Proving that $N(R)\times N(S)= N(R\times S)$

Question

Let $R,S$ be rings with $1_R,1_S$ respectively. Thanks to this answer, I realised that I probably make a mistake in proving that $$N(R) \times N(S)=N(R\times S),$$ where $N(R)$ is the nilradical (which is defined to be the sum of all nilpotent ideals of $R$).

But lets take it step by step:

For the $``\supseteq"$: We know that the ideals of $R\times S$ have the form $I\times J$, where $I$ is an ideal of $R$ and $J$ is an ideal of $S$. Then, by the definition of the nilradical, we have that $N(R)$ has the form $$N(R\times S)=(I_1\times J_1)+(I_2\times J_2) + \dotsb = (I_1+I_2+\dotsb)\times (J_1+J_2+\dotsb),$$ where $I_i\times J_i$ are nilpotent ideals of $R\times S$ respectively. Then, $I_i\times J_i$ is nilpotent $\iff (I_i\times J_i)^t=0$, for some $t\in \Bbb Z^+ \iff I_i^t\times J_i^t=0$. Thus, $I_i,J_i$ are nilpotent ideals. Thus, $I_1+I_2+\dotsb \subseteq N(R)$ and $J_1+J_2+\dotsb \subseteq N(S)$ and we are done.

For the $``\subseteq"$: I have written that if $N(R)=I_1+I_2+\dotsb$ and $N(S)=J_1+J_2+\dotsb$, then $$N(R)\times N(S)=\underbrace{(I_1+I_2+\dotsb)}_{I}\times \underbrace{(J_1+J_2+\dotsb)}_{J},$$ and then $I,J$ are nilpotent ideals, as a sum of nilpotents, which is not correct. But to finish my thought, if $I,J$ are nilpotents with $ I^n=J^m=0$, then $(I\times J)^{\max(m,n)}=I^{\max(m,n)}\times J^{\max(m,n)}=0$, so $I\times J$ is nilpotent and thus $N(R)\times N(S) \subseteq N(R\times S)$.

So:

1. Is the $``\supseteq"$ part correct?
2. Any help with the $``\subseteq"$? Does this syllogism help?

Answer 1

Let me try to provide you with a full answer. First, in order to fix our terminology let us agree to ad-hoc call these radicals you are interested in nilpotency radicals (this being a term I just coined up for usage here; in general ring theory there are lower and upper nilradicals, together with some other interesting radicals, so one should strive to be specific about the objects handled). Let us also agree to write $\mathrm{Np}(A)$ for the nilpotency radical of given ring $A$.

You make a statement that could be generalized to arbitrary finite direct products of rings, but it suffices to treat the elementary case of just two rings:

Proposition. For any two rings $A, B$ one has $\mathrm{Np}(A \times B)=\mathrm{Np}(A) \times \mathrm{Np}(B)$.

Proof: The inclusion from left to right is the easier one, as you noticed: consider a nilpotent bilateral (I do not like the English terminology ''two-sided'' so I will instead adopt a more latin one) ideal $J \subseteq A \times B$. By the general structure theorem describing ideals in finite direct products, we know there must exist bilateral ideals $I \subseteq A, I' \subseteq B$ such that $J=I \times I'$.

Since in general it is the case that $$(I \times J).(I' \times J')=(I.I') \times (J.J')$$ for any pairs of ideals $I, I' \subseteq A, J, J' \subseteq B$ (where by $I.I'$ I am expressly referring to the ideal-product of the two, calculated in the multiplicative monoid of bilateral ideals, rather then their subset-product $II'=\{xy\}_{x \in I\\ y \in I'}$ calculated in the multiplicative monoid of subsets of $A$) and thus consequently that

$$(I \times J)^{\underline{n}}=I^{\underline{n}} \times J^{\underline{n}} \tag{npot}$$

for any ideals $I \subseteq A, J \subseteq B$ and any $n \in \mathbb{N}$ (the underlining of the exponent $n$ is syntax I expressly use to distinguish the calculation of $n$-th powers in the multiplicative monoid of bilateral ideals from mere cartesian products), the fact that $J$ considered in the above paragraph is nilpotent tells you that $I, I'$ must also be, to the effect that $I \subseteq \mathrm{Np}(A), I' \subseteq \mathrm{Np}(B), J \subseteq \mathrm{Np}(A) \times \mathrm{Np}(B)$. This establishes the inclusion ''$\subseteq$''.

