Proving x ∉ (x,y)^n for any n ∈ N in F[x,y] Field

[Bashyboy]

Homework Statement

Consider $F[x,y]$, where $F$ is some field. I've been working on a problem all day and I'm having trouble with this last step. I am trying to show that $x \notin(x,y)^n$ for any $n \in \Bbb{N}$.

Homework Equations

The Attempt at a Solution

Note that $(x,y)^n = [(x)+(y)]^n = \sum_{i=0}^n (x)^i y^{n-i}$. I tried working with this, but I couldn't get anywhere. I could really use a hint; I don't want to have this become a problem that takes two days...

 

[fresh_42]

I don't get it. Why isn't $x \in (x) \subseteq (x,y) = (x,y)^1$ as I can write it $x=x^1+0\,$?
What does it mean an element is not an ideal?

 

[Bashyboy]

I don't get it. Why isn't $x \in (x) \subseteq (x,y) = (x,y)^1$ as I can write it $x=x^1+0\,$?
What does it mean an element is not an ideal?

Sorry, I forgot to mention to stipulate that $n \ge 2$.

 

[fresh_42]

Then under the assumption $x \in (x,y)^n$ for an $n>1$ we can write $x= \sum_{n_i+m_i > 1} f_ix^{n_i}y^{m_i}$ with $f_i \in \mathbf{F}\; , \;f_i = 0$ almost all. Now the substitution $y=1$ is a ring homomorphism and we get a representation $x \in \mathbf{F}[x]$.

 

[mathwonk]

try showing all elements of (x,y)^n have degree ≥ n.