Question

Question 14: If P is a prime ideal in a commutative ring R, is the ideal PXPa prime ideal in ? [Hint: ExerciseRXR 13.]

Short answer

Answer

The ideal PXP is not a prime ideal inRXR .

Step-by-step solution

Statement of theorem 6.14 

Theorem 6.14states that if Pis an ideal in a commutative ring Rwith identity. Then, P would be a prime idealsuch that the quotient ring R/P would be an integral domain.

 Step 2: Is the ideal PXP a prime ideal in RXR

Exercise 13states that when I would be an ideal in a ring R, then is an ideal in by exercise 8 of section 6.1. Show that is isomorphic to .

Also, $\left(\mathrm{\mathbb{Z}}\times \mathrm{\mathbb{Z}}\right)/\left(\left(2\right)\times \left(2\right)\right)\cong {\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$from exercise 13, it is observed that ${\mathrm{\mathbb{Z}}}_2\times {\mathrm{\mathbb{Z}}}_2$would not be an integral domain because $\left(\left[1\right],\left[0\right]\right)\left(\left[0\right],\left[1\right]\right)=\left(\left[0\right],\left[0\right]\right)$.

As a result, the product of prime ideals might not be the prime ideal according to theorem 6.14.

Hence, $P\times P$the ideal is not a prime ideal in $R\times R$.