# Primary decomposition of an ideal and its extension

I'm trying to solve a problem in Sharp's Steps in Commutative Algebra, to be precise Exercise 4.22 which states the following:

Let $$f:R \rightarrow S$$ be a surjective homomorphism of commutative rings.

Let $$I,Q_1,Q_2,...,Q_n,P_1,...,P_n$$ be ideals of $$R$$ all of which contain $$\ker f$$. Show that

$$I=Q_1 \cap\dots\cap Q_n ~~~~\text{with}~~ \sqrt {Q_i}= P_i~~~~\text{for}~~i=1,2,...,n$$ is a primary decomposition of $$I$$ if and only if

$$I^e=Q_1 ^e \cap \dots\cap Q_n ^e ~~~~\text{with}~~ \sqrt {(Q_i^e)}= P_i^e~~~~\text{for}~~i=1,2,...,n$$ is a primary decomposition of $$I^e$$, and that, when this is the case, the first of these is minimal iff the second is.

Deduce that $$I$$ is a decomposable ideal of $$R$$ iff $$I^e$$ is a decomposable ideal of $$S$$.

This is the first time I've been studying commutative algebra and I have a really hard time. Any help will be appreciated, thanks in advance.

Considering that $$I, Q_1, \dots, Q_n, P_1, \dots, P_n$$ are ideals of $$R$$ that contain $$\ker f,$$ the cosets $$\bar I = I / \ker f, \bar Q_i = Q_i / \ker f,$$ and $$\bar P_i = P_i / \ker f$$ are ideals of $$R / \ker f$$ for each integer $$1 \leq i \leq n$$ by the Fourth Isomorphism Theorem AKA the Correspondence Theorem.

We claim first that $$I = Q_1 \cap \cdots \cap Q_n$$ is a primary decomposition of $$I$$ with $$\sqrt{Q_i} = P_i$$ for each $$i$$ if and only if $$\bar I = \bar Q_1 \cap \cdots \cap \bar Q_n$$ is a primary decomposition of $$\bar I$$ with $$\sqrt{\bar Q_i} = \bar P_i$$ for each $$i.$$

Proof. Observe that we have $$i \in I$$ if and only if $$i + \ker f \in \bar I.$$ Evidently, for any element $$i \in I,$$ it follows that $$i + \ker f \in \bar I$$ by definition. Conversely, for any element $$j \in \ker f,$$ we have that $$j \in I$$ so that $$i = (i - j) + j$$ is an element of $$I$$ for any $$i \in I.$$ But this implies that $$i \in I$$ whenever $$i + \ker f \in \bar I.$$ Consequently, we have that $$I = Q_1 \cap \cdots \cap Q_n$$ if and only if $$\bar I = \bar Q_1 \cap \cdots \cap \bar Q_n.$$

We have also that $$r \in \sqrt{Q_i}$$ if and only if $$r + \ker f \in \sqrt{\bar Q_i}.$$ By definition, we have that $$r \in \sqrt{Q_i}$$ if and only if $$r^n \in Q_i$$ if and only if $$r^n + \ker f \in \bar Q_i$$ by the above, hence it suffices to prove that $$(r + \ker f)^n = r^n + \ker f.$$ But this is clear by the Binomial Theorem since all of the terms $$r^k$$ with $$0 \leq k \leq n - 1$$ have a factor of $$\ker f.$$

Our proof is complete once we establish that $$\sqrt{\bar Q_i} = \overline{\sqrt{Q_i}}.$$ But this follows from the above, as we have that $$r + \ker f \in \sqrt{\bar Q_i}$$ if and only if $$r^n + \ker f = (r + \ker f)^n \in \bar Q_i$$ if and only if $$r^n \in Q_i$$ if and only if $$r \in \sqrt{Q_i}$$ if and only if $$r + \ker f \in \overline{\sqrt{Q_i}}.$$ We conclude that $$\sqrt{Q_i} = P_i$$ if and only if $$\sqrt{\bar Q_i} = \bar P_i$$ for each $$i.$$ QED.

By the First Isomorphism Theorem, there exists a unique isomorphism $$\varphi : R / \ker f \to S$$ such that $$f = \varphi \circ \pi,$$ where $$\pi$$ is the canonical surjection $$\pi : R \to R / \ker f.$$ Consequently, the extension of any ideal $$J$$ of $$R$$ is given by $$f(J) = \varphi \circ \pi(J) = \varphi(\bar J).$$ By the above result, we conclude that $$I = Q_1 \cap \cdots \cap Q_n$$ if and only if $$\bar I = \bar Q_1 \cap \cdots \cap \bar Q_n$$ if and only if $$\varphi(\bar I) = \varphi(\bar Q_1 \cap \cdots \cap \bar Q_n) = \varphi(\bar Q_1) \cap \cdots \cap \varphi(\bar Q_n)$$ (by injectivity of $$\varphi$$) if and only if $$f(I) = f(Q_1) \cap \cdots \cap f(Q_n).$$ We have also that $$\sqrt{Q_i} = P_i$$ if and only if $$\sqrt{\bar Q_i} = \bar P_i$$ if and only if $$\varphi(\sqrt{\bar Q_i}) = \varphi(\bar P_i)$$ (by injectivity of $$\varphi$$) if and only if $$f(\sqrt{Q_i}) = f(P_i)$$ for each integer $$1 \leq i \leq n.$$

We turn our attention to the minimality assertion. By a minimal primary decomposition of $$I,$$ we mean that $$I = Q_1 \cap \cdots \cap Q_n$$ with $$\sqrt{Q_i}$$ distinct and $$\cap_{j \neq i} Q_j \not \subseteq Q_i$$ for each integer $$1 \leq i \leq n.$$ We have already seen that $$\sqrt{Q_i}$$ are distinct if and only if $$\sqrt{\bar Q_i}$$ are distinct if and only if $$\varphi(\sqrt{\bar Q_i})$$ are distinct (by injectivity of $$\varphi$$) if and only if $$f(\sqrt{Q_i})$$ are distinct. Likewise, we have that $$\cap_{j \neq i} Q_j \not \subseteq Q_i$$ if and only if $$\cap_{j \neq i} \bar Q_j = \overline{\cap_{j \neq i} Q_j} \not \subseteq \bar Q_i$$ if and only if $$\cap_{j \neq i} \varphi(\bar Q_j) = \varphi(\cap_{j \neq i} \bar Q_j) = \varphi(\overline{\cap_{j \neq i} Q_j}) \not \subseteq \varphi(\bar Q_i)$$ (by injectivity of $$\varphi$$) if and only if $$\cap_{j \neq i} f(Q_j) \not \subseteq f(Q_i).$$

Ultimately, an ideal $$I$$ of $$R$$ has a primary decomposition if and only if $$\bar I$$ has a primary decomposition if and only if $$\varphi(\bar I)$$ has a primary decomposition if and only if $$f(I)$$ has a primary decomposition in $$S.$$

• I understand more clearly when I write the solution on paper. Thank you so much for telling so clearly. Jun 12, 2020 at 21:35
• Perfect. Glad I could help. Jun 12, 2020 at 21:53