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For instance, it seems that any polynomial in tessarines/bicomplex numbers has roots in tessarines.

So, they seem to be more algebraically closed despite being not a field.

On the other hand, split-complex numbers are not algebraically closed, and their algebraic closure seems to be tessarines.

So, why only a field can be algebraically closed?

UPDATE Is there a notion of rings that are algebraically closed except for the polynomials with coefficients that are divisors of zero?

Anixx
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2 Answers2

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Take a non invertible element $a$, and consider for example the polynomial $ P (X) = aX - e$ where e is the unit element

AymaneMaaitat
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  • Thanks. But is there a notion of rings that are algebraically closed except for the polynomials with coefficients that are divisors of zero? – Anixx Apr 16 '21 at 15:29
  • @Anixx That seems to be a completely different question. Note that the coefficient $a$ in the polynomial $P(X)$ in this answer is not necessarily a zero-divisor, it's just non-invertible. – Alex Kruckman Apr 16 '21 at 15:35
  • Yes, but algebras that have non-zero-divisor non-invertible elements are intuitively not closed anyway. – Anixx Apr 16 '21 at 15:36
  • What about division algebras that are not fields (e.g. quaternions)? – Anixx Apr 16 '21 at 15:37
  • https://math.stackexchange.com/questions/1071874/is-there-an-algebraic-closure-for-the-quaternions – saulspatz Apr 16 '21 at 15:39
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    @Anixx I think you need to clarify your question (by editing the question), rather than shifting the goalposts in the comments. – Alex Kruckman Apr 16 '21 at 15:40
  • @AlexKruckman edited – Anixx Apr 16 '21 at 15:43
  • @saulspatz thanks, I think that part is answered by the link. – Anixx Apr 16 '21 at 15:48
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For a non-commutative case, we would have to consider non-commutative polynomials? With indeterminate $X$, for degree $1$ they would look like $$ a + bXc + dXe + fXg = 0 $$ and quadratics even worse with terms like $aXbXc$.

GEdgar
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