# Is the ideal prime in $\mathbb{C}[x_1,\dotsc,x_n]$?

I'm trying to find some examples of Complete Intersections which are also UFDs and while trying to do so I thought of this question (because I have been using Macaulay2 and I don't know how to use this software for determining prime ideals in polynomial rings over $$\mathbb{C}$$).

Consider the polynomial ring $$\mathbb{Q}[x_1,\dotsc,x_n]$$. Let $$f_1,\dotsc,f_k\in\mathbb{Q}[x_1,\dotsc,x_n]$$ be irreducible polynomials. Suppose the ideal generated by these polynomials $$I=(f_1,\dotsc,f_k)$$ is a prime ideal in $$\mathbb{Q}[x_1,\dotsc,x_n]$$. Also, assume each of these $$f_i$$'s are irreducible in $$\mathbb{C}[x_1,\dotsc,x_n]$$ as well. Then is it true that $$I\subseteq\mathbb{C}[x_1,\dotsc,x_n]$$ is also a prime ideal, i.e., $$\mathbb{C}[x_1,\dotsc,x_n]/I$$ is an integral domain? If it's not true, then will this statement be true if $$f_i$$'s were homogeneous?

## 1 Answer

I believe the answer is no, in large part because an ideal does not have canonical generators, and so assuming that some specific generators are irreducible isn't that powerful. For example, take $$n=2$$ and $$f_1(x_1,x_2)=x_2+x_1^2+1$$ and $$f_2(x_1,x_2)=x_2$$. In both cases, $$I=(x_2+x_1^2+1,x_2)=(x_1^2+1,x_2)$$. Then $$\Bbb Q[x_1,x_2]/I \cong \Bbb Q[x_1]/(x_1^2+1) \cong \Bbb Q[i]$$ is a domain, hence $$I$$ is prime there; however, $$\begin{multline*} \Bbb C[x_1,x_2]/I \cong \Bbb C[x_1]/(x_1^2+1) = \Bbb C[x_1]/((x_1+i)(x_1-i)) \\ \cong \Bbb C[x_1]/(x_1+i) \oplus C[x_1]/(x_1-i) \cong \Bbb C\oplus\Bbb C \end{multline*}$$ is not a domain, hence $$I$$ is not prime there. (I suspect that this example can be made homogeneous as well and still exhibit the same difference of behavior.)

• Thanks for your answer. Will the above statement be true in the special case- $\mathbb{Q}[x_1,\dotsc,x_8]/(x_1x_2+x_3x_4+x_5x_6, x_1x_2+2x_3x_4+x_7x_8)$? Nov 10, 2021 at 9:11
• My hunch is yes, but it's only a guess. Nov 10, 2021 at 17:48