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Let $R$ be a ring, then

For $R[x]/\langle x-1\rangle \cong R$, we define the map, $\varphi$ : $R[x]\rightarrow R$, defined by $\varphi(f) =f(1)$

For $R[x]/\langle x\rangle \cong R$, we define the map, $\varphi$ : $R[x]\rightarrow R$, defined by $\varphi(f) =f(0)$

I'm going to show that $R[x]/\langle x^2 -x\rangle \cong R$, help me find an isomorphism if it exists.

In the general case, I'm looking for a technique to define the map.

Thanks in advance...

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  • $\begingroup$ The last example is not correct, so there is no way to define the map. $\endgroup$ – Tobias Kildetoft Apr 27 '15 at 9:37
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    $\begingroup$ In the last example now, $R$ must have some non-zero element $a$ with $a^2 = 0$ to which $x$ must be sent. $\endgroup$ – Tobias Kildetoft Apr 27 '15 at 9:45
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    $\begingroup$ There is still no good reason why there should be an isomorphism. $\endgroup$ – Tobias Kildetoft Apr 27 '15 at 9:54
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    $\begingroup$ now the quotient ring on the left has at least 4 idempotents (1, 0, x, 1-x) but it is possible that R only has two, preventing an isomorphism. $\endgroup$ – rschwieb Apr 27 '15 at 9:56
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    $\begingroup$ @rschwieb And in fact the quotient is isomorphic to $R\times R$, so it will be rare that it is isomorphic to $R$ (though not impossible). $\endgroup$ – Tobias Kildetoft Apr 27 '15 at 10:01
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Such an isomorphism might exist for some special rings $R$, but it doesn't in general. If we assume that $R$ is commutative with $1$ (which is sensible, to avoid issues with the definition of the polynomial ring), then by the Chinese Remainder Theorem we have $$ R[x]/\langle x^2 - x \rangle \simeq R[x]/\langle x - 1 \rangle \times R[x]/\langle x \rangle \simeq R \times R $$ because $x - (x - 1) = 1$ implies that $\langle x \rangle + \langle x - 1 \rangle = R$.

For an explicit counterexample, suppose that $R$ is a domain, e.g. $R = \Bbb{Z}$. Then $$ R[x]/\langle x^2 - x \rangle \not\simeq R $$ because the images of $x$ and $x - 1$ in the quotient are non-zero and zero-divisors.

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