Let \(K,L\) be fields, and give \(K\times L\) componentwise addition and multiplication. Find all prime ideals of \(K\times L\).
Solution
Let \[ I\coloneqq K\times\{0\}, \qquad J\coloneqq\{0\}\times L. \]
1. The ideals \(I\) and \(J\) are maximal, hence prime.
The second coordinate projection \[ \pi_L:K\times L\longrightarrow L \] is surjective with kernel \(I\), so \[ (K\times L)/I\cong L. \] Since \(L\) is a field, \(I\) is maximal. Similarly, the first coordinate projection has kernel \(J\) and quotient isomorphic to \(K\), so \(J\) is maximal.
2. Every prime ideal \(P\subseteq K\times L\) is equal to \(I\) or\(J\).
In \(K\times L\), \[ (1,0)(0,1)=(0,0)\in P. \] Since \(P\) is prime, either \((1,0)\in P\) or \((0,1)\in P\).
If \((1,0)\in P\), then for every \(a\in K\), \[ (a,0)=(a,0)(1,0)\in P, \] so \(I\subseteq P\). By maximality of \(I\) from step 1 and properness of the prime ideal \(P\), this forces \(P=I\).
If \((0,1)\in P\), the same argument gives \(J\subseteq P\), hence \(P=J\).
3. The prime ideals of \(K\times L\) are exactly\[ \boxed{K\times\{0\},\ \{0\}\times L}. \]
Step 1 shows that both displayed ideals are prime, and step 2 shows that there are no others.
4. Q.E.D.
Step 3 gives the requested classification.