As for the reverse one, let us now consider an arbitrary element $u=(x,y) \in \mathrm{Np}(A) \times \mathrm{Np}(B)$. It is essential to understand the following ''finitary'' characterization of the nilpotency radical. Allow me first of all to write $\mathscr{Np}(A)$ for the set of all nilpotent bilateral ideals of a given ring $A$; for arbitary set $M$, let us agree to denote by $\mathscr{F}(M)$ the set of all finite subsets of $M$. With these conventions in place we have the description:

$$\mathrm{Np}(A)=\bigcup_{\mathscr{H} \in \mathscr{F}(\mathscr{Np}(A))} \sum_{I \in \mathscr{H}}I \tag{undir}$$

in other words the nilpotency radical is the union over the (upward-directed) family of all finite sums of nilpotent bilateral ideals of $A$.

With this in mind, we realize that there must exist finite subsets $\mathscr{K} \subseteq \mathscr{Np}(A), \mathscr{H} \subseteq \mathscr{Np}(B)$ such that $$x \in \sum_{I \in \mathscr{K}}I\\ y \in \sum_{J \in \mathscr{H}}J$$

The relation (npot) mentioned above ensures that $\mathscr{Np}(A \times B)=\mathscr{Np}(A) \times \mathscr{Np}(B)$. Thus, it follows easily that $$K=\sum_{I \in \mathscr{K}}(I \times \{0_B\})+\sum_{J \in \mathscr{H}}(\{0_A\} \times J) \subseteq \mathrm{Np}(A \times B)$$

since all the individual products $I \times \{0_B\}, \{0_A\} \times J$ are nilpotent ideals in the direct product; furthemore, since we also have

$$K=\left(\sum_{I \in \mathscr{K}}I\right) \times \{0_B\}+\{0_A\} \times \left(\sum_{J \in \mathscr{H}}J\right)$$

it is clear that $u=(x, 0_B)+(0_A, y) \in K \subseteq \mathrm{Np}(A \times B)$, settling the problem of the reverse inclusion. $\Box$

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Justification of relation (undir)

For given ring $A$ and arbitrary subset $X \subseteq A$ let us write $(X)_{\mathrm{b}}$ for the bilateral ideal generated by $X$ and let $\mathscr{Id}_{\mathrm{b}}(A)$ denote the set of all bilateral ideals of $A$.

Let us also recall that for any finite subset $\mathscr{M} \subseteq \mathscr{Id}_{\mathrm{b}}(A)$ (equivalently, for $\mathscr{M} \in \mathscr{F}(\mathscr{Id}_{\mathrm{b}}(A))$) one has the equivalent descriptions:

$$\left(\bigcup \mathscr{M}\right)_{\mathrm{b}}=\sum_{I \in \mathscr{M}}I \tag{*}$$

where the right-hand side term in the above is understood as a finite sum calculated in the commutative monoid of all subsets of $A$, namely $(\mathscr{P}(A), +)$, the addition $+: \mathscr{P}(A) \times \mathscr{P}(A) \to \mathscr{P}(A)$ being given by $X+Y=\{x+y\}_{x \in X\\ y \in Y}$ (the canonical extension of the addition on $A$ to the powerset $\mathscr{P}(A)$).

By definition we have $$\mathrm{Np}(A)=\left(\bigcup \mathscr{Np}(A)\right)_{\mathrm{b}}$$

Let us denote the right-hand side term in relation (undir) by $R$; by the previous observation (*) it follows that for any $\mathscr{H} \in \mathscr{F}(\mathscr{Np}(A)) \subseteq \mathscr{F}(\mathscr{Id}_{\mathrm{b}}(A))$ (in general $M \subseteq N$ immediately entails $\mathscr{F}(M) \subseteq \mathscr{F}(N)$) we have:

$$\sum_{I \in \mathscr{H}}I=\left(\bigcup \mathscr{H}\right)_{\mathrm{b}} \subseteq \left(\bigcup \mathscr{Np}(A)\right)_{\mathrm{b}}=\mathrm{Np}(A)$$

so clearly $$R \subseteq \mathrm{Np}(A) \tag{1}$$

We want to argue that at this point it suffices to show that $R$ defined precisely as the right-hand side of (undir) is itself a bilateral ideal. Let us first postpone the task of proving this and notice that once this fact is established, since it is obvious that for any $I \in \mathscr{Np}(A)$ one automatically has $\{I\} \in \mathscr{F}(\mathscr{Np}(A))$ and trivially $I=(I)_{\mathrm{b}}=(\bigcup \{I\})_{\mathrm{b}}$ so $I \subseteq R$. Since $R$ thus includes all nilpotent ideals, it follows that it also includes their union $$R \supseteq \bigcup \mathscr{Np}(A)$$

and as $R$ is established to be a bilateral ideal it follows that

$$R \supseteq \left(\bigcup \mathscr{Np}(A)\right)_{\mathrm{b}}=\mathrm{Np}(A) \tag{2}$$

From relations (1) and (2), by virtue of the axiom of extensionality (and its immediate corollary) we can infer that $R=\mathrm{Np}(A)$.

