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Thread: Infinite roots

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    Infinite roots

    Prove that a polynomial $\displaystyle f \in K[X,Y]$ has infinite roots in $\displaystyle K^2$ if $\displaystyle K$ is algebraically closed.

    thanks!
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  2. #2
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    Quote Originally Posted by roporte View Post
    Prove that a polynomial $\displaystyle f \in K[X,Y]$ has infinite roots in $\displaystyle K^2$ if $\displaystyle K$ is algebraically closed.

    thanks!
    If $\displaystyle f \in K[X,Y]$ is a non-constant polynomial then it factors into $\displaystyle f = \prod_i l_i$ where $\displaystyle l_i = a_iX+b_iY+c_i$ ($\displaystyle a_i,b_i$ not both zero) are linear polynomials. Thus, it suffices to prove that $\displaystyle l_i$ has infinitely many roots. WLOG say $\displaystyle a_i\not = 0$ then $\displaystyle X = -a_i^{-1}b_iY - a_i^{-1}c_i $ and so for any value of $\displaystyle Y\in K$ we can find a value of $\displaystyle X\in K$ so that $\displaystyle a_i X + b_i Y + c_i = 0$. Thus, the number of solutions to the equation $\displaystyle l_i = 0$ is infinite if $\displaystyle |K| = \infty$. Thus, the last remaining step is to show that $\displaystyle K$ is infinite. But if $\displaystyle |K| = n$ then consider the polynomial $\displaystyle X^n - X + 1 \in K[X]$, this polynomial has no zeros in $\displaystyle K$. Therefore then $\displaystyle K$ cannot be algebraically closed - a contradiction. Therefore $\displaystyle |K| = \text{ ME }$ and that completes the proof.
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