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I'm after a ring R which is finite and associative, but with non-transitive ideals. That is, there exists some J an ideal of R, and I an ideal of J, such that I is not an ideal of R... Plenty of examples of non-transitivity in infinite settings, but haven't found one that is finite...

Pearson
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  • Related (not duplicate): https://math.stackexchange.com/questions/144022/subideals-of-an-ideal – Watson Dec 22 '16 at 08:42

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A simple example is $R=\mathbb{F}_4[x]/(x^2)$, $J=(x)$, and $I=\{0,x\}$. You can get many similar examples by taking $J$ to be any ideal in a finite ring such that $J^2=0$ and $I$ to be an additive subgroup of $J$ that is not closed under multiplication by general elements of $R$.

Eric Wofsey
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  • Thank you! I may be missing something completely obvious so please forgive me, but it seems here that I = {0,x} cannot be an ideal of J since I is not even a subring? – Pearson Dec 22 '16 at 06:41
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    The product of any two elements of $J$ is $0$, so any additive subgroup of $J$ is an ideal. – Eric Wofsey Dec 22 '16 at 06:42
  • Ah yes, I do remember this result now. However, sorry to be painfully annoying... but I can't get my head around why {0, x} is an additive 'subgroup'... it doesn't appear to be closed under addition since 2x and 3x should be included? – Pearson Dec 22 '16 at 06:49
  • $2x=0$, since you're taking polynomials over $\mathbb{F}_4$ and $2=0$ in $\mathbb{F}_4$. – Eric Wofsey Dec 22 '16 at 06:49
  • I thought this would mean 4x = 0, since this field has 4 elements? Also, if that is the case, how is I different to J? J = (x) would then only have elements 0 and x too? – Pearson Dec 22 '16 at 06:53
  • The field $\mathbb{F}_4$ can be defined explicitly as $\mathbb{F}_2[y]/(y^2+y+1)$. So $0=2$ in $\mathbb{F}_4$, and you can call its four elements $0,1,y,$ and $y+1$. The ideal $J$ contains $yx$ and $(y+1)x$ in addition to $x$ and $0$. – Eric Wofsey Dec 22 '16 at 06:57
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Let $R$ be the ring of upper triangular $3\times 3$ matrices over a finite field, i.e. $$ R=\left\{\begin{pmatrix}a&b&c\\d&e&f\\0&0&g\end{pmatrix}: a,b,c,d,e,f,g\in F\right\}. $$ Let $J$ be the ideal $$ J=\left\{\begin{pmatrix}0&0&a\\0&0&b\\0&0&0\end{pmatrix}: a,b\in F\right\} $$ and let $I$ be $$ I=\left\{\begin{pmatrix}0&0&a\\0&0&0\\0&0&0\end{pmatrix}: a\in F\right\} $$

Nothing special
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TomGrubb
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