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

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In summary, the conversation discusses the ideal (x) in k[x,y] and how it is a primary ideal since it is a prime ideal. The conversation then goes on to discuss how to show that (x) is prime in k[x,y] by using the fact that $x|f(x,y) \iff f(0,\alpha) = 0,\ \forall \alpha \in k$. This leads to the conclusion that $x$ is a prime element of $k[x,y]$ and therefore generates a prime ideal.
  • #1
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Example (2) on page 682 of Dummit and Foote reads as follows:

------------------------------------------------------------------------

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

... ... etc

----------------------------------------------------------------------------

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]
 
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  • #2
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.
 
  • #3
Thanks Deveno.

Just working through your post now

Peter
 
  • #4
Peter said:
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
 
  • #5


Hello Peter,

Thank you for your question. To show that (x) is prime in k[x,y], we need to show that it satisfies the definition of a prime ideal in k[x,y].

First, let's recall the definition of a prime ideal in a ring R. An ideal P in R is said to be prime if for any two elements a and b in R, if their product ab is in P, then either a or b must be in P.

Now, let's apply this definition to the ideal (x) in k[x,y]. Let a and b be two elements in k[x,y] such that ab is in (x). This means that ab is a polynomial in k[x,y] with a term involving x. Since x is the only variable in (x), and all the terms in ab involve x, we can conclude that either a or b (or both) must involve x.

In other words, if ab is in (x), either a or b (or both) must be in (x). This satisfies the definition of a prime ideal, and thus, we can conclude that (x) is prime in k[x,y].

I hope this helps. Let me know if you have any further questions.

Best,
 

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

What is a prime ideal in k[x,y]?

A prime ideal in k[x,y] is a subset of the polynomial ring k[x,y] that satisfies the following properties:

  • It is an ideal, meaning it is closed under addition and multiplication by elements of k[x,y].
  • It does not contain the constant term 1.
  • For any two polynomials f and g in the ideal, their product fg is also in the ideal.
  • If f is in the ideal and g is any polynomial in k[x,y], then the product fg is also in the ideal.

In other words, a prime ideal in k[x,y] is a special type of ideal that is closed under multiplication and does not contain any unnecessary elements.

How is a prime ideal different from a regular ideal?

A prime ideal is a special type of ideal that is closed under multiplication and does not contain any unnecessary elements, while a regular ideal is simply closed under addition and multiplication. This means that every prime ideal is a regular ideal, but not every regular ideal is a prime ideal.

How do I know if a given ideal in k[x,y] is prime?

There are several ways to determine if an ideal in k[x,y] is prime. One method is to use the ideal's generators and check if they satisfy the properties of a prime ideal. Another method is to use the quotient ring k[x,y]/I, where I is the ideal in question. If the quotient ring is an integral domain, then the ideal is prime.

Can a prime ideal in k[x,y] be generated by one polynomial?

Yes, a prime ideal in k[x,y] can be generated by one polynomial. This is because every prime ideal in k[x,y] is a principal ideal, meaning it can be generated by a single element. However, not every principal ideal in k[x,y] is prime.

What is the geometric interpretation of a prime ideal in k[x,y]?

Geometrically, a prime ideal in k[x,y] corresponds to a subset of the xy-plane where all the points lie on a single curve. This curve is known as the zero locus of the prime ideal. In other words, the prime ideal represents the set of all polynomials that vanish at every point on the curve.

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