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We know in a commutative ring, if the only ideals are trivial and the whole ring, then the ring is a field, which is proved by every ideal is contained in a maximal ideal, which is proved by Zorn's lemma.

But in non-commutative case, we can find counter-examples that the only ideals are trivial and the whole ring, but there are elements in the ring that do not have multiplicative inverse.

My question is: where does the proof for commutative rings break down in the non-commutative ring?

I think the only possibility is some ideals may not be contained in a maximal ideal, is that true?

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    $\begingroup$ The first point is not a consequence of Zorn's lemma! It's a straight forward proof from scratch. No axiom of choice needed. $\endgroup$ – Olivier Bégassat Dec 19 '14 at 19:08
  • $\begingroup$ @OlivierBégassat what do you mean exactly? $\endgroup$ – annimal Dec 19 '14 at 19:10
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    $\begingroup$ with no two-sided ideals other than $0, R$ there are examples like matrix rings over division rings, etc. see en.wikipedia.org/wiki/Artin%E2%80%93Wedderburn_theorem en.wikipedia.org/wiki/Simple_ring $\endgroup$ – yoyo Dec 19 '14 at 19:13
  • $\begingroup$ In a noncommutative ring you have to distinguish between left, right, and two-sided ideals. So you first have to take care of this. What do you even mean by a maximal ideal in a noncommutative ring? Do you take it to be left, right,... ideal? $\endgroup$ – Mister Benjamin Dover Dec 19 '14 at 19:13
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    $\begingroup$ @annimal To explain Olivier's comment, consider the ideal generated by $a$ for any $a \ne 0$. Then this ideal is not trivial, so it must be the whole ring, therefore $1$ is contained in the ideal, therefore $ab = 1$ for some $b$, therefore $a$ is invertible. $\endgroup$ – 6005 Dec 22 '14 at 8:13
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Let $R$ be a ring with unity, take $r\in R \setminus \{0\}$, and suppose that $(r) = R$ as ideals.

In the commutative case, this means that $r$ is invertible. Why? Because every element of $(r)$ is of the form $ar$, so we have $ar=1$ for some $a\in R$.

But in the non-commutative case, if we are looking at two-sided ideals, the elements of $(r)$ include all elements of the form $\sum_{i=1}^n a_i r b_i$ (and more, if we don't assume $R$ unital). So the statement $(r)=R$ is not so strong anymore.

Consider the example of a $2\times 2$ matrix ring over a field. A rank $1$ matrix is not invertible, but it generates the whole ring as an ideal, because we can produce all rank $1$ matrices by multiplying, and then easily write any matrix as a sum of rank $1$ matrices.

On the other hand, if we assume that there are no non-trivial left ideals, for example, then a similar theorem still holds: a left- or right-simple unital ring is a division ring. But again, this is because principal left and right ideals have a nice structure, while principal two-sided ideals don't.

There is still much to be said about simple rings, but this takes more complicated forms, like Artin-Wedderburn.

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For a counterexample take the first Weyl algebra $\mathbf C[p,q]$ where $pq-qp=1$, which is a subalgebra of ${\rm End}_{\mathbf C}\mathbf C[X]$. Here $p$ acts on polynomials by $p f(X)=f'(X)$ and $q$ acts by $qf(X)=Xf(X)$. It should be checked $q,p$ are not invertible, yet $pq-qp=1$ means $(p)=(q)=1$. The point is that in noncommutative cases, even principal ideals are not that simple: $(a)$ is equal to all possible sums $\sum\limits_{i=1}^n a_i ab_i$ and a relation of the form $\sum\limits_{i=1}^n a_i ab_i=1$ doesn't imply invertibility.

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    $\begingroup$ Doh: it looks like you read the title without reading the post. My question is: where does the proof for commutative rings break down in the non-commutative ring? $\endgroup$ – rschwieb Dec 20 '14 at 15:27
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    $\begingroup$ @rschwieb I have corrected the title of the post. I still think examples are valuable. $\endgroup$ – Pedro Tamaroff Dec 20 '14 at 17:40

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