1. ## algebraic geometry

Hello,

Let $\displaystyle k$ be a field of characteristic 2. Then find the ideal of $\displaystyle X = V(t_1^2 + t_2^2 + t_3^2) \subseteq \mathbb{A}_k^3$.

I'm not sure how to solve this. I mean, since $\displaystyle k$ is of characteristic two, I can see that$\displaystyle (t_1 + t_2 + t_3)^2 = t_1^1 + t_2^2+t_3^2$, so that $\displaystyle I_X = \sqrt{\langle(t_1 + t_2 + t_3)^2\rangle}$. Is it just what we'd think it would be, i.e. $\displaystyle I_X = \langle t_1 + t_2 + t_3 \rangle$ ? If so, why is it true? But how do I find this radical ideal? And is this the best way to work out what the ideal is?

Thanks for any help

2. ## Re: algebraic geometry

There is an 'easy' way to work out the radical of an ideal in case this is principal (which in your case it is):

Let $\displaystyle J=\langle f \rangle$ where $\displaystyle f=\prod_{i=1}^n f_i^{k_i}$ is its factorization into irreducibles, then $\displaystyle \sqrt{J} = \langle \prod_{i=1}^n f_i \rangle$

To prove this let $\displaystyle g\in \sqrt{J}$ then there exists $\displaystyle m$ such that $\displaystyle g^m \in J$ ie. $\displaystyle g^m = pf$ for some polynomial $\displaystyle p$, by the uniqueness of a factorization in a polynomial ring (over a field at least) the $\displaystyle f_i$ must appear in the factorization of $\displaystyle g$, and so $\displaystyle g\in \langle \prod_{i=1}^n f_i \rangle$. On the other hand if $\displaystyle g\in \langle \prod_{i=1}^n f_i \rangle$ then $\displaystyle g^{\prod_{i=1}^n k_i} \in J$.

3. ## Re: algebraic geometry

Thanks. Is there some way to ensure that the polynomial I have is definitely irreducible?

4. ## Re: algebraic geometry

Originally Posted by slevvio
Hello,

Let $\displaystyle k$ be a field of characteristic 2. Then find the ideal of $\displaystyle X = V(t_1^2 + t_2^2 + t_3^2) \subseteq \mathbb{A}_k^3$.

I'm not sure how to solve this. I mean, since $\displaystyle k$ is of characteristic two, I can see that$\displaystyle (t_1 + t_2 + t_3)^2 = t_1^1 + t_2^2+t_3^2$, so that $\displaystyle I_X = \sqrt{\langle(t_1 + t_2 + t_3)^2\rangle}$. Is it just what we'd think it would be, i.e. $\displaystyle I_X = \langle t_1 + t_2 + t_3 \rangle$ ? If so, why is it true? But how do I find this radical ideal? And is this the best way to work out what the ideal is?

Thanks for any help
$\displaystyle I(V(J))=\sqrt{J}$ is not always true if your base field is not algebraically closed. is $\displaystyle k$ algebraically closed?

5. ## Re: algebraic geometry

Apologies, the underlying assumption we use is that k is algebraically closed - forgot to mention this

6. ## Re: algebraic geometry

well then, the degree of each $\displaystyle t_i$ in $\displaystyle f =t_1+t_2+t_3$ is one and so $\displaystyle f$ has no non-trival factorization, i.e. $\displaystyle f$ is irreducible. thus $\displaystyle (f)$ is prime and so $\displaystyle \sqrt{(f)}=(f)$. also $\displaystyle \sqrt{J^m} = \sqrt{J}$, for any ideal $\displaystyle J$ and any integer $\displaystyle m \geq 1$. thus $\displaystyle I(V((f^2)))=\sqrt{(f^2)}=\sqrt{(f)^2}=\sqrt{(f)}=( f)$.

7. ## Re: algebraic geometry

Let's say that we have a field $\displaystyle k$ which is not necessarily algebraically closed. Is it still true that $\displaystyle V(I) = V(J) \iff \sqrt{I} = \sqrt{J}$? For example, if this is true then I think that $\displaystyle \sqrt{\langle x-1, x^2 + y^2 -1 \rangle} = \langle x-1, y \rangle$ over any field k, but I am not too sure.

8. ## Re: algebraic geometry

Originally Posted by slevvio
Let's say that we have a field $\displaystyle k$ which is not necessarily algebraically closed. Is it still true that $\displaystyle V(I) = V(J) \iff \sqrt{I} = \sqrt{J}$? For example, if this is true then I think that $\displaystyle \sqrt{\langle x-1, x^2 + y^2 -1 \rangle} = \langle x-1, y \rangle$ over any field k, but I am not too sure.
the answer to your first question is no. for example consider $\displaystyle I =(x^2+y^2)$ and $\displaystyle J=(x^4+y^4)$ over $\displaystyle \mathbb{A}^2(\mathbb{Q})$.
the answer to your second question is yes, although you should have mentioned if you were considering $\displaystyle \mathbb{A}^2(k)$ or $\displaystyle \mathbb{A}^3(k),$ etc.
you always need to give two important piece of infiormation: your ground field and the dimension.

9. ## Re: algebraic geometry

Ok, thanks. I should have written that I was working in affine 2-space. Unfortunately I worked out that result by assuming the result in the first question, so how would I calculate it properly. Do you have any recommendations of books I can read on this subject, because I am having real difficulty trying to work out radical ideals

10. ## Re: algebraic geometry

well, because $\displaystyle x^2-1 \in (x-1)$, we have $\displaystyle (x-1,x^2+y^2-1)=(x-1,y^2)$. now clearly $\displaystyle (x-1,y) \subseteq \sqrt{(x-1,y^2)}$ and since $\displaystyle (x-1,y)$ is a maximal ideal of $\displaystyle k[x,y]$, we must have $\displaystyle (x-1,y)=\sqrt{(x-1,y^2)}$.

it is a good exercise to prove that the same result holds, i.e. $\displaystyle \sqrt{(x-1,y^2)}=(x-1,y)$, in affine $\displaystyle n$-space for any $\displaystyle n \geq 2.$

note that the ideal $\displaystyle (x-a_1, \ldots , x_n-a_n)$, for any field k and any $\displaystyle a_i \in k$, is always a maximal ideal of $\displaystyle k[x_1, \ldots , x_n]$ because the map $\displaystyle \varphi: k[x_1, \ldots . x_n] \longrightarrow k$ defined by $\displaystyle \varphi(f(x_1, \ldots, x_n))=f(a_1, \ldots , a_n)$ is an onto ring homomorphism and $\displaystyle \ker \varphi = (x_1-a_1, \ldots , x_n - a_n).$ if $\displaystyle k$ is algebraically closed, then the converse is also true, by Nullstellensatz, i.e. every maximal ideal of $\displaystyle k[x_1,x_2, \ldots , x_n]$ is in the form $\displaystyle (x-a_1, \ldots , x_n-a_n)$ for some $\displaystyle a_i \in k.$

there are algorithms for finding the radical of an ideal in polynomial rings but they are not easy and i don't think you'll need them. just simplify your ideal first and use the properties of radicals and remember what i just said about maximal ideals.

11. ## Re: algebraic geometry

In this case, as in many, you do not need to actually compute the radical of $\displaystyle \mathfrak{p}$. We have

$\displaystyle V=V(x^2+y^2+z^2)=V((x+y+z)^2)=V(x+y+z)$.

But $\displaystyle x+y+z$ irreducible $\displaystyle \Rightarrow \mathfrak{p}$ prime
$\displaystyle \Rightarrow\, I(V(\mathfrak{p}))=\mathfrak{p}$.