Units, nilpotents, and idempotents lift from $A/\mathfrak{N}$ to $A$.

Proof: Units and nilpotents are obvious. In fact they lift to any of their representatives. For idempotents, if $x^2=x\in A/\mathfrak{N}$, then $(1-x)x=0 \in A/\mathfrak{N}$, so $(1-x)^kx^k=0\in A$ for sufficiently large $k$. And $(1-x)^k+x^k=1-x+x=1\in A/\mathfrak{N}$, so lifts to a unit $(1-x)^k+x^k$. Moreover, its inverse $u=1\in A/\mathfrak{N}$. So $(ux)^k(u(1-x))^k=0,ux^k+u(1-x)^k=1\in A$ and $ux=x,u(1-x)=1-x\in A/\mathfrak{N}$.

This can be interpreted by sheaf theory, which is to be discussed in later posts.

Prime ideals of $A_1\times...\times A_n$ is of the form $A_1\times...\times p_i\times ... \times A_n$, where $p_i$ is a prime ideal of $A_i$. What about countable products? (Profinite exists. Boolean Ring)

Proof: Multiplying by $(0,...,1,...,0)$ we see $I=I_1\times...\times I_n$. Then $(A_1\times...\times A_n)/I=A_1/I_1\times...\times A_n/I_n$. It is a domain iff $n-1$ factors are $0$ and the other is a domain. Actually the index set does not matter, as this is a product. Direct sums are of interest, and we will discuss it later. The projection onto each factor corresponds geometrically to inclusion into the disjoint union. Multiplication by $(0,...,1,...,0)$ means restrict the function to $i$-th component. The above demonstrates that ideals of a product works independently on factors, and so the subset is irreducible, iff it is restricted in one part, and irreducible there.

Let $f:A\rightarrow B$ be surjective. Then $f(\mathfrak{R}(A))\subseteq \mathfrak{R}(B)$. The inclusion may be strict. What about $\mathfrak{N}$?

If $A$ is semilocal then the above is an equality.

Proof: Since $1+f^{-1}(b) a$ is invertible, so is $1+b f(a)$ for all $b\in B$. Let $f$ be the quotient map from a domain $A$ by some principle ideal generated by a power. Then $\mathfrak{R}\supseteq \mathfrak{N}\supsetneq (0)=f(\mathfrak{R}(A))$. For non-surjective morphisms, the two thing may have no relation at all. For example, let $A$ be a local domain and $f$ the embedding into $B$, its field of fractions. Then $f(\mathfrak{R}(A))=\mathfrak{R}(A)$ is very large but $\mathfrak{R}(B)=0$. Since prime ideals always pull back, we always have $f(\mathfrak{N}(A))\subseteq \mathfrak{N}(B)$. For Jacobson radicals, the reason actually is the same since when $f$ is surjective, maximal ideals pull back. This is like saying, if a function vanishes on every closed point, then it vanishes on every closed point of a closed subset. If it vanishes on every point, then its pullback vanishes on every point. In the polynomial case, since $\mathfrak{N}=\mathfrak{R}$, this reduces to trivial intuition. Denote the kernel by $I$ and the collection of maximal ideals $\mathcal{M}$. It is equivalent to $\cap_{\mathfrak{m} \in\mathcal{M}}\mathfrak{m} + I=\cap_{\mathfrak{m}\supseteq I}\mathfrak{m}$. Passing to $A/\cap_{\mathfrak{m} \in\mathcal{M}} \mathfrak{m}\cong \prod_{\mathfrak{m} \in\mathcal{M}} A/\mathfrak{m}$, it is equivalent to $I=\cap_{\mathfrak{m}\supseteq I}\mathfrak{m}$. This is a product of fields, so by 2. above, all ideals are products of the whole field or $0$. $I$ has $0$ in the components of $\mathfrak{m}\supseteq I$ while $k_i$ otherwise, which is exactly equal to $\cap_{\mathfrak{m}\supseteq I}\mathfrak{m}$. This does not work when $|\mathcal{M}|$ is infinite, because then Chinese remainder theorem does not hold. Continuing the discussion of a., this is saying if in addition closed points are finite, then a function vanishing on a subset of them must be induced by some function vanishing on all of them. Taking the example of $\mathbb{Z}$, $p$ vanishes on the single point $\mathrm{Spec}(\mathbb{Z}/p^2\mathbb{Z})$, but cannot be induced by some elements vanishing on all of $\mathrm{Spec}\mathbb{Z}$: such elements must be $0$. This happens because we fail to let it vanish at all other primes simultaneously: infinite product does not make sense. However in $\mathrm{Spec}(\prod_{p=2,3,5,...}\mathbb{Z}/p^2\mathbb{Z})$, this holds, as we can always pull back to $(2,3,5,...)$.

