A primality test for submodules using Grobner basis

In this paper, an algorithm is presented to check if a submodule of the free module R [X] is prime, using Gröbner Basis.


Introduction
Gianni, Trager and Zacharias [3] present an algorithm to check whether an ideal of the polynomial ring R [X] is prime, where R is a commutative Noetherian ring with unit and X is the set of indeterminates x 1 , . . .x n .In this article, the Preliminary Results section presents some conclusions from that paper.The Primality Test section includes some important definitions and basic affirmations to enunciate and prove lemmas and theorems in support of the new algorithm for prime submodules.Finally, in the Examples section this algorithm is illustrated.

Preliminary results
Let R be a commutative Noetherian ring with identity.R [X] s is a free module where X is the set of indeterminates x 1 , . . .x n .Definition 2.1.[4] A submodule N of a module M over a ring R is said to be a prime submodule if N 6 = M , and r ∈ R and m ∈ M satisfy rm ∈ N , so that r ∈ (N : M ) or m ∈ N .Lemma 2.2.[5] A submodule N of an R-module M is prime if and only if P = (N : M ) is a prime ideal of R and the RP -module MN is torsion-free.
Example 2.3.If N is a submodule of an R-module M such that (N : M ) is a maximal ideal of R, then N is a prime submodule.Also, if N is a maximal submodule of an R-module M , then N is a prime submodule.Moreover, let R be an integral domain and let N be a submodule of an Rmodule M such that (N : M ) = 0.In this case, N is a prime submodule.
The primality test of Gianni, Trager and Zacharias for ideals of a polynomial ring is directly quoted below from [3].Algorithm 2.4.ALGORITHM P T (R; x; I) Primality test for ideals.

INPUT:
Ring R; OUTPUT: TRUE if I is prime, otherwise FALSE.
Step 1: If n = 0 then I ⊂ R is prime the return TRUE otherwise return FALSE.
Step 4: Step 5: Step 6: If f is not irreducible over K 0 then return FALSE.
Step 7: Compute Step 8: If I ec ⊂ I 0 then return TRUE, otherwise return FALSE.
The proofs of the statements below, for Lemma 2.6, Corollary 2.7 and Proposition 2.8 can be found in [6].
Lemma 2.6.Let V ⊂ S be multiplicatively closed subsets of R, and N ⊂ R [X] s be a submodule.If Corollary 2.7.Let S ⊂ R be a multiplicatively closed set and N ⊂ R [X] s be a submodule.If a ∈ S exists such that Proposition 2.8.Let R be an integral domain, hpi ⊂ R be a principal prime ideal, and

Primality test
In this section, lemmas and theorems that justify the new algorithm for prime submodules are stated, as well as some explanations.At the end of this section, the algorithm is presented.
Corollary 3.1.Let R be an integral domain and K be the quotient field of R. Then for a submodule s can be found.
The proofs for the following results correspond to the ones used for ideals.
It is straightforward to prove that φ and Φ are well-defined, so their proofs are not presented here.
Given ker (Φ) = T ((N : R [X] s ) s ), and using Lemmas 3.2 and 3.3, N is a prime submodule if and only if T (N ) is also prime.
Finally, if N is a prime submodule, then (N : R [X] s ) is a prime ideal of R [X] and the homomorphism ) ∩ R turns out to be prime.2 Lemma 3.5 is provided by [6].Nevertheless, due to its importance, its proof is analyzed here.Lemma 3.5.Let R be an integral domain, N be a submodule of R [X] s , and hpi ⊂ R be a principal prime ideal.Thus it is possible to find g ∈ R−hpi such that ´.
Proof.Recall from Proposition 2.8 that it is possible to find a ∈ R−hpi, and therefore ´.
Corollary 3.6.Let R be an integral domain with quotient field K and N be an R [X] s -submodule.Therefore, it is possible to find Proof.This is merely a direct application of Lemma 3.5 with p = 0. 2 Lemma 3.7.Let R be an integral domain with quotient field K and Lemma 3.8.Let R be an integral domain with quotient field K, and where a ∈ R − {0}.Therefore, (3.1) Lemma 3.9.Let R be an integral domain with quotient field K, and Theorem 3.10.Let R be an integral domain with quotient field K, and Proof.This follows immediately from Lemmas 3.7, 3.8 and 3.9. 2 Corollary 3.11.Let N be a submodule of R [x] s .Thus N is a prime submodule if and only if (ii) Letting R 0 = R ((N : R [x] s ) ∩ R), K 0 be the quotient field of R 0 and N 0 be the image of Proof.

Examples
Two examples are presented below to illustrate how the algorithm works.Firstly, the algorithm is used to show a submodule that is not prime, and secondly, it is used to identify a prime submodule.