Consider the ideal $I=(ag-ec-1,ah+bg-cf-de)$ of $R=K[a,b,c,d,e,f,g,h]$.
Is $I$ prime when $K=\overline{\mathbb{F}}_2$ is the algebraic closure of a field of 2 elements?
Can computers answer this question? Using existing software?
Apparently if $K$ is a field of characteristic 0, or $K$ is $\mathbb{Z}/p\mathbb{Z}$ for $p < 30$, then $I$ is prime.
If $K$ is a finite field, then presumably $I$ is prime in $R$ iff $I+(F)$ is prime in $(\mathbb{Z}/p\mathbb{Z})[a,b,c,d,e,f,g,h,x]$ where $K=\mathbb{Z}[x]/(F,p)$. If so, then $I$ remains prime over fields of size $2^n$ for $n \leq 8$ and $n=60$, $90$, and $100$.
I find it a little hard to believe that $I$ is not prime over $K=\overline{\mathbb{F}}_2$ if it is prime over all these smaller fields. Certainly I find it hard to believe it is not prime when it is given as an exercise in a book assuming no algebraic geometry background.
Here is how to ask Singular over $\mathbb{C}$:
LIB "primdec.lib"; ring R=0,(a,b,c,d,e,f,g,h),lp; ideal I=(a*g - e*c - 1, a*h + b*g - c*f - d*e); def S=absPrimdecGTZ(I); setring S; absolute_primes;
And then over a finite field of size 4:
LIB "primdec.lib"; ring R=2,(a,b,c,d,e,f,g,h,x),lp; ideal I=(a*g - e*c - 1, a*h + b*g - c*f - d*e); primdecGTZ( I + ideal( x^2+x+1 ) );
If the first line of the answer has a [1] in it and there is no [2] at the same indention level, then the ideal is primary. If the ideals listed in the next [1] and [2] are the same, then it is prime.