# how to determine whether an ideal is prime or not by an algorithm

Given polynomials $f_{1},\cdots,f_{n}\in \mathbb{C}[x_{1},\cdots,x_{m}]$, do we have an algorithm to determine whether the ideal $I=(f_{1},\cdots,f_{n})$ is prime ideal or not? Of course, we assume the polynomials are irreducible.

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Macaulay2 has a function isPrime which tells you if an ideal is or not prime. –  Mariano Suárez-Alvarez Apr 7 '11 at 15:59
Why do you assume that the polynomials are irreducible? –  Qiaochu Yuan Apr 7 '11 at 16:36
to rule out the trivial case that its factors might not be in the ideal. –  Jiang Apr 7 '11 at 17:28
It just doesn't seem to me that that assumption really helps. –  Qiaochu Yuan Apr 8 '11 at 4:36

Let $R$ be a Noetherian ring and let $I$ be an ideal in $R[x]$. Then the following facts hold:

• $I$ is prime in $R[x]$ $\Longleftrightarrow$ $I\cap R$ is prime in $R$ and $\overline{I}$ is prime in $R/(R\cap I)$.

• If $R$ is an integral domain and $I \cap R=0$, then $I$ is prime in $R[x]$
$\Longleftrightarrow$ $I K[x]$ is prime in $K[x]$ and $I=IK[x] \cap K[x]$. Here $K$ denotes the fraction field of $R$.

Using the above to successively eliminate variables, this shows that one can reduce the problem of checking primiality to the one-variable case, where many efficient methods are known. I think this is also how the Grobner basis works, since these can algorithmically compute the elimination ideals above.

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Eliminate, eliminate, eliminate the eliminators of elimination theory... =) –  Harry Gindi Apr 8 '11 at 6:05

There is such a test. Some explanation can be found in: "An introduction to Gröbner bases, By William Wells Adams, Philippe Loustaunau", or the original article (http://portal.acm.org/citation.cfm?id=65034) the above text is based on. See also the singular manual.

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