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Let $A$ be a unitary ring, $R$ a subring and $I$ an ideal of $A$. It is easy to prove that $R\cap I$ is an ideal of $R$ and that $R+I$ is a sub ring of $A$, but how do I show the following isomorphism? $$\frac{R}{R\cap I} \cong \frac{R+I}{I}$$

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Hint: Construct a suitable homomorphism $f: R \to \frac{R+I}{I}$, note down its kernel and use the 1st isomorphism theorem. – Sayantan Jan 24 '13 at 0:37
2nd hint: the only reasonable choice for $f$ is obvious.... – David Wheeler Jan 24 '13 at 14:24
@Sayantan I'm writing a solution, but I have a doubt, isn't $\frac{R+I}{I} = \frac{R}{I}$? – Temitope.A Jan 24 '13 at 21:30
@DavidWheeler We define $f:r\mapsto r+I$ the projection of R onto the ideal I. The nucleus is the set of elements in the subring R that are also in I. From which the isomorphism. – Temitope.A Jan 24 '13 at 21:33
@Temitope.A : No it isn't. Take $A = \mathbb{Z}$, $R= 2\mathbb{Z}$ and $I = 3\mathbb{Z}$. Then $R+I = \mathbb{Z}$. Since $I \nsubseteq R$, it doesn't make any sense to write $R/I$. But as $I$ is an ideal of $R+I$, it makes perfect sense to write $(R+I)/I$. – Sayantan Jan 25 '13 at 3:39
up vote 1 down vote accepted

Or, to give form to Sayantan's idea in the opposite direction, define

$$f: R+I\to R/(R\cap I)\;\;,\;\;f(r+i):=r+(R\cap I)$$

and extend the definition by linearity (to finite sums and etc.). What's the above homorphism's kernel and image? And now apply the first isomorphism theorem.

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This suggestion works, but requires an extra step (compared to Sayantan's idea) because you need to prove that the function above is well-defined: if $r+i = r' + i'$, why is $f(r+i) = f(r'+i')$? – Manny Reyes Jan 24 '13 at 3:34
Good point: $$r+i=r'+i'\Longrightarrow r-r'=i'-i\in R\cap I\Longrightarrow r+ (R\cap I)=r'+(R\cap I)$$ Thanks. – DonAntonio Jan 24 '13 at 3:58

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