Question

$p$ is an odd prime and $a\in\mathbb{Z}$ with $\gcd (a,p)=1$ implies $\exists x\in\mathbb{Z}:x^2\equiv a\bmod p$ iff $a^\frac{p-1}{2}\equiv 1\mod p$

score 556 · Answer 1 · 2026-03-25 23:24:24

556

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$p$ is an odd prime and $a\in\mathbb{Z}$ with $\gcd (a,p)=1$ implies $\exists x\in\mathbb{Z}:x^2\equiv a\bmod p$ iff $a^\frac{p-1}{2}\equiv 1\mod p$

Published on 25 Mar 2026 - 23:24

score 175 · Answer 2 · 2026-03-28 08:49:15

175

Views

A finite $\mathbb{Z}$-module whose submodules are totally ordered by inclusion.

Published on 28 Mar 2026 - 8:49

score 1k · Answer 3 · 2014-06-09 15:26:51

1k

Views

The Chinese remainder theorem for modules.

Published on 09 Jun 2014 - 15:26

score 154 · Answer 4 · 2014-06-10 10:22:22

154

Views

Show that $\operatorname{Hom}_R(E, F)$ is an $R$-module in a natural way.

Published on 10 Jun 2014 - 10:22

score 66 · Answer 5 · 2026-03-28 08:32:17

66

Views

An inverse limit of a certain inverse system

Published on 28 Mar 2026 - 8:32

score 769 · Answer 6 · 2026-03-25 03:17:12

769

Views

Alternative Proof why algebraic groups are smooth

Published on 25 Mar 2026 - 3:17

score 1.2k · Answer 7 · 2014-06-11 02:56:17

1.2k

Views

Jacobson radical of a finitely generated $\mathbb Z$-algebra is nilpotent

Published on 11 Jun 2014 - 2:56

score 83 · Answer 8 · 2014-06-11 15:31:16

83

Views

"Finitely generated as an $R$-module"

Published on 11 Jun 2014 - 15:31

score 220 · Answer 9 · 2014-06-12 15:03:21

How can one show that ${\rm Hom}\Bigl(\prod\limits_{i\geqslant 1} \Bbb Z,\Bbb Z\Bigr)$ has cardinality less than $2^{\mathfrak c}$?

score 369 · Answer 10 · 2014-06-13 00:27:00

369

Views

Minimal set of generators VS Length of a module.

Published on 13 Jun 2014 - 0:27

score 561 · Answer 11 · 2014-06-13 01:53:08

561

Views

Free module finite generated with a submodule not finite generated.

Published on 13 Jun 2014 - 1:53

score 150 · Answer 12 · 2026-03-25 21:45:00

150

Views

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Published on 25 Mar 2026 - 21:45

score 237 · Answer 13 · 2014-06-13 16:59:32

237

Views

free $\mathbb Z$-module of rank $2$

Published on 13 Jun 2014 - 16:59

score 53 · Answer 14 · 2014-06-13 18:39:33

53

Views

Show that if $M$ is a R-module free, for all $r \in R$ and $m \in M$ such that $r \neq 0$ e $m \neq 0$ we have $rm \neq 0$.

Published on 13 Jun 2014 - 18:39

score 82 · Answer 15 · 2026-03-25 15:48:38

82

Views

Dimension of a certain vector space

Published on 25 Mar 2026 - 15:48

List Question

$p$ is an odd prime and $a\in\mathbb{Z}$ with $\gcd (a,p)=1$ implies $\exists x\in\mathbb{Z}:x^2\equiv a\bmod p$ iff $a^\frac{p-1}{2}\equiv 1\mod p$

A finite $\mathbb{Z}$-module whose submodules are totally ordered by inclusion.

The Chinese remainder theorem for modules.

Show that $\operatorname{Hom}_R(E, F)$ is an $R$-module in a natural way.

An inverse limit of a certain inverse system

Alternative Proof why algebraic groups are smooth

Jacobson radical of a finitely generated $\mathbb Z$-algebra is nilpotent

"Finitely generated as an $R$-module"

How can one show that ${\rm Hom}\Bigl(\prod\limits_{i\geqslant 1} \Bbb Z,\Bbb Z\Bigr)$ has cardinality less than $2^{\mathfrak c}$?

Minimal set of generators VS Length of a module.

Free module finite generated with a submodule not finite generated.

Associated primes and Noetherian condition

free $\mathbb Z$-module of rank $2$

Show that if $M$ is a R-module free, for all $r \in R$ and $m \in M$ such that $r \neq 0$ e $m \neq 0$ we have $rm \neq 0$.

Dimension of a certain vector space

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