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I am not taking algebraic geometry or trying to prove anything. I'm just looking for a simple intuitive understanding of Bézout's theorem for the case when one of the curves is a constant function. From https://en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem#Examples we have the example: "Two distinct non-parallel lines (in the same plane) always meet in exactly one point. Two parallel lines intersect at a unique point that lies at infinity."

But take y = 1 and y = x. These two lines meet once at (1,1), but the product of the degrees of these curves is 0*1 = 0.

I feel like I'm missing something very simple or don't understand the hypotheses of the theorem... can someone please help explain? Thanks!

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The line described by $y=1$ is the zero set of a certain polynomial in the two variables $x$ and $y$, namely $y-1$. This polynomial has degree $1$, which is the exponent of $y$ in the first term. So, by definition, the degree of that curve is $1$.

On the other hand, the line described by $y=1$ is also the graph of a certain polynomial in the single variable $x$, namely $1$. This polynomial has degree $0$. But that isn't the polynomial relevant to Bézout's theorem!

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  • $\begingroup$ Got it, does this mean the fundamental theorem of algebra is a special case of Bézout's theorem? From wikipedia we have: en.wikipedia.org/wiki/B%C3%A9zout%27s_theorem#Examples "The special case where one of the curves is a line can be derived from the fundamental theorem of algebra." But it doesn't seem FTA is being used to show a special case, it seems the FTA itself is the special case. If you take the polynomials y - 0 = 0 and y - ax^n - bx^(n-1) - ... = 0 then Bézout's theorem would say these two curves meet at 1*n places in C, counting multiplicity, which is FTA, no? $\endgroup$ – hli Aug 14 at 18:25
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    $\begingroup$ One result is Bézout's theorem on a line. Another result is FTA. Given either result, you can easily derive the other. Which one comes first is a matter of tactics, that's all. $\endgroup$ – Chris Culter Aug 14 at 21:48
  • $\begingroup$ Oh I see, thank you! $\endgroup$ – hli Aug 14 at 21:50
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The equation $y=1$ is linear, as is any equation $$ax+by+c=0,$$ where $a,b,c$ are constants. These are all of first degree since the operations on the variables involve nothing more than scalar multiplication and addition.

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