0

I'm trying to prove the title statement. As usual in these cases, I'm using Zorn's lemma. This is my advance:

Consider a nonzero right ideal $I$ of $A$.

Let $G = \{ C \in P(A) : \text{C is a right ideal of A} \land \{0\} \neq C \subset I\}$. Suppose $B$ is a chain of G. Then, if $ B = \phi$, $I$ is trivially a lower bound; by the other side, if $B \neq \phi$, Consider $\bigcap B$. It is a right ideal, but I can't see that it is not the zero ideal. With that detail at hand, I can apply Zorn's lemma, and it will cast the desired minimal right ideal.

Edit: $\mathbb{F}$ is a field so that A is an algebra over a field $\mathbb{F}$. On the other hand, This is a detail that my professor let me for solving, which appear in his proof of the following result:

$\textbf{Proposition:}$Let $A$ be a finite-dimensional simple and associative algebra over a field $\mathbb{F}$, then every nonzero right ideal contains a nonzero idempotent element.

John Mars
  • 982
  • 2
    It's preferred on MSE to make the body of your question self-contained rather than just to refer to the title. As for the maths, the claim is surely wrong. E.g., in $\Bbb{R}[x]$, if $I$ is any non-zero ideal, $xI$ is an ideal properly contained in $I$, so there are no minimal non-zero ideals. – Rob Arthan Mar 17 '21 at 23:12
  • @RobArthan I'm a beginner in F-algebras. Could you give me an explanation of your counterexample? – John Mars Mar 17 '21 at 23:35
  • I've written it up in more detail as an answer. Please comment on the answer if it doesn't make sense. – Rob Arthan Mar 17 '21 at 23:43
  • I don't understand the down-votes and the vote to close on the grounds that your question lacks details or clarity: you could maybe say what $F$ is and where you got the problem from. – Rob Arthan Mar 17 '21 at 23:48
  • 1
    The assumption that $A$ is finite-dimensional is crucial. – Eric Wofsey Mar 18 '21 at 00:09
  • @EricWofsey Could you give me a hint about how to use it here? – John Mars Mar 18 '21 at 00:11
  • 1
    I've updated my answer to address the case when $A$ is finite-dimensional and the claim becomes true. – Rob Arthan Mar 18 '21 at 00:33
  • @RobArthan When a user states a question that appears wrongly stated since it is more or less obviously missing a crucial hypothesis, it is natural to cast a 'lacks details' vote and/or comments asking for clarification. It is possible to provide answers in the form of "well your question's wrong" and then later provide addenda to answer the corrected question, but that's a bit of back-and-forth that's worth avoiding in a lot of cases, I think. – rschwieb Mar 18 '21 at 13:28
  • Here is a related question: Rings in which every ideal contains a minimal ideal – rschwieb Mar 18 '21 at 13:36

1 Answers1

2

Consider the $\Bbb{R}$-algebra $A = \Bbb{R}[x]$ comprising polynomials in the variable $x$ with coefficients in the real numbers $\Bbb{R}$. (As $A$ is commutative, we don't have to worry about the distinction between right ideals and left ideals.) If $I$ is a non-zero ideal in $A$, then $xI$, the ideal comprising all polynomials $xf$, where $f \in I$, is properly contained in $I$ (because, if $f$ is a polynomial in $I$ of minimal degree, then $f \not\in xI$). So in $A$, there are no minimal non-zero ideals - any non-zero ideal contains a smaller one. Hence the statement you are trying to prove does not hold in $A$, without some additional restrictions.

Edit: the above relates to the question before the edit telling us that $A$ was expected to be finite-dimensional over $F$. If you know that $A$ is finite-dimensional, any ideal is a finite-dimensional vector space over $F$. So, if $I = I_1 \supset I_2 \supset \ldots$ is a maximal strictly descending sequence of right ideals in $A$, then it is also a strictly decreasing sequence of finite-dimensional vector spaces over $F$. Putting $d_i = \dim_F(I_i)$, the $d_i$ are a strictly decreasing sequence of natural numbers. So the sequence $d_i$ must be finite and must have a least non-zero element (given that $I \neq \{0\}$ so $d_1 > 0$). If $d_k$ is that least non-zero element, then $I_k$ is a minimal non-zero right ideal contained in $I$.

Rob Arthan
  • 51,538
  • 4
  • 53
  • 105
  • I see. The minimal degree in $I$ was the detail that I didn't deduce from your comment. – John Mars Mar 18 '21 at 00:19
  • Have you an example of a simple associative non-finite-dimensional F-algebra for which this result fails? In your first example, The $\mathbb{R}$-algebra is not simple... – John Mars Mar 18 '21 at 14:28
  • I don't understand: a simple algebra has no non-trivial ideals. – Rob Arthan Mar 18 '21 at 15:04
  • Indeed, but this result applies to one-side ideals. Then I'm looking for an example where the algebra is simple but not finite-dimensional, and the result fails. – John Mars Mar 18 '21 at 15:09
  • Good point. How about an $\Bbb{N}\times\Bbb{N}$ matrix ring, with the right ideals $I_n$ comprising the matrices whose first $n$ rows are $0$? – Rob Arthan Mar 18 '21 at 15:17
  • Do you mean $M_{\mathbb{NxN}}{(R)}$, where R is a ring? – John Mars Mar 18 '21 at 15:39
  • 1
    Yes, with $R$ a field. – Rob Arthan Mar 18 '21 at 15:44
  • @JohnMars who about the first Weyl algebra $A_1(K)$ over a field $K$, which is a simple, infinite-dimensional non-commutative $K$-algebra generated by $x$ and $y$ subject to $xy=yx+1$. Moreover, $A_1(K)$ is a domain, meaning that it has no zero divisors and it is well-known that if a domain contains a minimal left (or right) ideal, then the ring is a division ring (see for example exercise 1.19. in Lam's book "First Course in Noncommutative rings"). Since $A_1(K)$ is not a division ring it can't have a minimal right or left ideal. See https://en.wikipedia.org/wiki/Weyl_algebra for $A_1(K)$. – Christian Lomp Oct 11 '24 at 21:06