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This is an exercise on Page 8 of Hartshone's Algebraic Geometry:

Give an example of an irreducible polynomial $f \in \mathbb R[x,y]$, whose zero set $Z(f)$ in $\mathbb A_{\mathbb R}^2$ is not irreducible.

I think such an example must come from the fact that $\mathbb R$ is not algebraically closed. But I have no idea as to finding a concrete one.

Thanks very much.

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    Is the empty set reducible?2012-07-16
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    Very similar to Mariano's example is the polynomial $f(x, y) = y^2 + x^2 (x - 1)^2$ given in Exercise 1.26 in William Fulton's Algebraic Curves book.2012-07-16
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    Dear @ PseudoNeo, no the empty set is explicitly said to be *non* irreducible in EGA and De Jong's Stack Project, as well as in many other sources.2012-07-16
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    @GeorgesElencwajg: As I wrote “reducible”, I guess your “no” is a “yes“? One of those days, I'll write a linguistics thesis on the mathematicians' jargon. A language where “non irreducible” or “isn't nondegenerate” are common phrases clearly deserves more attention.2012-07-16
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    Dear @Pseudoneo, you are quite right: my "no" is a "yes" and it *is* strange that one says that the empty set is not irreducible rather than reducible! In the same vein algebraic geometers used to say that a variety was nonsingular (and even not nonsingular to emphasize that a singularity was present!) but fortunately Zariski coined the word regular to replace nonsingular in his magnificent creation of a purely algebraic definition for the concept. And if you ever write that linguistics thesis, I'd love to see a copy :-)2012-07-16
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    @GeorgesElencwajg: I don't know about this particular use, but conceivably the empty set is considered not irreducible but not reducible either. I'm pretty certain that in a UFD (or integral domain) the invertible elements (and zero) are considered to be neither irredicible nor composite, which is not too different a situation.2012-07-16

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$f(x,y)=(x^2-1)^ 2+y^2$ works, as far as I can tell.

To show this is irreducible: if $f$ is a product of two non-constant factors $g$ and $h$, the product of the leading forms of $g$ and $h$ is $x^4$. We can assume then that these leading forms are either $x$ and $x^3$ or $x^2$ and $x^2$. In the first case, one of the factors is linear and this is impossible because it would have infinitely many zeroes, and $f$ does not. The factors must then both have $x^2$ as initial form, and $g$ is then of the form $x^2+ax+by+c$ with $a$, $b$ and $c$ reals. If $b$ is not zero, this has infinitely many zeroes and so then so has $f$: we know this is not true; it follows that $g$ depends only on $x$; the same works for $h$. This is absurd.

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    Dear Mariano, wouldn't it be simpler to notice that $f$ seen as a polynomial in $y$ has no root in $\mathbb R[x]$, because the sum of two squares (not both zero) in $\mathbb R[x]$ is non zero ?2012-07-16
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    @GeorgesElencwajg, sure it is! I have spent too much time recently thinking about the shape of polynomials :)2012-07-17
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    Thank you very much for the example. As to the the proof of the irreducibility of $f$, can I do it like this? Firts, factor $f$ as $f=f_1f_2$ in $\mathbb C[x,y]$ where $f_1=(x-1)+iy,f_2=(x-1)-iy$. Clearly both $f_1$ and $f_2$ are irreducible and none of them is in $\mathbb R[x,y]$. Then conclude that $f$ is irreducible in $\mathbb R[x,y]$. Is there anything wrong?2012-07-22
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    I am pretty curious about how did you come up with this example? Could you tell me about that?2017-11-07
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Added: This answer interprets "irreducible" to mean with respect to the classical (analytic) topology on $C(\mathbb{R})$, not the (relative) Zariski topology.

Consider the elliptic curve $y^2 = x^3 + Ax + B$, for $A,B \in \mathbb{R}$ with $\Delta = 4A^3 + 27B^2 \neq 0$.

Then the polynomial $f(x,y) = x^3 + Ax + B - y^2$ is irreducible. However the real zero set of $f$ has two components iff $x^3 + Ax + B$ has three real roots (see e.g. this picture for a good idea of what's going on here). Thus taking for instance $A = -1, B = 0$ gives an answer to the question.

If you want to think about things this way, a natural followup result is Harnack's Theorem: if $C/\mathbb{R}$ is a smooth, geometrically integral projective curve of genus $g$, then the real locus $C(\mathbb{R})$ has at most $g+1$ connected (hence irreducible, by smoothness) components, and all numbers of components between $0$ and $g+1$ are possible. Thus the above example is in some sense the simplest.

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    Why is this curve not connected for the Zariski topology ? Doesn't Harnack theorem involves transcandental topology ?2012-07-16
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    @user10676: thanks, you're absolutely correct. I added a warning to this effect.2012-07-16
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    Thank you very much for the example and the deeper insight.2012-07-22