I was thinking about how a chain of irreducible polynomials in $K[X_1,\ldots,X_n]$, where $K$ is a field, behave with respect of being prime. What I mean is the folowing:

If $\{f_1,\ldots,f_n\}$ is a sequence of irreducible polynomials in $K[X_1,\ldots,X_n]$, I know that the ideal generated by $(f_i)$ is prime for every $i = 1,\ldots,n$. But what I can have from the ideals of the form $I_k = (f_1,f_2,\ldots,f_k)$? Are these ideals $I_k$ prime?

I'm asking this because if this is true there is a very simple answer for my question here.

Thank you for the help!!


2 Answers 2


There's no reason for this to be true. For example, take $f_1 = y^2 - x^3, f_2 = y^2 + x^3$ in $k[x, y]$. Then $(f_1, f_2) = (y^2, x^3)$ is not even radical. Geometrically, the problem is that the two curves $f_1 = 0$ and $f_2 = 0$ intersect with multiplicity at the origin. This doesn't happen generically, but a generic problem is that two curves will intersect at more than one point in general, and then the corresponding ideal will still fail to be prime. For example, take $f_1 = y$ and $f_2 = y - (x^2 - x)$. Then $(f_1, f_2) = (y, x^2 - x)$ has the property that $k[x, y]/(f_1, f_2) \cong k \times k$ has nontrivial idempotents.


Let us think about $\Bbb{R}[x]$. Two irreducible polynomials are $-x^2-1$ and $x^4 + 1$. $-x^2-1$ has $i^{4k+1}$ as a root, and $x^4+1$ has $i^{3k}$ as root. Both aren't present in $\Bbb{R}$

The ideal $(-x^2-1,x^4 + 1)$ is not prime.

This is because, on adding the two polynomials, we get $x^4-x^2=x^2(x^2-1)$.

Both $x^2$ and $(x^2-1)\notin (-x^2-1,x^4 + 1)$


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