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I'm trying to understand what the author of this book means in this paragraph:

He wants to prove that the polynomial $X^2+Y^2-1$ is irreducible using geometrical tools. I have the following doubts:

  1. What he means by Since every line has at least three points on it ?

  2. Why the fact that $X^2+Y^2-1=F(X,Y)\cdot G(X,Y)$, where F, G are lines implies that the circle $X^2+Y^2-1=0$ is made up of two lines?

I would really grateful if anyone could help me.

Thanks a lot.

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What book are you quoting? – Georges Elencwajg Aug 26 '13 at 11:16
@GeorgesElencwajg Conics and Cubics. A Concrete Introduction to Algebraic Curves by Robert Byx. – user42912 Aug 26 '13 at 16:03
Thanks for the reference, user42912. – Georges Elencwajg Aug 26 '13 at 16:42
@GeorgesElencwajg you're welcome :) – user42912 Aug 26 '13 at 21:27
  1. A line in $k^2$ has the form $aX+bY+c=0$ where $a$ and $b$ cannot both be zero. We may assume without loss of generality that $a\ne 0$, so that we have $X=-a^{-1}bY-a^{-1}c$. It follows that for any $\alpha\in k$, the point $$(-a^{-1}b\alpha-a^{-1}c,\alpha)$$ lies on the line. It follows that the number of points on any line is in bijective correspondence with the field $k$. So in order for a line to have at least three points, we should assume that $k$ is not the field of $2$ elements, which is not a very strict assumption.
  2. If $X^2+Y^2-1=F(X,Y)\cdot G(X,Y)=0$, then notice by the zero product property that any point on the circle must satisfy at least one of the linear polynomials $F(X,Y)$ and $G(X,Y)$. It is also true that any point satisfying the polynomials $F$ and $G$ will lie on the circle. It follows that the circle must be the union of two lines (a contradiction).
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Dear Jared, You could note that in char. $2$ one has $X^2 +Y^2 - 1 = (X +Y - 1)^2$, and so in fact the equation does become reducible. (So checking that the intersection of a circle and a line has at most two points is not quite trivial, and fails in char. $2$.) Regards, – Matt E Aug 26 '13 at 2:59
I forgot to mention $k=\mathbb R$ – user42912 Aug 26 '13 at 3:07

The "proof" is absolute nonsense:

If you replace the word "circle" by "degenerate conic" (= union of two lines, non necessarily distinct) the "argument" still applies, which proves that it is false, since degenerate conics do exist!
And actually the circle is reducible over an algebraically closed field of characteristic $2$, and such a field is necessarily infinite, so that lines have infinitely many points.
This confirms that the purported proof is no proof at all.

The mistake is the statement "a line and a circle have at most two points in common": this is false if the circle contains a line and since what we want to prove is that the circle contains no line we may not assume it in the proof: the argument is circular (pun intended!)

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Dear Georges, The OP clarified in comments to other answers that the ground field is $\mathbb R$, so I think one should interpret "circle" to literally mean a Euclidean circle, rather than a general (possibly degenerate) conic. Best wishes, – Matt E Aug 26 '13 at 11:59
Dear @Matt, I agree but the proof doesn't use that the ground field is $\mathbb R$ nor anything else on the ground field, so that it cannot be correct. The assertion that "a line and a circle have at most two points in common" seems to me to beg the question. – Georges Elencwajg Aug 26 '13 at 12:13
I think the author just want to show how an Algebra problem can be viewed as a Geometrical problem and vice-versa in the introduction chapter. I think the author didn't intend to prove formally this problem. – user42912 Aug 26 '13 at 16:06
Dear user42912, as an algebraic geometer I totally adhere to this view of algebra and geometry helping each other. As a caveat, however, let me emphasize that some very obvious geometric facts can only been proved by amazingly difficult algebra. For example the fact that, in a variety of dimension $n$, the subvariety cut out by a single regular function has dimensionon $n-1$ requires a very hard theorem of commutative algebra, namely Krull's Hauptidealsatz (=principal ideal theorem). – Georges Elencwajg Aug 26 '13 at 17:02

If the field $k$ is the finite field of 2 elements then this argument cannot work - there are lines in $\mathbb{F}_2[x,y]$ whose locus includes only two points, e.g. $x+y=1$, and indeed $x^2 + y^2 - 1 = (x+y+1)^2$ over $\mathbb{F}_2$. So, let's assume $\text{char } k$ is at least 3.

In that case, a line of the form $ax+by = c$ has at least 3 solutions $(x_0,y_0)$. Since our field $k$ has at least 3 different elements, we can substitute each of these in for x or y (a or b could be 0) and solve for the remaining variable. This gives us three solutions to the equation $ax+by=c$.

If $x^2 + y^2 - 1 = F(x,y) \cdot G(x,y)$, then any solution $(x_0,y_0)$ to the equation $x^2 + y^2 - 1 = 0$ must also then satisfy $F(x,y) = 0$ or $G(x,y) = 0$. This means that the circle $x^2 + y^2 - 1 = 0$ is actually a union of two lines.

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I forgot to mention $k=\mathbb R$ – user42912 Aug 26 '13 at 3:07
Just one question since he works in $\mathbb R$, why he didn't say that the line of the form $ax+by=c$ has infinite solutions, instead of saying that it has at least 3 solutions which is a particular case of the fact of the line has infinite points? Did you follow me? – user42912 Aug 26 '13 at 3:19
In terms of style, the author may simply want to give a very general proof (one that applies when $k$ is not the real numbers). We also don't need to consider the whole infinite set of solutions to finish the proof, so the author might just be trying to keep things as simple as possible. It is not as important here that our field is infinite as that its characteristic is not 2. – Thom Tyrrell Aug 26 '13 at 5:19
yes, indeed. Thank you very much for your help! – user42912 Aug 26 '13 at 5:58
Dear Thom, if a circle is a union of two lines there is no contradiction with Bézout because Bézout requires of the two curves considered should have no irreducible component in common, which is exactly what we are trying to prove. Moreover a circle is the union of two (identical) lines over any field of characteristic $2$ and not only over $\mathbb F_2$. If in particular such a field is algebraically closed, Bézout can be invoked and thus cannot prove that the circle is irreductible, since it is reducible! – Georges Elencwajg Aug 26 '13 at 17:07

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