I am currently studying for my first exam in my second course of abstract algebra and cannot seem to understand the reasoning in the following explanation. In finding a maximal ideal of $\mathbb{Z}$x$\mathbb{Z}$, why are we able to state in computing the factor ring (in the following picture) that the left hand side is equal to the right hand side? I guess i’m not understanding how that type of computation is possible.
Let $\mathbb{Z}\times 3\mathbb{Z}$ be an ideal of $\mathbb{Z}\times\mathbb{Z}$. Then compute the factor ring
$$\begin{align*}
(\mathbb{Z}\times\mathbb{Z})/(\mathbb{Z}\times 3\mathbb{Z}) &\cong (\mathbb{Z}/\mathbb{Z})\times(\mathbb{Z}/3\mathbb{Z})\\
&\cong \{0\}\times\mathbb{Z}_3\\
&\cong \mathbb{Z}_3
\end{align*}$$
Since $3$ is a prime number, the ring $\mathbb{Z}_3$ is a field, and hence the factor ring $(\mathbb{Z}\times\mathbb{Z})/(\mathbb{Z}\times 3\mathbb{Z})$ is a field.
Best Answer
I think the following lemma solves your question:
Lemma: Let $R,S$ be rings and $I$ an ideal of $R$ and $J$ an ideal of $S$. Then
$$(R\times S)/(I \times J) \cong R/I \times S/J$$
Proof: Define a map $\psi: R \times S \to R/I \times S/J: (r,s) \mapsto (r+I, s+J)$. Clearly this is a surjection with kernel $I \times J$ and the isomorphism follows from the first isomorphism theorem of rings. $\quad \square$