I am to find a basis of the following ideal:
$$\sqrt{<x^5-2x^4+2x^2-x, \quad x^5-x^4-2x^3+2x^2+x-1>}$$
Truth be told, I'm not entirely confident of my solution. I will present it and then ask the pertinent questions:
First of all, note that $$x^5-2x^4+2x^2-x=x(x-1)^3(x+1),$$ and $$x^5-x^4-2x^3+2x^2+x-1=(x-1)^3(x+1)^2$$
According to a theorem in the textbook of which I'm doing these problems, if $I$ is an ideal in $k[x_1,x_2,...,x_k]$, then $\sqrt{I}=I(V(I))$.
Using this theorem, I found that $V(I)$ is $\{-1,1\}$. Thus, $$I(V(I))=(x-1)(x+1)=x^2-1.$$
Is this correct? I think this is the right theorem to use, and I think what I provided was indeed a basis of some sort; can someone confirm/correct this?
Thank you in advance
Your result is correct. Another way to get there:
Since we are in $k[x]$, which is a principal ideal domain, we know that the ideal is generated by the greatest common divisor of $x(x-1)^3(x+1)$ and $(x-1)^3(x+1)^2$, which is $(x-1)^3(x+1)$. To get the radical of the ideal generated by this polynomial, you simply have to strip all the multiplicities in the prime factorization. This also yields $(x-1)(x+1)$.