Let \(R\) and \(S\) be commutative rings, and \(f:R\rightarrow S\) a ring homomorphism.
Show that if \(I\) is a prime ideal of \(S\), then \[ f^{-1}(I)=\{r\in R:f(r)\in I\} \] is a prime ideal of \(R\).
Let \(N\) be the set of nilpotent elements of \(R\): \[ N=\{r\in R:r^m=0\text{ for some }m\geq 1\}. \] \(N\) is called the nilradical of \(R\). Prove that it is an ideal which is contained in every prime ideal.
Part (a) lets us define a function \[ \begin{aligned} f^*:\{\text{prime ideals of }S\} &\rightarrow \{\text{prime ideals of }R\},\\ I &\mapsto f^{-1}(I). \end{aligned} \] Let \(N\) be the nilradical of \(R\). Show that if \(S=R/N\) and \(f:R\rightarrow R/N\) is the quotient map, then \(f^*\) is a bijection
Solution
1. Part (a): Proof that \(f^{-1}(I)\) is a prime ideal of \(R\):
1.1. Let \(I \subset S\) be a prime ideal.Consider the composite homomorphism \(\pi \circ f: R \xrightarrow{f} S \xrightarrow{\pi} S/I\), where \(\pi\) is the canonical projection.
composition of ring homomorphisms.
1.2. The kernel of \(\pi \circ f\) is:\[ \ker(\pi \circ f) = \{r \in R \mid f(r) \in I\} = f^{-1}(I). \]By the First Isomorphism Theorem for rings, \(R / f^{-1}(I) \cong \operatorname{Im}(\pi \circ f) \subseteq S/I\).
First Isomorphism Theorem.
1.3. Because \(I\) is a prime ideal in the commutative ring \(S\), the quotient \(S/I\) is an integral domain.Any subring of an integral domain is an integral domain, so \(R / f^{-1}(I)\) is an integral domain.Therefore \(f^{-1}(I)\) is a prime ideal of \(R\).
characterization of prime ideals via quotient rings being integral domains.
2. Part (b): Proof that the nilradical \(N\) is an ideal contained in every prime ideal:
2.1. \(0 \in N\) since \(0^1 = 0\).If \(x \in N\) with \(x^m = 0\) and \(r \in R\), then \((rx)^m = r^m x^m = r^m \cdot 0 = 0\) (using commutativity of \(R\)), so \(rx \in N\).
commutativity of \(R\).
2.2. If \(x, y \in N\) with \(x^m = 0\) and \(y^n = 0\), consider \((x + y)^{m + n - 1}\).By the binomial theorem in a commutative ring:\[ (x + y)^{m + n - 1} = \sum_{k=0}^{m + n - 1} \binom{m + n - 1}{k} x^k y^{m + n - 1 - k}. \]For each term, either \(k \ge m\) (so \(x^k = 0\)) or \((m + n - 1 - k) \ge n\) (so \(y^{m+n-1-k} = 0\)).Thus \((x + y)^{m + n - 1} = 0\), so \(x + y \in N\).Therefore \(N\) is an ideal of \(R\).
binomial theorem.
2.3. Let \(\mathfrak{p} \subset R\) be any prime ideal, and let \(x \in N\) with \(x^m = 0\).Since \(0 \in \mathfrak{p}\), we have \(x^m \in \mathfrak{p}\).By primality of \(\mathfrak{p}\), \(x^m \in \mathfrak{p} \implies x \in \mathfrak{p}\).Thus \(N \subseteq \mathfrak{p}\) for every prime ideal \(\mathfrak{p}\).
primality of \(\mathfrak{p}\).
3. Part (c): Proof that \(f^*: \operatorname{Spec}(R/N) \to \operatorname{Spec}(R)\) is a bijection:
3.1. By the Correspondence Theorem for rings, the quotient map \(f: R \to R/N\) induces an inclusion-preserving bijection between ideals of \(R/N\) and ideals of \(R\) containing \(\ker(f) = N\):\[ J \mapsto f^{-1}(J), \qquad \mathfrak{a} \mapsto f(\mathfrak{a}) = \mathfrak{a}/N. \]
Correspondence Theorem for ideals.
3.2. By Part (a), if \(J\) is a prime ideal of \(R/N\), then \(f^{-1}(J)\) is a prime ideal of \(R\).
Part (a).
3.3. By Part (b), every prime ideal \(\mathfrak{p} \in \operatorname{Spec}(R)\) contains \(N = \ker(f)\).For any such prime \(\mathfrak{p}\), the quotient isomorphism:\[ (R/N) / (\mathfrak{p}/N) \cong R/\mathfrak{p} \]shows that \(\mathfrak{p}/N\) is a prime ideal of \(R/N\) because \(R/\mathfrak{p}\) is an integral domain.
Third Isomorphism Theorem for rings.
3.4. The inverse map is \(\mathfrak{p} \mapsto \mathfrak{p}/N\), with \(f^{-1}(\mathfrak{p}/N) = \mathfrak{p}\) and \(f(f^{-1}(J)) = J\).Thus \(f^*\) is a bijection between \(\operatorname{Spec}(R/N)\) and \(\operatorname{Spec}(R)\).
two-sided inverse.
4. Conclusion:Parts (a), (b), and (c) are fully proved. Q.E.D.
1 through 3.