1. ## Varieties

Let $\displaystyle V$ be the $\displaystyle y$-axis contained in $\displaystyle \mathbb{R}^2$ ($\displaystyle V=\mathbb{V}(x)$).

In lecture, my professor said $\displaystyle V$ is irreducible (which makes sense), but I don't know how to show it. One hint he gave us is that "a prime ideal is radical" i.e. if $\displaystyle I$ is prime then $\displaystyle I = \text{Rad}(I)$.

Any insight is much appreciated.

2. Originally Posted by mathman88
Let $\displaystyle V$ be the $\displaystyle y$-axis contained in $\displaystyle \mathbb{R}^2$ ($\displaystyle V=\mathbb{V}(x)$).

In lecture, my professor said $\displaystyle V$ is irreducible (which makes sense), but I don't know how to show it. One hint he gave us is that "a prime ideal is radical" i.e. if $\displaystyle I$ is prime then $\displaystyle I = \text{Rad}(I)$.

Any insight is much appreciated.
well, a variety is irreducible iff its ideal is prime. now it's easy to see that in your example the ideal of $\displaystyle V$ is $\displaystyle \langle x \rangle$ and clearly $\displaystyle \mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y]$ is an integral domain. thus $\displaystyle \langle x \rangle$ is a prime ideal.

3. Originally Posted by NonCommAlg
well, a variety is irreducible iff its ideal is prime. now it's easy to see that in your example the ideal of $\displaystyle V$ is $\displaystyle \langle x \rangle$ and clearly $\displaystyle \mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y]$ is an integral domain. thus $\displaystyle \langle x \rangle$ is a prime ideal.
Expanding on this, what if the field wasn't $\displaystyle \mathbb{R}$ but arbitrary, call it $\displaystyle F$.

My professor said $\displaystyle V$ is irreducible when $\displaystyle F$ is algebraically closed. How would that be shown? Also what would be an example of a non algebraically closed field where $\displaystyle V$ is reducible?

Thanks!

4. So in general, here's what I'm thinking:

We have that $\displaystyle F$ is an algebraically closed field.

Now $\displaystyle V$ is irreducible $\displaystyle \iff \mathbb{I}(V)$ is prime. But by Hilbert's Nullstellensatz we see that $\displaystyle \mathbb{I}(V) = \text{Rad}(x)$.

So $\displaystyle V$ is irreducible $\displaystyle \iff \text{Rad}(x)$ is prime.

I'm stuck after this. What is $\displaystyle \text{Rad}(x)$ explicitly? Also I'm able to show if $\displaystyle A$ is a prime ideal then $\displaystyle A = \text{Rad}(A)$, but does the other direction work too?

Thanks!

5. Originally Posted by mathman88
So in general, here's what I'm thinking:

We have that $\displaystyle F$ is an algebraically closed field.

Now $\displaystyle V$ is irreducible $\displaystyle \iff \mathbb{I}(V)$ is prime. But by Hilbert's Nullstellensatz we see that $\displaystyle \mathbb{I}(V) = \text{Rad}(x)$.

So $\displaystyle V$ is irreducible $\displaystyle \iff \text{Rad}(x)$ is prime.

I'm stuck after this. What is $\displaystyle \text{Rad}(x)$ explicitly? Also I'm able to show if $\displaystyle A$ is a prime ideal then $\displaystyle A = \text{Rad}(A)$, but does the other direction work too?

Thanks!
$\displaystyle I=\langle x \rangle$ is prime since $\displaystyle F[x,y]/\langle x \rangle \cong F[y]$ is an integral domain, so $\displaystyle \text{Rad}(x) = \langle x \rangle$ by the theorem you said you knew how to prove.

Thus $\displaystyle V$ is irreducible $\displaystyle \iff \langle x \rangle$ is prime...

6. Thanks!

Now what about that field where $\displaystyle V$ is reducible? I'm having trouble picturing what that would be.

7. I think $\displaystyle \mathbb{F}_2$ will do the trick.