Fulton 1 - Lemmas

Lemma 1. Suppose $R,S,T$ are groups (or rings, modules) and $\pi:R\to S$ , $\varphi:R\to T$ are group (or ring, module) homomorphisms such that $\ker\pi\subset\ker\varphi$ . Then there exists an induced homomorphism $\widetilde{\varphi}:\pi(R)\to T$ such that $\widetilde{\varphi}\circ\pi=\varphi$ . Moreover, $\ker\widetilde{\varphi}=\pi(\ker\varphi)$ .

Suppose $a,b\in R$ such that $\pi(a)=\pi(b)$ . Then $a-b\in\ker\pi\subset\ker\varphi$ . So $a+\ker\varphi=b+\ker\varphi$ and $\varphi(a)=\varphi(b)$ . Thus the map $\widetilde{\varphi}:\pi(R)\to T$ defined by $\pi(a)\mapsto\varphi(a)$ is well defined homomorphism.

Lemma 2. If $R$ is a PID then the prime ideals of $R[X]$ are precisely those of the form $(0)$ , $(F(X))$ where $F$ is irreducible, and $(P, F(X))$ where $P$ is a prime ideal and $F(X) + P$ is irreducible over $(R/P)[X]$ .

Let $I$ be a non-zero prime ideal of $R[X]$ .

Suppose $J=I\cap R\ne \{0\}$ . Then $R[X]/I\cong (R[X]/J[X])/(I/J[X])\cong(R/J)[X]/(I/(J[X])).$ Since $R/J$ is a field, $(R/J)[X]$ is a Euclidean Domain. Thus $I/(J[X])=(F(X)+J[X])$ for some $F\in R[X]$ , of minumum degree, such that $F+J[X]$ is irreducible over $(R/J)[X]$ . Therefore $I=(J, F(X))$ .

Suppose $I\cap R=\{0\}$ . Let $F\in I$ be of minimum degree. Suppose $G\in I$ such that $\text{gcd}(F,G)=1$ . By Gauss’ Lemma $\text{gcd}(F,G)=1$ in $K[X]$ where $K$ is the field of fractions of $R$ . Since $K[X]$ is a Euclidean domain, there exist $A,B\in K[X]$ such that $A(X)F(X)+B(X)G(X)=1\in K[X]$ . If we let $\alpha\in R\setminus\{0\}$ be a common denominator for the coefficients of $A,B$ then $\alpha A(X)F(X)+\alpha B(X)G(X)=\alpha$ . Therefore $\alpha\in I$ , a contradiction. Thus $F\mid G$ for all $G\in I$ . So $I=(F)$ .