Solutions to the exercises of *Introduction to Commutative Algebra, Atiyah*, Chapter 8.

## Artin Rings

Exercise *symbolic power* of $p_{i}$ (Chapter 4, Exercise 13). Show that for each $i=1,…,n$ there exists an integer $r_{i}$ such that $p_{i}⊆q_{i}$.

**Proof.**Since $A$ is Noetherian and $r(q_{i})=p_{i}$, it follows that there exists $r_{i}$ such that $p_{i}⊆q_{i}$. Taking $S_{p_{i}}$ on both side, we have $p_{i}⊆S_{p_{i}}(q_{i})=q_{i}$.

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Suppose $q_{i}$ is an isolated primary component. Then $A_{p_{i}}$ is an Artin local ring, hence if $m_{i}$ is its maximal ideal we have $p_{i}=0$ for all sufficiently large $r$, hence $q_{i}=p_{i}$ for all large $r$.

**Proof.**If $q_{i}$ is an isolated primary component, then by definition $p_{i}$ is a minimal prime ideal. Hence $A_{p_{i}}$ is an Artin local ring. By (8.6), $m_{i}=0$ for all sufficiently large $r$, hence $q_{i}=0_{c}=(m_{i})_{c}=p_{i}$ for all large $r$.

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If $q_{i}$ is an embedded primary component, then $A_{p_{i}}$ is *not* Artinian, hence the powers $m_{i}$ are all distinct, and so the $p_{i}$ are all distinct. Hence in the given primary decomposition we can replace $q_{i}$ by any of the infinite set of $p_{i}$-primary ideals $p_{i}$ where $r≥r_{i}$, and so there are infinitely many minimal primary decompositions of $0$ which differ only in the $p_{i}$-component.

**Proof.**If $q_{i}$ is an embedded primary component, then $p_{i}$ is not minimal. Hence $A_{p_{i}}$ is not Artinian. By (8.6) again, the powers $m_{i}$ are all distinct, so $p_{i}=(m_{i})_{c}$ are all distinct. Hence replacing $q_{i}$ by $p_{i}$ where $r≥r_{i}$ gives another primary decomposition of $0$, which is also minimal by the 1st uniqueness theorem of primary decompositions.

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Exercise

1. | $A$ is Artinian; |

2. | $Spec(A)$ is discrete and finite; |

3. | $Spec(A)$ is discrete. |

**Proof.** i) $⟹$ ii) $⟹$ iii) is clear.

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Exercise

1. | $A$ is Artinian; |

2. | $A$ is a finite $k$-algebra. |

**Proof.**

i) $⟹$ ii): | If $A$ is Artinian, then $A$ is a finite direct product of Artin local rings by (8.7). Clearly each multiplicand is also a finitely generated $k$-algebra. So it reduces to the case that $A$ is local. Let $m$ be the maximal ideal of $A$, then $L=A/m$ is a finite extension of $k$ by the Hilbert Nullstellensatz. If $m_{n}=0$, then $A=m_{0}⊇m⊇m_{2}⊇⋯⊇m_{n−1}⊇0$; each $m_{t}/m_{t+1}$ is a $L$-vector space of finite dimension since $A$ is Noetherian. Hence $dim_{L}(A)=∑_{t=0}dim_{L}(m_{t}/m_{t+1})$ is finite. Therefore $dim_{k}(A)$ is also finite. |

ii) $⟹$ i): | Every ideals of $A$ is also a $k$-vector subspace of $A$. Since $A$ is a finite-dimensional $k$-vector space, it satisfies d.c.c on ideals. $□$ |

Exercise

1. | $f$ is finite; |

2. | the fibres of $f_{∗}$ are discrete subspaces of $Spec(B)$; |

3. | for each prime ideal $p$ of $A$, the ring $B⊗_{A}k(p)$ is a finite $k(p)$-algebra ($k(p)$ is the residue field of $A_{p}$); |

4. | the fibres of $f_{∗}$ are finite. |

Prove that i) $⟹$ ii) $⟺$ iii) $⟹$ iv).

**Proof.**

i) $⟹$ iii): | If $f$ is finite, then also is $1⊗f:k(p)→B⊗_{A}k(p)$. Therefore $B⊗_{A}k(p)$ is a finite $k(p)$-algebra. |

ii) $⟺$ iii): | Let $C$ denote $B⊗_{A}k(p)$. The fibres of $f_{∗}$ are $f_{∗−1}(p)≃Spec(C)$. Since $f$ is of finite type, there is a surjective homomorphism $A[x_{1},…,x_{n}]→B$. Tensoring with $k(p)$ gives a surjective homomorphism $k(p)[x_{1},…,x_{n}]→C$. Therefore $C$ is a finitely generated $k(p)$-algebra, so Noetherian. By Exercise 2, $Spec(C)$ discrete if and only if $C$ is Artinian. By Exercise 3 it’s equivalent to that $C$ is a finite $k(p)$-algebra. |

iii) $⟹$ iv): | Under the same notation above, iii) implies that $C$ is Artinian. Hence $Spec(C)$ is finite. $□$ |

If $f$ is integral and the fibres of $f_{∗}$ are finite, is $f$ necessarily finite?

**Proof.**No. If $K$ is a field and $L/K$ is an algebraic extension, and $f:K→L$ the inclusion map, then $f$ is integral (algebraic), the fibres of $f_{∗}$ are finite (since $Spec(L)$ and $Spec(K)$ are one-point spaces). But $f$ is not necessarily finite, for example if $K=Q$ and $L=Qˉ $.

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Exercise

**Proof.** Let $A$ and $A_{′}$ denote the coordinate ring of $X$ and $L$ respectively. Then $A_{′}$ can be considered as the subring of $A$. By the Nullstellensatz, $X≃Max(A)$ and $L≃Max(A_{′})$. And if $f:A_{′}→A$ denotes the inclusion map, then by Chapter 5, Exercise 16, $X→L$ is the induced mapping of $f$. Since $A$ is integral over $A_{′}$, $f$ is finite. By Exercise 8.4, the fibres of $f_{∗}$ are finite.

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Exercise

**Proof.**Let $B=(A/q)_{p}$. Then primary ideals lying between $q$ and $p$ correspond to ideals of $B$. Clearly $B$ is an Artinian local ring. Then the result follows immediately from (6.8).

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