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Is there an algorithm to determine if we have been given a ring $A$ and its ideal $I$, whether or not $I$ is a maximal ideal of $A$? I found that sometimes proving that ideal is maximal might be tricky, like in here and a general algorithm is missing for example in Sage.

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    $\begingroup$ In what form would you produce the ring and the ideal as input to such an algorithm? $\endgroup$ – Hagen von Eitzen Dec 11 '14 at 16:51
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    $\begingroup$ It depends on how the ideal is given... For instance, suppose we consider the ideal $I = \{ f \in \mathbb Q[x] \mid f(\zeta_3) = 0 \} \unlhd \mathbb Q[x]$, where $\zeta_3 = \sum_n n^{-3}$ is Apéry's constant. $\endgroup$ – Myself Dec 11 '14 at 16:53
  • $\begingroup$ Oh well. I think my question was stupid as there are uncountable many rings but the tape in Turing machine has countable length. So never mind for the general case. But it would be nice to see at least some general method to prove if some ideal is maximal or not. The form if input is not important at the moment, as I would like to learn first to just find out a way to prove that ideals are maximal in general case. $\endgroup$ – algebrastudent Dec 11 '14 at 16:56
  • $\begingroup$ I'm pretty sure there is an example where whether or not an ideal is maximal is undecidable. $\endgroup$ – Matt Samuel Dec 11 '14 at 16:58
  • $\begingroup$ well, this the beauty of algebra. Different tricks and approaches for different problems. Algorithms kills all the beauty. $\endgroup$ – Bhaskar Vashishth Dec 11 '14 at 17:01
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Let $R$ be the $\mathbb Q$ vector space generated by all ordinals of cardinality less than $c$ together with the ordinal $c$. Let multiplication be given by intersection. Let $I$ be the ideal generated by all countable ordinals. Since the continuum hypothesis is undecidable, it is undecidable whether $I$ is a maximal ideal.

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  • $\begingroup$ Lol! Did you just come up with this or do you know who did? $\endgroup$ – Myself Dec 11 '14 at 17:08
  • $\begingroup$ I just came up with it. $\endgroup$ – Matt Samuel Dec 11 '14 at 17:08
  • $\begingroup$ Wow. This is the kind of answer that makes me wish I could favorite answers. $\endgroup$ – Alex Wertheim Dec 11 '14 at 17:09
  • $\begingroup$ Actually what exactly is abelianized ordinal sum? $\endgroup$ – Myself Dec 11 '14 at 17:12
  • $\begingroup$ Take the semigroup given by ordinal sum and abelianize it. $\endgroup$ – Matt Samuel Dec 11 '14 at 17:13

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