Timeline for Relative Dickson (trace) criterion for Jacobson radical?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| May 28, 2019 at 0:07 | vote | accept | darij grinberg | ||
| May 27, 2019 at 9:20 | answer | added | Aurel | timeline score: 5 | |
| May 26, 2019 at 23:05 | comment | added | darij grinberg | @Aurel: Oh, of course -- I was just being stupid. | |
| May 26, 2019 at 22:57 | comment | added | Aurel | You are welcome! About two-sidedness, maybe we don't have the same definition of $\mathbb{K}$-algebra: do you impose that the map $\mathbb{K}\to A$ has image in the center of $A$? I do, and in this case the two-sidedness follows from $J(\mathbb{K})A = AJ(\mathbb{K})$. | |
| May 26, 2019 at 22:52 | comment | added | darij grinberg | @Aurel: Ah, thank you! I don't see why $J\left(\mathbb{K}\right) A$ is a two-sided ideal of $A$, but I do see that it is a right ideal of $A$, which is enough for me. So the "$\Longleftarrow$" direction of Conjecture 2 is proven. | |
| May 26, 2019 at 21:44 | comment | added | Aurel | Let $a$ be in $J(\mathbb{K})A$. Then the matrix of left multiplication by $1-a$ is congruent to the identity $\bmod J(\mathbb{K})$, so its determinant is $1 \bmod J(\mathbb{K})$, so this determinant is invertible in $\mathbb{K}$ and hence the matrix itself, hence $1-a$, is invertible. Since $J(\mathbb{K})A$ is a two-sided ideal of $A$ this proves the inclusion. | |
| May 26, 2019 at 21:19 | comment | added | darij grinberg | @Aurel: Why is $J\left(\mathbb{K}\right) A \subseteq J\left(A\right)$ ? | |
| May 26, 2019 at 20:47 | comment | added | Aurel | We have $J(\mathbb{K})A \subset J(A)$, so we have a surjection $f \colon \overline{A}\to A/J(A)$ and and therefore $f(J(\overline{A})) = 0$, i.e. $\overline{a}\in J(\overline{A}) \Rightarrow a\in J(A)$. | |
| May 26, 2019 at 13:22 | history | asked | darij grinberg | CC BY-SA 4.0 |