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I was browsing through some algebraic geometry problems and came across this :

(1) Let $k$ be an algebraically closed field, and $f \in k\left[x_1, \ldots, x_n\right]$ an irreducible polynomial. Prove that the algebraic set $V(f)$ is irreducible.

(2) This is not true if $k$ is not algebraically closed: give an example of an irreducible polynomial $f \in \mathbb{R}[x, y]$ such that $V(f) \subseteq \mathbb{A}_{\mathbb{R}}^2$ is not irreducible.

So I was surprised that part (2) was possible, because I thought that the multivariate polynomial rings are UFDs so all irreducible elements are prime and we know that the vanishing locus of a prime ideal is irreducible set so how is it possible to have an irreducible polynomial $f \in \mathbb{R}[x, y]$ such that $V(f) \subseteq \mathbb{A}_{\mathbb{R}}^2$ is not irreducible? I don' think any of the relevant theorems involved like the Hilbert basis theorem requires us to have an algebraically closed field.

Bill
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1 Answers1

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Note: If $I:=(f_1,..,f_k) \subseteq \mathbb{R}[x_1,..,x_n]$ is any ideal it follows the polynomial

$$F:=f_1^2+\cdots +f_k^2$$

have the same zero set as $I$: There is an equality $X:=V(I)=V(F)$. Hence any real algebraic variety may be defined as the zero set of a hypersurface, and $F$ is irreducible in many cases.

hm2020
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