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My problem: Let $A$ and $B$ be two rings, let $I$ be an ideal of $A$ and $J$ an ideal of $B.$ Prove that $I \times J$ is an ideal of $A \times B$ and $\dfrac{A \times B}{I \times J} \cong \dfrac{A}{I} \times \dfrac{B}{J}.$

Let $(i_1, j_1) \in I \times J$ and $(i_2, j_2) \in I \times J$ be arbitrary. Thus $i_1, i_2 \in I$, $j_1, j_2 \in J.$ We have $(i_1,j_1) + (i_2,j_2) = (i_1+i_2,j_1+j_2) \in I \times J$ since $i_1+i_2 \in I$ and $j_1+j_2 \in J$ because $I$ and $J$ are ideals and are closed under addition. Now let $(i,j) \in I \times J$ and $(a,b) \in A \times B$ be arbitrary. Thus $i \in I, j \in J.$ We have $(a,b)\cdot(i,j) = (ai, bj).$ Since $I$ is an ideal in $A$ and $i \in I$, we have $ai \in I.$ Similarly, since $J$ is an ideal in $B$ and $j \in J$, we have $bj \in J.$ Therefore $(a,b)(i,j) =(ai, bj) \in I \times J.$ A similar argument shows that $(i, a)(j, b) = (ij, ab) \in I \times J.$ Thus $I \times J$ is an ideal of $A \times B.$

How can I establish the isomorphism?

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Consider the ring homomorphism $f=(p_I,p_J)\colon A\times B \rightarrow A/I\times B/J$ deduced from the canonical homomorphisms: $\,p_I\colon A\rightarrow A/I$ and $\,p_J\colon B \rightarrow B/J$.

It is surjective (product of surjections). What is its kernel?

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The natural candidate for the isomorphism is the map $A/I \times B/J \to (A \times B)/(I \times J)$ given by $(a + I,b + J) \mapsto (a,b) + I \times J$. You should prove that this is indeed an isomorphism.

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