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Questions tagged [decidability]

11 questions from the last 365 days
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3 votes
1 answer
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Is decomposability of integer polynomials over the rational numbers an undecidable problem?

By a decomposition of a polynomial $F(x)$ over a field $K$ we mean writing $F(x)$ as $$ F(x)=G(H(x)) \quad(G(x), H(x) \in K[x]), $$ which is nontrivial if $\operatorname{deg} G(x)>1$ and $\...
SARTHAK GUPTA's user avatar
1 vote
0 answers
151 views

Decide if a group is abelian

Let $G = \langle X: R\rangle$ be a finitely presented group. The following problem seems very natural to me, yet I cannot find any reference for it: Decide if $G$ is abelian or not. With a reduction ...
user540172's user avatar
2 votes
0 answers
78 views

Is this variant of post correspondence problem undecidable?

The post correspondence problem, as defined by wikipedia, is undecidable. The problem is defined as follows. Let $A$ be an alphabet with at least two symbols. The input of the problem consists of ...
dips_123's user avatar
1 vote
0 answers
119 views

What is the minimal length of an undecidable sentence in those ZFC related theories?

If we measure the length of a sentence by the number of occurrences of atomic subformulas in it. So, for example in set theory written in $\sf FOL(\in)$, the length of a sentence is the number of ...
Zuhair Al-Johar's user avatar
10 votes
0 answers
274 views

Open problems in complete theories

It is well-known that every complete recursively enumerable first-order theory is decidable. Does that mean that such theories are "trivial", or are there still interesting open problems ...
user avatar
14 votes
2 answers
851 views

Examples of finitely presented subgroups of $\operatorname{GL}(n,\mathbb{Z})$ with unsolvable decision problems

$\DeclareMathOperator\GL{GL}\DeclareMathOperator\Aut{Aut}$Does there exist a finitely presented subgroup of $\GL(n,\mathbb{Z})$ for which it is known that the conjugacy problem is unsolvable (if yes, ...
Mapy Duq's user avatar
  • 143
10 votes
1 answer
418 views

Is the theory of $(\operatorname{Ord}, <)$ decidable?

Over a sufficiently strong second-order theory (such as Morse-Kelley set theory), it's possible to define the truth of all first-order formulae in the language of sets. This allows us to construct the ...
Jade Vanadium's user avatar
1 vote
1 answer
127 views

Can we effectively define a theory of all upward absolute sentences over theories of hereditarily bounded sets?

Lets' take the theory of hereditarily bounded sets[1][2]. We know that every $S_k$ where $k \in \omega$ is decidable and has as its canonical model the set ${\sf H}_k$ of all sets hereditarily of size ...
Zuhair Al-Johar's user avatar
5 votes
1 answer
197 views

Can one compute the automorphism group of a curve of genus >1?

Given a sufficiently nice perfect field $k$ and a smooth projective curve $C$ of genus $g_C>1$ over $k$, can one compute the automorphism group ${\rm Aut}(C)$? It is known that ${\rm Aut}(C)$ is ...
Arno Fehm's user avatar
  • 2,051
14 votes
3 answers
2k views

Guaranteed correct digits of elementary expressions

Let $E$ be the set of elementary expressions, defined as the smallest set of mathematical expressions such that $1 \in E$, and if $a,b \in E$ then $a + b$, $a - b$, $ab$, $a/b$, $\exp(a)$, $\log(a)$ ...
rosan98's user avatar
  • 361
8 votes
1 answer
1k views

Why don't Zeilberger and Gosper's algorithms contradict Richardson's theorem?

Richardson's theorem proves that whether an expression A is equal to zero is undecidable. A is in this case an expression, constructed from $x,e^x,\sin(x)$ and the constant function $\pi$ and $\ln(2)$ ...
Martin Clever's user avatar