ad + bc)x + bdx^2 +\langle x^2 \rangle = ac + (ad + bc)x + \langle x^2 \rangle, \tag 4$

since

$bdx^2 \in \langle x^2 \rangle \Longrightarrow bdx^2 + \langle x^2 \rangle = \langle x^2 \rangle; \tag 5$

thus, multiplication in $\Bbb Z_6[x]/ \langle x^2 \rangle$ is given by

$(a + bx) \cdot (c + dx) \equiv ac + (ad + bc)x \; \mod \langle x^2 \rangle; \tag 6$

from this formula it is easy to see that "$\cdot$" is not a component-wise operation in $\Bbb Z_6[x]/\langle x^2 \rangle$.

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