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My algebra book introduces sum, intersection and product of ideals (in non-commutative rings), and then says that all three operations are commutative and associative, without proof.

I see no reasons why the product of ideals should be commutative, but I wasn't able to find a counterexample either.

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Have you tried proving that? That is, by the definitions and step by step. –  Asaf Karagila Jun 23 '11 at 18:18
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Implicit in your question is that the other operations are commutative and all are associative. This is true. Be careful that one-sided ideals can act a bit strangely. –  Jack Schmidt Jun 23 '11 at 18:40
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@Asaf: probably the OP did try to prove it, but because it's false I imagine he had some trouble... –  Pete L. Clark Jun 23 '11 at 21:27
    
@Pete: When I get stuck and go seek the advice of others, be it online or in person, I usually mention what I had tried (unless I have a very good reason not to do so). –  Asaf Karagila Jun 23 '11 at 21:31
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@Asaf: The OP said "I see no reasons why the product of ideals should be commutative", which to me at least implies that he looked for a proof. Also to me, your comment sounds like it might be suggesting "It's very straightforward to prove, just follow your nose". Since in fact the statement is false, this seems a bit misleading. –  Pete L. Clark Jun 23 '11 at 21:35
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2 Answers 2

up vote 11 down vote accepted

It is not true that ideal product is commutative in non-commutative (associative, unital) rings. An easy example is given by triangular rings (see for instance pp 17-22, especially p. 18 of Lam's First Course in Non-commutative Rings).

You can easily construct a finite counterexample with 8 elements: Let a be the ring of upper-triangular matrices; let i be the (two-sided) ideal of a consisting of strictly upper triangular matrices, and let j be the (two-sided) ideal consisting of matrices whose second row is zero. Then ij = 0, but ji = i. This works over any field, in particular the field with 2 elements.

You can verify this in GAP:

gap> a:=AlgebraWithOne(GF(2),[[[0,1],[0,0]],[[1,0],[0,0]]]*One(GF(2)));
<algebra-with-one over GF(2), with 2 generators>
gap> i:=Ideal(a,[[[0,1],[0,0]]]*One(GF(2)));
<two-sided ideal in <algebra-with-one of dimension 3 over GF(2)>, (1 generators)>
gap> j:=Ideal(a,[[[1,0],[0,0]],[[0,1],[0,0]]]*One(GF(2)));;
gap> ij:=Ideal(a,Concatenation(List(i,x ->List(j,y->x*y))));;
gap> ji:=Ideal(a,Concatenation(List(i,x ->List(j,y->y*x))));;
gap> Dimension(ij);Dimension(ji);
0
1
gap> ji = i;
true
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Your answer would have been nice even without the verification in GAP. But I installed GAP now and tried your code. Nice. –  Thomas Klimpel Jun 23 '11 at 21:21
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For another example, consider the ring $\mathbb{Z}\langle x,y\rangle$ be the ring of polynomials with integer coefficients in two noncommuting variables $x$ and $y$ (this is the same as the integral semigroup ring of the free monoid of rank $2$). It's almost like the usual polynomial ring, except that the monomials are given by words in $x$ and $y$; so that, for example, the weight two monomials are $x^2$, $xy$, $yx$, and $y^2$, all distinct; the weight three monomials are $x^3$, $x^2y$, $xyx$, $yx^2$, $xy^2$, $yxy$, $y^2x$, and $y^3$, all distinct; etc.

Let $I = (x)$, the principal ideal generated by $x$. It consists of all elements of the form $$\sum_{i=1}^n p_i(x,y)xq_i(x,y)$$ with $p_i,q_i\in\mathbb{Z}\langle x,y\rangle$, $n\geq 0$; i.e., all polynomials in which every monomial has an $x$ in it, somewhere. Likewise, $J=(y)$ consists of all polynomials in which every monomial has a $y$ in it somewhere.

Now, clearly, $xy\in IJ$. But it cannot be in $JI$, because every monomial in every nonzero element of $JI$ must have an $x$ that follows its first $y$, which is not the case for $xy$.

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Right, this is a "universal" counterexample. –  Pete L. Clark Jun 23 '11 at 21:26
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