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i understand that if two rings are isomorphic, and one ring is an integral domain, so must the other be.
however, consider two rings, both commutative rings with unity. is it possible that one ring contains zero divisors and one does not while there exists an ring homomorphism between the two? there could not be a isomorphism between the two rings because there would be no one to one or onto mapping between the two rings. but could there be an operation preserving mapping between an integral domain and a commutative ring with unity and with zero divisors? clearly if such a mapping did exists it would not seem to be one to one or onto, but does this rule out the potential of a homomorphism existing between the two?

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Given $R$ any ring, $R[x]/(p(x))$ has zero divisors if $p(x)$ is not irreducible, and there is an inclusion $R\hookrightarrow R[x]/(p(x))$ which is a ring homomorphism. –  anon Apr 27 '13 at 23:51

3 Answers 3

The quotient map $f: \mathbb{Z} \to \mathbb{Z}/(4)$ maps an integral domain $\mathbb{Z}$ to a ring with a (nilpotent) zerodivisor $\mathbb{Z}/(4)$. Now, the quotient map $g: \mathbb{Z}/(4) \to (\mathbb{Z}/(4))/(2) = \mathbb{Z}/(2)$ maps this ring to the field $\mathbb{Z}/(2)$.

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Given any unital ring $R$ (with multiplicative identity $1_R$, say), there is a unique ring homomorphism $\Bbb Z\to R$ (take $1\mapsto 1_R$ and "fill in the blanks" from there).

This may be an injective map, even if $R$ has zero divisors. For example, take $$R=\Bbb Z[\epsilon]:=\Bbb Z[x]/\langle x^2\rangle.$$ Surjective examples are easy to come by. You are of course correct that such maps cannot be bijective if $R$ has zero divisors.

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Consider the map \begin{align*} \phi : k&\hookrightarrow k[X,Y]/(XY)\\ a&\mapsto a + (XY), \end{align*} where $k$ is a field. This is the inclusion of $k$ into a quotient of the polynomial ring over $k$ in two variables. We see that this is a ring homomorphism, as $$\phi(ab) = (ab + (XY)) = (a + (XY))(b + (XY)) = \phi(a)\phi(b).$$ So we have an injection of a field (not just an integral domain, and not just a homomorphism) into a commutative ring with zero divisors: $$(X + (XY))(Y + (XY)) = XY + (XY) = 0 + (XY) = (XY).$$ Like xyzzyx said in his answer, you can also find a surjective map from an integral domain to a commutative ring with zero divisors (eg. $\pi : \Bbb{Z}\twoheadrightarrow\Bbb{Z}/n\Bbb{Z}$, where $n$ is composite), so we can indeed find injective or surjective maps from an integral domain to a ring with zero divisors.

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