Ideals of direct products of rings

TL;DR

This paper investigates when ideals in direct products of rings can be decomposed into ideals of factors, providing necessary and sufficient conditions and characterizations for commutative rings and their varieties.

Contribution

It extends known results from commutative unitary rings to general commutative rings, offering new criteria and characterizations for ideal decomposability in various varieties.

Findings

Ideal decomposability does not always hold in general commutative rings.

Necessary and sufficient conditions for ideal decomposability are established.

A Mal'cev type condition characterizes decomposability in varieties of commutative rings.

Abstract

It is known that an ideal of a direct product of commutative unitary rings is directly decomposable into ideals of the corresponding factors. We show that this does not hold in general for commutative rings and we find necessary and sufficient conditions for direct decomposability of ideals. For varieties of commutative rings we derive a Mal'cev type condition characterizing direct decomposability of ideals and we determine explicitly all varieties satisfying this condition.

Equations

$$\begin{array}[]{c|cccc}+&0&a&b&c\\ \hline\cr 0&0&a&b&c\\ a&a&0&c&b\\ b&b&c&0&a\\ c&c&b&a&0\end{array}$$

$$(R_1\times \{0\})\cap ((\{0\}\times R_2)+I)$$

$$=(R_1\times \{0\})\cap (I_1\times R_2)=I_1\times \{0\}\subseteq I,$$

$$((R_1\times \{0\})+I)\cap (\{0\}\times R_2)$$

$$=(R_1\times I_2)\cap (\{0\}\times R_2)=\{0\}\times I_2\subseteq I.$$

$$(a,0)$$

$$(0,b)$$

$$((R_1\times \{0\})+I)\cap ((\{0\}\times R_2)+I)=$$

$$=((R_1\times \{0\})+(I_1\times I_2))\cap ((\{0\}\times R_2)+(I_1\times I_2))=(R_1\times I_2)\cap (I_1\times R_2)=$$

$$=I_1\times I_2=I.$$

$$I\vee J$$

$$I\wedge J$$

$$(ax,by)=(a,b)(dfx+cey)+((ac)(e(x-y)),(bd)(-f(x-y)))\in I(a,b)$$

$$I(2,2)=(2,2)\mathbb{Z}+(4\mathbb{Z}\times 4\mathbb{Z})=(4\mathbb{Z}\times 4\mathbb{Z})\cup ((2+4\mathbb{Z})\times (2+4\mathbb{Z}))$$

$$I(x,x)$$

$$=\{(nx+xr(x),nx+xs(x))\mid n\in \mathbb{Z}\text{ and }r(x),s(x)\in F(x)\}$$

$$(c,d)=(c,e)+(a,b)(0,b^{-1}(d-e))\in I$$

Full text

AMS Subject Classification: 13A15, 16R40, 08B05

Keywords: Commutative ring, ring ideal, direct product, directly decomposable ideal, Mal’cev condition, variety of commutative rings

Recall that an ideal of a ring $\mathbf{R}=(R,+,\cdot)$ is a non-empty subset $I$ of $R$ such that if $a,b\in I$ then $a-b\in I$ and $ar,ra\in I$ for every $r\in R$. For other concepts used here the reader is referred to any monograph on rings, see e.g. [2] and [5]. Ideals play a crucial role in the theory of rings since the kernels of homomorphisms are ideals and rings can be factorized by means of ideals.

It is an elementary fact that in rings, ideals and congruences are in a one-to-one correspondence. Hence, if $\mathbf{R}=(R,+,\cdot)$ is a ring then the ideal lattice $\operatorname{\mathbf{Id}}\mathbf{R}:=(\operatorname{Id}\mathbf{R},\subseteq)$ and the congruence lattice $\operatorname{\mathbf{Con}}\mathbf{R}:=(\operatorname{Con}\mathbf{R},\subseteq)$ are isomorphic. Hence $\operatorname{\mathbf{Id}}\mathbf{R}$ is a modular bounded lattice with the least element $\{0\}$ and the greatest element $R$ where the supremum and infimum of ideals $I_{1},I_{2}$ are given by $I_{1}\vee I_{2}=I_{1}+I_{2}$ and $I_{1}\wedge I_{2}=I_{1}\cap I_{2}$, respectively.

