What the heck is a monomorphism?

\( \newcommand{\Z}{\mathbb{Z}} \newcommand{\Zi}{\mathbb{Z} [i]} \newcommand{\Zmp}{\mathbb{Z}_p [x]/(x^2+1)} \newcommand{\Rij}{R/(I \cap J)} \newcommand{\Rx}{R/I \times R/J} \)

Problem 1

Let \(R\) be a ring with identity and let \(I, J\) be ideal of \(R\) such that \(I + J = R\). Show that there is an isomorphism \[ R/(I \cap J) \cong R/I \times R/J \]


Claim. The map \(\phi: \Rij \rightarrow \Rx\) given by \[ a+ I\cap J \mapsto (a+I, a+J) \] is an isomorphism.

Proof. First, we must show that \(\phi\) is well defined. Say that we have \(a+I\cap J = b+I\cap J\). Then \(b=a+k\), for some \(k \in I\cap J\). (This is true because \(a-b \in I\cap J\).) Then apply \(\phi\) to \(a,b\). We obtain \((a+I, a+J)\) and \((a+k+I, a+k+J)\). Since \(k \in I\) and \(k \in J\), \((a+k+I, a+k+J) = (a+I, a+J)\). So \(\phi(a+I\cap J) = \phi(b+I\cap J)\). Thus, the map is well defined.

To show that \(\phi\) is an isomorphism, we must show that it is a bijective homomorphism. First, we show that \(\phi\) is a homomorphism.

Under addition: \[ \phi(a+I\cap J) + \phi(b+I\cap J) \] \[ = (a+I, a+J) + (b+I, b+J) \] \[ = (a+b+I, a+b+J) \] \[ = \phi(a+b+I\cap J) \] \[ = \phi(a+I\cap J + b+I\cap J) \]

Under multiplication: \[ \phi(a+I\cap J) \phi(b+I\cap J) \] \[ = (a+I, a+J) (b+I, b+J) \] \[ = (ab+I, ab+J) \] \[ = \phi(ab + I\cap J) \] \[ = \phi( (a+I\cap J) (b+I\cap J) ) \]

We must now show that \(\phi\) is surjective and injective.