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This is the question and its answer:

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I understood the ordinary way of solution, but there is another method that is stated in the lines 2 and 3 from below, could anyone explain it in details for me please? why the ring $\mathbb{Z_5}$ is isomorphic to our ring? and I know that in the definition of the field every nonzero element has a multiplicative inverse, how does this help here?

Thanks!

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    $\begingroup$ It's a subtle thing, but you have to understand that there is an isomorphism between $\mathbb Z_{10}$ and $\mathbb Z_{5}\times\mathbb Z_2$, with the set $S$ going to the elements $(k,0)$ where $k\in\mathbb Z_5$. But the set of those elements in the product ring is essentially the same as $\mathbb Z_5$. $\endgroup$ Sep 7 '17 at 18:45
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Note that $\phi\colon x+5\Bbb Z\mapsto 6x+10\Bbb Z$ is a rng homomorphism. Indeed, it is clearly well-defined, compatible with addition, and - as $6x\cdot 6y=36xy\equiv 6xy\pmod{10}$ - also compatible with multiplication. Just verify that it is injective ($6x$ can only be a multiple of $10$ if $x$ is a multiple of $5$) and that its image is $S$, and we are done.

Now if you replace $(S,+,\cdot)$ with the field $(\Bbb F_5,+,\cdot)$, claims A,B,C,E are trivially true and D trivially false.

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  • $\begingroup$ What is $\mathbb{F_{5}}$? $\endgroup$
    – user426277
    Sep 8 '17 at 9:15
  • $\begingroup$ What is $\mathchoice$ ? $\endgroup$
    – user426277
    Sep 8 '17 at 9:16
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The "multiplication by $6$" $$\mathbb{Z}_5 \to \mathbb{Z}_{10}, \quad x \to 6x$$ is a ring isomorphism between $\mathbb{Z}_5$ and your subring $S$.

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This is a more theoretical answer.

Chinese remainder theorem means that $x\bmod 10$ is determined by $x\bmod 2$ and $x\bmod 5$. This can be broadened to see that:

$$\mathbb Z_{10}\cong \mathbb Z_2\times \mathbb Z_5$$

With $a+10\mathbb Z\mapsto (a+2\mathbb Z,a+5\mathbb Z$) and inverse $$(b+2\mathbb Z,c+5\mathbb Z)\to (6c-5b)+10\mathbb Z$$

That $6c-5b$ is the result of applying Chinese remainder theorem to $x\equiv b\pmod{2},\, x\equiv c\pmod{5}$.

That product formula is what is going on here.

Now, the subset of elements $(0,c)$ in $\mathbb Z_2\times\mathbb Z_5$ act like a ring, but with a different unit from the parent ring, because it is exactly isomorphic to $\mathbb Z_5$. And, in $\mathbb Z_{10}$ this is exactly the elements of your $S$ - the elements $a+\mathbb Z_{10}$ such that $a+2\mathbb Z=0+2\mathbb Z$ are the elements with $a$ divisible by $2$.

More generally, if $R$ is a commutative ring with identity, and $e_1,\dots,e_n$ are elements such that:

$$e_ie_j=\begin{cases}0&i\neq j\\ e_i&i=j\end{cases}\\ e_1+e_2+\cdots e_n=1$$

Then each of $e_iR$ is a ring with $e_i$ an identity, and:

$$R\cong e_1R\times e_2R\times\cdots\times e_nR$$

with map: $r\mapsto (e_1r,e_2r,\dots,e_nr)$ and inverse map: $(r_1,\dots,r_n)\mapsto r_1+r_2+\cdots+r_n$.

In the case of $R=\mathbb Z_{10}$ you have $e_1=5,e_2=6$.

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Consider the isomorphism $\varphi: \{0, 2, 4, 6, 8\} \to \mathbb{Z}_5$ given by $0 \to 0, 2 \to 2, 4\to4,6\to1,8\to3$.

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  • $\begingroup$ You are correct. I will change my answer $\endgroup$
    – AsafHaas
    Sep 7 '17 at 18:50

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