An integral domain $A$ is a UFD iff both of the following are satisfied:

• Every irreducible element is prime.
• Principle ideals satisfy A.C.C.

Proof: For UFDs, it is crystal clear that these are satisfied. Conversely, we can easily split $a$ into a finite product of irreducible elements, by A.C.C.. The product is unique because irreducibles are prime. We should care about the cases when irreducible element is not prime.

Let $\{P_{\lambda}\}_{\lambda\in\Lambda}$ be a non-empty totally ordered (by inclusion) family of prime ideals. Then $\cap P_{\lambda}$ is prime. Thus for any ideal $I$, there is some minimal prime ideal containing $I$.

Proof: If $ab\in\cap P_{\lambda}$, then for all $\lambda$, either $a,b$ is in $P_{\lambda}$. So the one of the collections of primes containing $a$ and $b$ respectively is not bounded below. Thus either of $a,b$ is in the intersection. The corollary then follows from Zorn's lemma.

Let $A$ be a ring, $P_1,...,P_r$ ideals. Suppose $r-2$ of them are prime. Then if $I\subseteq \cup_i P_i$, then $\exists i:I\subseteq P_i$.

Proof: This is mysterious. Proof is not hard, but I do not know why. I will write when I know its meaning or usage.

In a ring $A$, if every ideal $I\subsetneq \mathfrak{N}$ contains a nonzero idempotent, then $\mathfrak{N}=\mathfrak{R}$.

Proof: Notice when $A$ is reduced, this amounts to say if every ideal contains a nonzero idempotent, then $\mathfrak{R}=0$: If $a\ne 0$, then $(a)$ contains a nonzero idempotent $e=ka$, with $e(1-e)=0$, so $1-ka$ is not a unit, and $a\notin R$. The general case follows by passing to $A/\mathfrak{N}$. But this is more like an awkward exercise.

A local ring contains no idempotents $\ne 0,1$.

Proof: Otherwise it would split as a direct product. By 2. above, it has at least two maximal ideals. Geometrically, a local picture cannot be a disjoint union.

The ideal $\mathfrak{Z}$ of zero-divisors is a union of prime ideals.

Proof: Non-zero-divisors form a multiplicative set: If $a,b$ are not zero-divisors, and $a b x=0$, we have $b x=0$ and $x =0$. The primes in the localization with respect to this set corresponds exactly to primes consisting of zero-divisors. Everything is clear. This is similar to the case of non-nilpotent elements is out of some prime ideals, or that localization with respect to a prime ideal is local.

The topics are from Matsumura, H. (June 30, 1989). "Chapter 1: Commutative Rings and Modules". Commutative Ring Theory. Cambridge University Press. p. 6. ISBN 978-0-521-36764-6. and Atiyah, M. F.; MacDonald, I. G. (February 21, 1994). "Chapter 1: Rings and Ideals". Introduction to Commutative Algebra. Westview Press. p. 11. ISBN 978-0-201-40751-8.

Friday, August 2, 2013