Let $K=\bar{K}$ a field, $I=(f_1,f_2,f_3,f_4)\subset K[x_0,x_1,x_2,x_3]$ the ideal where \begin{align} f_1 &= x_0x_3-x_1x_2,\\ f_2&=x_0^2x_2-x_1^3,\\ f_3&=x_1x_3^2-x_2^3,\\ f_4&=x_0x_2^2-x_1^2x_3. \end{align} Show $Rad(f_1,f_2,f_3)=I$.

It's easy to show that $f_4^2=x_2^3f_2+x_1^3f_3-2x_1^2x_2^2f_1$ and this implies $f_4\in Rad(f_1,f_2,f_3)$, hence $I\subset Rad(f_1,f_2,f_3)$.

For the reverse inclusion I was thinking about the Hilbert's Nullstellensatz which states $Rad(J)=\mathcal{I}(V(J))$. This lead to solve the system $f_1=f_2=f_3=0$ and I can't figure it out with calculations. What I can do?

Thanks in advance to those who can answer me.

  • $\begingroup$ Find a Groebner basis, then find $V(J)$. $\endgroup$
    – markvs
    Feb 5, 2022 at 17:27
  • $\begingroup$ @markvs thank you for your answer, but I'm not suppose to use Groebner basis, we have never seen them. $\endgroup$ Feb 5, 2022 at 17:34
  • 2
    $\begingroup$ Then just solve the system of equations. Consider two cases $x_3=0$ and $x_3\ne 0$. $\endgroup$
    – markvs
    Feb 5, 2022 at 17:37

1 Answer 1


I answer my own question to show how I solved this problem.

I proved the ideal $I$ is prime and hence radical.

Let's consider $\phi:K[x_0,x_1,x_2,x_3]\to K[t,y]$ with \begin{align} x_0 &\to t^4\\ x_1 &\to t^3y\\ x_2 &\to ty^3\\ x_3 &\to y^4 \end{align} It's immediate to see $I\subset \ker\phi$. Moreover, a generic polynomial in $K[x_0,x_1,x_2,x_3]/I$ can be written as $$ a(x_0,x_3)+b(x_0,x_3)x_1+c(x_0,x_3)x_1^2+d(x_0,x_3)x_2+e(x_3)x_2^2+I. $$ Applying $\phi$ to this, we obtain $$ a(t^4,y^4)+b(t^4,y^4)t^3y+c(t^4,y^4)t^6y^2+d(t^4,y^4)ty^3+e(y^4)t^2y^6 $$ and this is $0$ iff $a,b,c,d,e=0$, and this implies $\ker\phi = I$, and so $K[x_0,x_1,x_2,x_3]/I$ is isomorphic to a subring of $K[t,y]$ which is a domain. So $I$ must be prime.


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