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The polynomial $$f(x,y) = (x^2 − 1)^2 + (y^2 − 1)^2$$ is an example of an irreducible polynomial in $\mathbf{R}[x,y]$ which is irreducible but whose zero set has multiple components in the Zariski topology (since $V(f)$ is four distinct points). This is an answer to exercise 12 in section 1 of Hartshorne.

However, all the examples (of irreducible polynomials over $\mathbf{R}[x,y]$ with reducible zero sets) I can come up with look like this; i.e. their zero sets are just finite collections of points.

My question is this:

Is there an irreducible polynomial in $\mathbf{R}[x,y]$ whose zero set has multiple components (w.r.t the Zariski topology) that are not points?

NOTE: By "multiple components", I mean components in the Zariski topology. So you might guess something like $xy-1$ or $y^2-x^3+x$, since their zero sets each clearly have two components in the Euclidean topology. For one of those polynomials to be an answer, however, there would need to be some polynomial whose zero set is the left piece and another whose zero set is the right piece.

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It's not clear to me whether you want it to be the union of zero sets of two nontrivial polynomials, or you want it not to be such a union. –  Gerry Myerson Jan 27 '13 at 0:44
@GerryMyerson I edited the note; hopefully it is clearer now. I was just anticipating that people would answer with things that had multiple components in the Euclidean topology. Either of the cases you mention would be interesting (unless the zero set splits up as a collection of points, as in my first example.) –  Daenerys Naharis Jan 27 '13 at 3:40

2 Answers 2

up vote 2 down vote accepted

No. By Bézout's theorem, two polynomials $f, g \in \mathbb{R}[x, y]$ which are relatively prime have the property that their zero sets intersect in at most finitely many points. So if $f$ is irreducible and $g$ vanishes at infinitely many points where $f$ also vanishes, then $f | g$ and $g$ necessarily vanishes everywhere $f$ does.

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The zero set of $(x^2-2)^2+(y-2)^2-1$ is the union of two distorted circles. Wolfram Alpha says it is irreducible; and here is a picture of the curve.

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Note that your polynomial $f$ has degree $2$ in the variable $y$. If $p\in\mathbb R[x]$ satisfies $f\bigl(x,p(x)\bigr)=0$, then $p$ must have degree $2$, say $p(x)=ax^2+bx+c$; but comparing degree $4$ terms we get $1+a^2=0$, which is impossible. This shows in a elementary fashion that $f$ is irreducible in $\mathbb R[x,y]$. –  Matemáticos Chibchas Jan 26 '13 at 21:38
I'm not sure I understand OP's objection to elliptic curves such as $y^2=x^3-x$, but whatever it is I think it would apply to $(y-2)^2=1-(x^2-2)^2$ as well. –  Gerry Myerson Jan 27 '13 at 0:46
Yes this clearly has two components in the Euclidean topology, but the question is if you could demonstrate a polynomial which has the left circle as its zero set and a different polynomial with the right circle as its zero set (i.e. a union of closed sets in the Zariski topology) –  Daenerys Naharis Jan 27 '13 at 3:24
And you will only be satisfied if the answer is yes, if I understand what you've added to the statement of the question. You want an irreducible polynomial that cuts out a reducible variety. If you had such a thing, and viewed it over the complex numbers, what would be the consequences? –  Gerry Myerson Jan 27 '13 at 4:00

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