Having two rings $\mathbf{R}_{1}=(R_{1},+,\cdot)$ and $\mathbf{R}_{2}=(R_{2},+,\cdot)$, it is elementary that for $I_{1}\in\operatorname{Id}\mathbf{R}_{1}$ and $I_{2}\in\operatorname{Id}\mathbf{R}_{2}$ we have $I_{1}\times I_{2}\in\operatorname{Id}(\mathbf{R}_{1}\times\mathbf{R}_{2})$. On the other hand, if $I\in\operatorname{Id}(\mathbf{R}_{1}\times\mathbf{R}_{2})$ then there need not exist $I_{1}\in\operatorname{Id}\mathbf{R}_{1}$ and $I_{2}\in\operatorname{Id}\mathbf{R}_{2}$ with $I_{1}\times I_{2}=I$. If such ideals $I_{1},I_{2}$ do not exist, then $I$ is called skew. Otherwise, $I$ will be called directly decomposable. For commutative rings $\mathbf{R}_{1},\mathbf{R}_{2}$ we will derive conditions under which $\mathbf{R}_{1}\times\mathbf{R}_{2}$ has no skew ideals.

We say that a direct product of finitely many rings has directly decomposable ideals if every ideal of this product is a direct product of ideals of the corresponding factors.

For sets $M_{1},M_{2}$ and $i\in\{1,2\}$ let $\pi_{i}$ denote the $i$-th projection from $M_{1}\times M_{2}$ onto $M_{i}$.

It is easy to see that an ideal $I$ of $\mathbf{R}_{1}\times\mathbf{R}_{2}$ is directly decomposable if and only if $\pi_{1}(I)\times\pi_{2}(I)=I$ which is equivalent to $\pi_{1}(I)\times\pi_{2}(I)\subseteq I$.

Direct decomposability of ideals in commutative rings was used by the first and the third author in their study of complementation in ideal lattices, see [3].

Let $\mathbf{K}_{4}=(\{0,a,b,c\},+)$ denote the four-element Kleinian group whose operation table for $+$ looks as follows:

[TABLE]

We show an example of rings whose direct product contains a skew ideal.

In what follows, we derive necessary and sufficient conditions under which an ideal of the direct product of rings is directly decomposable. For this purpose, we borrow the method developed by Fraser and Horn ([4]) for congruences.

1. (i)

$I$* is directly decomposable.* 2. (ii)

$(R_{1}\times\{0\})\cap((\{0\}\times R_{2})+I)\subseteq I$* and** *

$((R_{1}\times\{0\})+I)\cap(\{0\}\times R_{2})\subseteq I$. 3. (iii)

If $(a,b)\in I$ then $(a,0),(0,b)\in I$. 4. (iv)

$((R_{1}\times\{0\})+I)\cap((\{0\}\times R_{2})+I)=I$.

If $I=I_{1}\times I_{2}$ then

[TABLE]

(ii) $\Rightarrow$ (iii):

If $(a,b)\in I$ then

[TABLE]

(iii) $\Rightarrow$ (i):

If $(a,b)\in\pi_{1}(I)\times\pi_{2}(I)$ then there exists some $(c,d)\in R_{1}\times R_{2}$ with $(a,d),(c,b)\in I$, hence $(a,0),(0,b)\in I$ which shows $(a,b)=(a,0)+(0,b)\in I$.

(i) $\Rightarrow$ (iv):

If $I=I_{1}\times I_{2}$ then

[TABLE]

(iv) $\Rightarrow$ (ii):

This follows immediately. ∎

The following result is already known but it follows easily from the equivalence of (i) and (iii) in Theorem 2.

[TABLE]

Hence, join and meet in $\operatorname{\mathbf{Id}}(\mathbf{R}_{1}\times\mathbf{R}_{2})$ are computed “component-wise” showing distributivity of $\operatorname{\mathbf{Id}}(\mathbf{R}_{1}\times\mathbf{R}_{2})$. ∎

Another application of Theorem 2 is the following example:

Denote by $\mathbb{Z}$ the ring of integers. As usually, for $a\in\mathbb{Z}$ put $a\mathbb{Z}:=\{ax\mid x\in\mathbb{Z}\}$. Of course, for each $a\in\mathbb{Z}$, $a\mathbb{Z}$ is a commutative ring which is unitary only in case $a\in\{-1,0,1\}$. The next theorem shows that, in general, the rings $a\mathbb{Z}\times b\mathbb{Z}$ contain skew ideals. Hence, the rather exotic ring $\mathbf{K}\times\mathbf{K}$ from Example 1 is not the only commutative ring possessing skew ideals.

