Fulton 1.4

Posted on November 15, 2017

1.22.* Let I be an ideal in a ring R, \pi:R\to R/I the natural homomorphism.

(a) Show that for every ideal J&39; of R/I, \pi^{-1}(J&39;)=R is an ideal of R containing I, and for every ideal J of R containing I, \pi(J)=J&39; is an ideal of R/I. This sets up a natural one-to-one correspondence between ideals of R/I and ideals of R that contain I.

Suppose J&39; is an ideal of R/I. Since \varphi is a ring homomorphism, \varphi^{-1}(J&39;)=J is an abelian group. Suppose r\in R, j\in J. Then observe \varphi(rj)=\varphi(r)\varphi(j)\in J&39;. Thus rj\in\varphi^{-1}(J&39;)=J. So J is an ideal.

Suppose J is an ideal of R. Note \varphi(J)=J&39; is an abelian group. Let j+I\in\varphi(J), r+I\in R/I. Observe since j\in J, rj\in J. Thus (j+I)+ (r+I)=rj+I=\varphi(rj)\in\varphi(J)=J&39;. Thus J&39; is an ideal.

(b) Show that J&39; is a radical ideal if and only if J is radical. Similarly for prime and maximal ideals.

Suppose J&39; is radical. Suppose r\in R such that r^n\in J. Then \varphi(r^n)= \varphi(r)^n\in J&39;. Thus \varphi(r)\in J&39;. Thus r\in\varphi^{-1}(J&39;)=J.

Conversely, suppose J is radical. Let r+I\in R/I such that r^n+I\in J&39;. Then r^n\in\varphi^{-1}(J&39;)=J and hence r\in J. So \varphi(r)=r+I\in J&39;.

Suppose J&39; is prime. Let a,b\in R such that ab\in J. Then \varphi(ab)= \varphi(a)\varphi(b)\in J&39;. So one of \varphi(a) or \varphi(b) must be in J&39;. Thus one of a or b must be in J.

Conversely, suppose J is prime. Let a+I,b+I\in R/I such that ab+I\in J&39;. Then ab\in J. So one of a or b must be in J and therefore one of a+I or b+I must be in J&39;.

Suppose J&39; is maximal. Suppose there exists a proper ideal K of R such that J\subsetneq K. Then by (a), \varphi(K) is a proper ideal of R/I properly containing J&39;, a contradiction. So J is maximal.

Conversely, suppose J is maximal. Suppose there exists a proper ideal K&39; of R/I such that J&39;\subsetneq K&39;. Then by (a), J&39;\subsetneq\varphi^{-1}(K&39;) \subsetneq R, a contradiction. So J&39; is maximal.

(c) Show that J&39; is finitely generated if J is. Conclude that R/I is Noetheerian if R is Noetherian. Any ring of the form k[X_1,\ldots,X_n]/I is Noetherian.

Suppose J is finitely generated by \{j_1,\ldots,j_n\}\subset J. Then for any j\in J we have j=r_1j_1+\ldots+r_nj_n, r_1,\ldots,r_n\in R. Thus for any j+I\in J&39; we have j+I=(r_1j_1+\ldots+r_nj_n)+I = (r_1+I)(j_1+I)+\ldots+(r_n+I)(j_n+I). That is, J&39; is generated by \{j_1+I,\ldots,j_n+I\}.

Therefore if R is Noetherian, so is R/I. Since k[X_1,\ldots,X_n] is Noetherian, k[X_1,\ldots,X_n]/I is Noetherian for any ideal I.