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Prime ideal (x) in k[x,y]

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Example (2) on page 682 of Dummit and Foote reads as follows:

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(2) For any field k, the ideal (x) in k[x,y] is primary since it is a prime ideal.

... ... etc

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Now if (x) is prime then obviously (x) is primary BUT ....

How do we show that (x) is prime in k[x, y]?

Would appreciate some help

Peter

[This has also been posted on MHF]
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
I'm a bit rusty on multi-variate polynomials, but this is what I think is true:

$x|f(x,y) \iff f(0,\alpha) = 0,\ \forall \alpha \in k$

Now suppose that $x|f(x,y) = g(x,y)h(x,y)$ with $x \not\mid g(x,y)$.

We have (for any $\alpha \in k$), $0 = f(0,\alpha) = g(0,\alpha)h(0,\alpha)$ with $g(0,\alpha) \neq 0$.

Since $k$ is a field (and thus an integral domain), it must be the case that for any such $\alpha$ (and thus all of them), that $h(0,\alpha) = 0$, showing that $x|h(x,y)$.

In short, $x$ is a prime element of $k[x,y]$ and thus generates a prime ideal.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks Deveno.

Just working through your post now

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thanks Deveno.

Just working through your post now

Peter
Just worked through your post

Really fundamental and interesting way to show (x) is a prime ideal in k[x,y] ....thanks

Peter