prove that $f$ has a root in $K$ if and only if there exists a point $(b,c)$ such that $(b,c)≠ (0,0)$ and $f^h(b,c)=0$ Given the following definition:
For any univariate polynomial $f= a_mx^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ of degree $m$ in the ring $K[x]$, its homogenization in the ring $K[x,y]$ is
$f^h=a_mx^m+a_{m-1}x^{m-1}y+...+a_1xy^{m-1}+a_0y^m$
My question: How do I prove that $f$ has a root in the field $K$ if and only if there exists a point $(b,c)$ such that $(b,c)≠ (0,0)$ and $f^h(b,c)=0$
I am having trouble understanding the homogenization definition, it's just the same as the polynomial with another variable? also is the question saying if there exists a point in the homogenization that equals zero then $f$ has a root? Any help would be great.
 A: I think the homogenisation of $f$ is fairly straightforward. You basically do what it says, and get a polynomial of two variables rather than one. For example, if $f(x) = 3x^2 + 2x - 6$, then $f^h(x, y) = 3x^2 + 2xy - 6y^2$. If $f$ has degree $m$, another way to write it is:
$$f^h(x, y) = y^mf(xy^{-1}).$$
Indeed, this expression is very helpful in order to prove the result. If $f(b, c) = 0$ and $(b, c) \neq (0, 0)$, then either $b = 0$ or $c = 0$. If $c \neq 0$, then we get
$$0 = f^h(b, c) = c^m f(bc^{-1}) \implies f(bc^{-1}) = 0,$$
since $c^m \neq 0$. If $c = 0$, then we can't really use the above expression, since it features $c^{-1}$. However, we have
$$0 = f^h(b, 0) = a_m b^m  + 0 + 0 + \ldots + 0.$$
Since $f$ is of degree $m$, we have $a_m \neq 0$, and hence $b^m = 0$. As we are in a field, the only way this can happen is if $b = 0$, and hence $(b, c) = 0$, against assumption. Thus, necessarily, $c \neq 0$, and we have a root.
The converse is even easier. If $f$ has a root $x$, then $f(x) = f^h(x, 1) = 0$.