Now on to the reason why $R$ is a bilateral ideal. In order to prove this, I think it to your benefit (when dealing with several other areas of mathematics even apart from ''pure algebra'', these are considerations fundamental to many areas of order theory, topology and analysis) to introduce some essential notions.

An ordered set is a pair $(A, R)$ where $R$ is an order relation on $A$; we will write equivalently:

$$(x,y) \in R \Leftrightarrow xRy \Leftrightarrow x \leqslant_R y$$

in the case of a fixed order relation $R$ (the first two manners of notation are more generally applied to any binary relation; the symbol $\leqslant_R$ refers strictly to the case of order relations).

An upward directed set is an ordered set $(A, R)$ such that for any two elements $x, y \in A$ there must exist a third one $z \in A$ such that $x, y \leqslant_R z$. The typical example for such objects is precisely the set $\mathscr{F}(A)$ of all finite subsets of a given set $A$, ordered by inclusion.

We proceed with a series of lemmas:

Lemma 1. Consider a ring $A$ and nonempty subset $\varnothing \neq \mathscr{M} \subseteq \mathscr{Id}_{\mathrm{b}}(A)$ which is upward directed when ordered by inclusion. Then the union $$\bigcup \mathscr{M} \in \mathscr{Id}_{\mathrm{b}}(A)$$

is itself a bilateral ideal (upward directed unions of ideals are ideals).

Proof: Since $\mathscr{M}$ is nonempty (and any ideal is a nonempty subset) the union $$J:=\bigcup \mathscr{M}$$ is nonempty. Let $x, y \in J$ be arbitrary: there will exist $I, I' \in \mathscr{M}$ such that $x \in I, y \in I'$; since $\mathscr{M}$ is upward directed under inclusion, there must exist $I'' \in \mathscr{M}$ such that $I, I' \subseteq I''$ and hence $x, y \in I''$; $I''$ being in particular an additive subgroup of $A$ it follows that $x-y \in I'' \subseteq J$. So far we have thus ascertained that $J$ is itself an additive subgroup of $A$.

As to $J$ being closed with respect to multiplication on the left, we have

$$AJ=A\left(\bigcup_{I \in \mathscr{M}}I\right)=\bigcup_{I \in \mathscr{M}}(AI) \subseteq \bigcup_{I \in \mathscr{M}}I=J$$

since all members $I \in \mathscr{M}$ are in particular left ideals which amongst others means that $AI \subseteq I$. Closure with respect to multiplication on the right is treated analogously, and we conclude that $J \in \mathscr{Id}_{\mathrm{b}}(A)$. $\Box$

Before presenting the next lemma, a few more preliminary notions. Let $X$ be a family of objects (in the axiomatic system I prefer and which I use to formalise mathematics, ultimately any ''object'' is a set, the two terms being synonymous) indexed by $I$ and let $R$ be an order relation on $I$; we say that $X$ is increasing with respect to $R$ if

$$(\forall i, j)(i \leqslant_R j \Rightarrow X_i \subseteq X_j)$$

With this in place we state

Lemma 2. Let $(I, R)$ be an upward directed set and $X$ be a family indexed by $I$ increasing with respect to $R$. Then the set of all its terms $\{X_i\}_{i \in I}$ is itself upward directed when ordered by inclusion.

Proof: Let us abbreviate $$\mathscr{A}:=\{X_i\}_{i \in I}$$ Conisdering two arbitrary elements $M, N \in \mathscr{A}$ there must exist $i, j \in I$ such that $M=X_i, N=X_j$; by upward directedness there must also exist $k \in I$ with $i, j \leqslant_R k$ and since the family $X$ is increasing we have $M, N \subseteq X_k \in \mathscr{A}$. $\Box$

In our particular case, we can apply lemmas 1 and 2 by realising that:

• $\mathscr{F}(\mathscr{Np}(A))$ is indeed upward directed under inclusion (this we already pointed out)
• the family defined by $$\left(\left(\bigcup \mathscr{H}\right)_{\mathrm{b}}\right)_{\mathscr{H} \in \mathscr{F}(\mathscr{Np}(A))}$$ is indeed increasing with respect to inclusion
• $R$ is by definition the union of the nonempty (because $\varnothing \in \mathscr{F}(M)$ is always nonempty for any set $M$) collection of ideals $$\left\{\left(\bigcup \mathscr{H}\right)_{\mathrm{b}}\right\}_{\mathscr{H} \in \mathscr{F}(\mathscr{Np}(A))}$$

collection which is thus upward directed under inclusion ensuring that the union $R$ itself is an ideal.