Problem:
Solution:
It is obvious that $ x \in I $ and $ y \in I $, if $ I $ were a principal ideal then there exists $ f \in I $ such that every element in $ p \in I $ can be written as $ p = f g_p $.
Now $ f g_x = x $ and $ f g_y = y $
Since $ \deg(x) = 1$, so either $ f $ has degree 1 and $ g_x $ has degree 0 or $ f $ has degree 0 and $ g $ has degree 1.
Suppose $ f $ has degree 1 so $ f = ax + by + c $ and therefore $ g_x = d $ has degree 0. We found $ fg_x = (adx + bdy + cd) = 0 $, so $ ad = 1 $ and hence $ d \ne 0 $, we also know $ b = c = 0 $, so $ f = ax $ and $ g_x = d $. However, $ f g_y = ax g_y = y $, there couldn't be a polynomial $ g_y $ that satisfy this, so $ f $ has to have degree 0.
Now $ f = e $ has degree 0, but then it implies $ e \in I $, but there is not way $ e \in I $ because any constant cannot be written as $ p(x, y)x + q(x, y)y $, so we conclude $ \langle x, y. \rangle $ cannot be a principal ideal.
Solution:
It is obvious that $ x \in I $ and $ y \in I $, if $ I $ were a principal ideal then there exists $ f \in I $ such that every element in $ p \in I $ can be written as $ p = f g_p $.
Now $ f g_x = x $ and $ f g_y = y $
Since $ \deg(x) = 1$, so either $ f $ has degree 1 and $ g_x $ has degree 0 or $ f $ has degree 0 and $ g $ has degree 1.
Suppose $ f $ has degree 1 so $ f = ax + by + c $ and therefore $ g_x = d $ has degree 0. We found $ fg_x = (adx + bdy + cd) = 0 $, so $ ad = 1 $ and hence $ d \ne 0 $, we also know $ b = c = 0 $, so $ f = ax $ and $ g_x = d $. However, $ f g_y = ax g_y = y $, there couldn't be a polynomial $ g_y $ that satisfy this, so $ f $ has to have degree 0.
Now $ f = e $ has degree 0, but then it implies $ e \in I $, but there is not way $ e \in I $ because any constant cannot be written as $ p(x, y)x + q(x, y)y $, so we conclude $ \langle x, y. \rangle $ cannot be a principal ideal.
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