[TABLE]

for all $x,y\in\mathbb{Z}$ proving direct decomposability of $I(a,b)$. Finally, assume $\gcd(c,d)\neq 1$. Suppose $I(a,b)$ to be directly decomposable. Then $(a,0)\in I(a,b)$ and hence there exist $g,h,i\in\mathbb{Z}$ with $(a,0)=(ag+ach,bg+bdi)$. From this we conclude $g=-di$ and hence $ch-di=1$ whence $\gcd(c,d)=1$, a contradiction. Hence $I(a,b)$ is not directly decomposable in this case. ∎

Although the rings of the form $c\mathbb{Z}$ are rings of integers, Theorem 6 yields the following result.

[TABLE]

of $2\mathbb{Z}\times 2\mathbb{Z}$ generated by $(2,2)$ is skew since $(0,0),(2,2)\in I(2,2)$, but $(0,2)\notin I(2,2)$. This is in accordance with Theorem 6.

Using Theorem 2 we can generalize the situation described in Example 5. Namely, we can derive a Mal’cev type condition characterizing varieties of commutative rings whose ideals are directly decomposable.

[TABLE]

of $\mathbf{F}(x)\times\mathbf{F}(x)$ generated by $(x,x)$. Since $I(x,x)$ is directly decomposable and $(0,0),$ $(x,x)\in I(x,x)$, we have $(x,0)\in I(x,x)$. Hence there exist some $n\in\mathbb{Z}$ and $r(x),s(x)\in F(x)$ with $(nx+xr(x),nx+xs(x))=(x,0)$ which implies $xt(x)=x$ with $t(x):=r(x)-s(x)$. Conversely, assume there exists a unary term $t$ satisfying $xt(x)\approx x$. Let $\mathbf{R}_{1}=(R_{1},+,\cdot),\mathbf{R}_{2}=(R_{2},+,\cdot)\in\mathcal{V}$ and $I\in\operatorname{Id}(\mathbf{R}_{1}\times\mathbf{R}_{2})$ and assume $(a,b)\in I$. Then $(a,0)=(a,b)(t(a),0)\in I$ and $(0,b)=(a,b)(0,t(b))\in I$. According to (iii) of Theorem 2, $I$ is directly decomposable. ∎

Using Theorem 9, we can explicitly describe all varieties of commutative rings having directly decomposable ideals.

However, we can also consider classes of commutative rings which do not form a variety. In this case we cannot apply Theorem 9 in order to prove that rings belonging to such a class have directly decomposable ideals. A typical example is the following:

[TABLE]

showing $I=J\times F$. Hence, $\mathbf{R}\times\mathbf{F}$ has directly decomposable ideals. By induction we obtain that $\mathbf{R}\times\mathbf{F}_{1}\times\cdots\times\mathbf{F}_{n}$ has directly decomposable ideals if $n\geq 1$ and $\mathbf{F}_{1},\ldots,\mathbf{F}_{n}$ are fields.

In particular, we can consider the case where $\mathbf{R}$ denotes the commutative ring $\mathbf{K}$ from Example 1. Then $\mathbf{K}\times\mathbf{F}$ has directly decomposable ideals despite the fact that $\mathbf{K}\times\mathbf{K}$ does not have this property. Similarly, if $c$ and $d$ are integers satisfying $\gcd(c,d)\neq 1$ then $(c\mathbb{Z}\times d\mathbb{Z})\times\mathbf{F}$, $c\mathbb{Z}\times\mathbf{F}$ and $d\mathbb{Z}\times\mathbf{F}$ have directly decomposable ideals though $c\mathbb{Z}\times d\mathbb{Z}$ does not have this property.

Acknowledgement. The authors thank the referee for his valuable suggestions which increased the quality of the paper.