MHB Is the Ideal (x) in k[x, y] a Prime Ideal?

[Math Amateur]

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]

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.

 

[Math Amateur]

Thanks Deveno.

Just working through your post now

Peter

 

[Math Amateur]

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