1. ## algebraic geometry

Hello,

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

I'm not sure how to solve this. I mean, since $k$ is of characteristic two, I can see that $(t_1 + t_2 + t_3)^2 = t_1^1 + t_2^2+t_3^2$, so that $I_X = \sqrt{\langle(t_1 + t_2 + t_3)^2\rangle}$. Is it just what we'd think it would be, i.e. $I_X = \langle t_1 + t_2 + t_3 \rangle$ ? 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 $J=\langle f \rangle$ where $f=\prod_{i=1}^n f_i^{k_i}$ is its factorization into irreducibles, then $\sqrt{J} = \langle \prod_{i=1}^n f_i \rangle$

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

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

I'm not sure how to solve this. I mean, since $k$ is of characteristic two, I can see that $(t_1 + t_2 + t_3)^2 = t_1^1 + t_2^2+t_3^2$, so that $I_X = \sqrt{\langle(t_1 + t_2 + t_3)^2\rangle}$. Is it just what we'd think it would be, i.e. $I_X = \langle t_1 + t_2 + t_3 \rangle$ ? 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
$I(V(J))=\sqrt{J}$ is not always true if your base field is not algebraically closed. is $k$ 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 $t_i$ in $f =t_1+t_2+t_3$ is one and so $f$ has no non-trival factorization, i.e. $f$ is irreducible. thus $(f)$ is prime and so $\sqrt{(f)}=(f)$. also $\sqrt{J^m} = \sqrt{J}$, for any ideal $J$ and any integer $m \geq 1$. thus $I(V((f^2)))=\sqrt{(f^2)}=\sqrt{(f)^2}=\sqrt{(f)}=( f)$.

7. ## Re: algebraic geometry

Let's say that we have a field $k$ which is not necessarily algebraically closed. Is it still true that $V(I) = V(J) \iff \sqrt{I} = \sqrt{J}$? For example, if this is true then I think that $\sqrt{\langle x-1, x^2 + y^2 -1 \rangle} = \langle x-1, y \rangle$ 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 $k$ which is not necessarily algebraically closed. Is it still true that $V(I) = V(J) \iff \sqrt{I} = \sqrt{J}$? For example, if this is true then I think that $\sqrt{\langle x-1, x^2 + y^2 -1 \rangle} = \langle x-1, y \rangle$ over any field k, but I am not too sure.
the answer to your first question is no. for example consider $I =(x^2+y^2)$ and $J=(x^4+y^4)$ over $\mathbb{A}^2(\mathbb{Q})$.
the answer to your second question is yes, although you should have mentioned if you were considering $\mathbb{A}^2(k)$ or $\mathbb{A}^3(k),$ 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 $x^2-1 \in (x-1)$, we have $(x-1,x^2+y^2-1)=(x-1,y^2)$. now clearly $(x-1,y) \subseteq \sqrt{(x-1,y^2)}$ and since $(x-1,y)$ is a maximal ideal of $k[x,y]$, we must have $(x-1,y)=\sqrt{(x-1,y^2)}$.

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

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

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 $\mathfrak{p}$. We have

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

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