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Let $\mathbb{F}$ be a field and let $\mathbb{K}$ be its algebraic closure. Let $p_1,p_2,...,p_k \in \mathbb{F}[x_1,...,x_n]$ be such that the ideal generated by these elements in the ring $\mathbb{F}[x_1,...,x_n]$ is proper. Does it hold that the ideal generated by $p_1,p_2,...,p_k$ in the ring $\mathbb{K}[x_1,...,x_n]$ is still proper?

The statement is true whenever $k=1$ (and $\mathbb{K}$ is not required to be the algebraic closure of $\mathbb{F}$): if the ideal generated by $p\in \mathbb{F}[x_1,...,x_n]$ in the ring $\mathbb{K}[x_1,...,x_n]$ is the whole ring, then there is $q \in \mathbb{K}[x_1,...,x_n]$ such that $pq=1$. Since $\mathbb{K}$ is a field and $p, q$ are invertible, it is the case that $p,q \in \mathbb{K}$. Therefore $p \in \mathbb{F}$ and then $q \in \mathbb{F}$.

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Yes, and this is just linear algebra. Suppose we are looking for polynomials $f_1,\dots,f_k$ of degree at most $m$ such that $\sum f_ip_i=1$. This equation can be expanded out into a system of linear equations in the coefficients of the $f_i$. If this system of linear equations has no solution in $\mathbb{F}$, it also has no solution in any extension of $\mathbb{F}$ (since we can determine whether there is a solution by Gaussian elimination, which uses only the field operations).

Or, from a more sophisticated perspective, let $R=\mathbb{F}[x_1,\dots,x_n]$, $S=\mathbb{K}[x_1,\dots,x_n]$, $I=(p_1,\dots,p_k)\subseteq R$, and $J=(p_1,\dots,p_k)\subseteq S$. We then have canonical isomorphisms $S\cong\mathbb{K}\otimes_\mathbb{F} R$ and $$S/J=S/IS\cong S\otimes_R R/I\cong \mathbb{K}\otimes_\mathbb{F} R/I.$$ If $I$ is proper, then $R/I$ is a nontrivial $\mathbb{F}$-vector space, so $S/J\cong \mathbb{K}\otimes_\mathbb{F} R/I$ is a nontrivial $\mathbb{K}$-vector space, so $J$ is proper.

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  • $\begingroup$ Very nice! Your "more sophisticated perspective" proof seems to work for any faithfully flat morphism of rings $\mathbb F\to \mathbb K$. Interesting. $\endgroup$ – Georges Elencwajg Jun 7 '18 at 19:41
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Suppose that $I=(p_1,...,p_k)$ is a proper ideal $\mathbb{F}[x_1,...,x_n]$. Let $J$ be a maximal ideal of $\mathbb{F}[x_1,...,x_n]$ containing $I$. Then $J=(q_1,...,q_m)$, for some $q_1,...,q_m$ in $\mathbb{F}[x_1,...,x_n]$, and we can consider the ideal of $\mathbb{K}[x_1,...,x_n]$ generated by $q_1,...,q_m$. Let it be $J'$.

Let $\Phi$ be the map $\mathbb{F}[x_1,...,x_n] \to \mathbb{K}[x_1,...,x_n]/J'$ such that $f \mapsto f + J'$. Since $J \subset J'$, $J \subset \ker(\Phi)$. Being $J$ maximal and being $\ker(\Phi)$ an ideal, $\ker(\Phi)=J$. Therefore there is an injective map $\mathbb{F}[x_1,...,x_n]/J \to \mathbb{K}[x_1,...,x_n]/J'$. Notice that $\mathbb{K}[x_1,...,x_n]/J'$ cannot be trivial, because $\mathbb{F}[x_1,...,x_n]/J$ is not trivial. Therefore $J'$ is proper. Since the ideal generated by $p_1,...,p_n$ in $\mathbb{K}[x_1,...,x_n]$ is contained in $J'$, it must be proper.

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