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Apr
4
answered If $F$ has characteristic $p$ and $f(x)=x^p-a\in F[x]$, then $f(x)$ is either irreducible over $F$ or $f(x)$ splits in $F$.
Apr
4
reviewed Edit Group having exactly one non trivial proper subgroup.
Apr
4
revised Group having exactly one non trivial proper subgroup.
fixed latex
Apr
4
revised Prove a map is closed
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Apr
3
revised Prove a map is closed
added 6 characters in body
Apr
3
answered Prove a map is closed
Apr
3
comment Every prime is maximal in a Jacobson ring?
@GeorgesElencwajg Prp. 1.11 in Atiyah–MacDonald (on p. 8) covers both the prime avoidance lemma and a similar statement roughly saying: »$\mathfrak p \supseteq \mathfrak a_1 ∩ … ∩ \mathfrak a_n ⇒ \mathfrak p \supseteq \mathfrak a_i$ for some $i$« with equality implying equality. I think this is what harajm is refering to.
Apr
3
comment Every prime is maximal in a Jacobson ring?
Don’t have a counterexample right now, but the prime avoidance lemma affects finite intersections only.
Apr
3
comment Give an example to show that the inequalities are strict inequalities
By the way, the wording of the (quoted) question is horrible in my opinion. All inequalites are in fact non-strict – just as stated – because for $(a_n) = (b_n) = 0$, all terms are zero, so equality holds everywhere, meaning you can’t replace »$≤$« by »$<$« without rendering the general statement false. What they should have asked for is an example for which all inequalities are strict inequalities.
Apr
3
answered F=F3[x]/(x3+2x-1) where F3 is the field with 3 elements then which of the following are correct
Apr
3
revised F=F3[x]/(x3+2x-1) where F3 is the field with 3 elements then which of the following are correct
formatting
Apr
3
revised Prove there are no other invariant subspaces
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Apr
3
comment Prove there are no other invariant subspaces
@MarianoSuárez-Alvarez True, and that argument would be that the $F[X]$-submodules of the former are exactly its ideals which are preserved under the given ring isomorphism. On the other hand, I don’t think that I hid the whole thing under «fairly easy to manage» – I hid it under «This answer … assumes familiarity with basic module theory». To me, the thing became clear once I did the translation. However, I gave an alternative argument for the final blow now. Do you think it’s okay?
Apr
3
revised Prove there are no other invariant subspaces
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Apr
2
answered Prove there are no other invariant subspaces
Apr
2
answered Use the the division algorithm and the fact that $I= (\alpha)$
Apr
2
comment prove $A$ is nilpotent if $tr(A^k)=0$ for all $k>0$
Have you searched this site for this question?
Mar
31
awarded  Nice Answer
Mar
31
comment Can flooring and ceiling thirds sum up to more than a whole?
By the way, wouldn’t a more elegant approach be to set the last column width to $x - 2·\lfloor x/3 \rfloor$ so the column widths will always add up to $x$?
Mar
31
answered Can flooring and ceiling thirds sum up to more than a whole?