Varieties

**mathman88** · April 19th 2010, 08:14 AM

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

In lecture, my professor said $V$ 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 $I$ is prime then $I = \text{Rad}(I)$ .

Any insight is much appreciated.

**NonCommAlg** · April 19th 2010, 09:41 AM

Originally Posted by mathman88

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

In lecture, my professor said $V$ 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 $I$ is prime then $I = \text{Rad}(I)$ .

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 $V$ is $\langle x \rangle$ and clearly $\mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y]$ is an integral domain. thus $\langle x \rangle$ is a prime ideal.

**mathman88** · April 19th 2010, 11:57 AM

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 $V$ is $\langle x \rangle$ and clearly $\mathbb{R}[x,y]/\langle x \rangle \cong \mathbb{R}[y]$ is an integral domain. thus $\langle x \rangle$ is a prime ideal.

Expanding on this, what if the field wasn't $\mathbb{R}$ but arbitrary, call it $F$ .

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

Thanks!

**mathman88** · April 19th 2010, 06:24 PM

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

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

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

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

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

Thanks!

**chiph588@** · April 19th 2010, 07:54 PM

Originally Posted by mathman88

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

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

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

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

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

Thanks!

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

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

**mathman88** · April 20th 2010, 08:34 AM

Thanks!

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

**chiph588@** · April 20th 2010, 09:43 PM

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

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