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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 $R[x]$
$\Longleftrightarrow$ $I K(R)[x]$ K[x]$is prime in$K(R)[x]$K[x]$ and $I=IK(R)[x]I=IK[x] \cap K(R)[x]$K[x]$. Here$K(A)$K$ 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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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 $R[x]$
$\Longleftrightarrow$ $I K(R)[x]$ is prime in $K(R)[x]$ and $I=IK(R)[x] \cap K(R)[x]$. Here $K(A)$ 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.