How can I prove that this ideal is radical?

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.

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

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.