Question

Question 17: Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$be a surjective homomorphism of commutative rings. If J is a prime ideal in S, and , prove that I is a prime ideal in R.$\mathbf{I}=\left\{\left.\mathbf{r}\in \mathbf{R}\right|\mathbf{f}\left(\mathbf{r}\right)\in \mathbf{J}\right\}$

Short answer

Answer:

It is proved that is a prime ideal in .

Step-by-step solution

Definition of prime ideal

An ideal P with commutative ring R is known as prime when$P\ne R$and when$bc\in P$ then $b\in Porc\in P$.

Show that is a prime ideal in R

There would be certain$s\in S-Js$ because is a prime ideal. There would be some$\overline{s}\in R$ in which $f\left(\overline{s}\right)=s$because f would be surjective and then $\overline{s}\in I$. As a result, $I\ne R$.

Assume that such that . It follows that$f\left(ab\right)=f\left(a\right)f\left(b\right)\in J$ . Either $f\left(a\right)\in J$or$f\left(b\right)\in J$ because J would be prime.

When $f\left(a\right)\in J$then $\left(b\right)\in I$. As a result, there is either $\left(a\right)\in Ior\left(b\right)\in I$ . Therefore, I would be a prime ideal in R.

Hence, it is proved that is I a prime ideal in R .