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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.

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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 $S.$

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

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