if V(f) is irreducible, then how to show that the polynomial f itself is irreducible? [closed]

Question

V(f) is the zero locus of the polynomial f in the polynomial ring k[x1, x2, ..., xn] with k an algebraically closed field.

If V(f) is irreducible, then how to show that 'f' is irreducible?

Answer 1

This is not true as stated. Let $n=1$ and consider $V(x_1^2)=\{0\}$. This subvariety is irreducible, but $x_1^2$ is reducible.

Answer 2

If $f=x_1^2$ then $V(f)$ is irreducible, but $f$ is not.

Answer 3

As pointed out, the statement is false as written. To get a true statement, you must assume also that $f$ is square-free (or equivalently that the scheme $V(f)$ is reduced). Then, if you assume that $f = gh$ has a non-trivial factorization, by the square-freeness, $g$ and $h$ are distinct polynomials even up to multiplying by a non-zero constant, so $V(f) = V(g) \cup V(h)$, contradicting the irreducibility of $V(f)$. Note that without the square-freeness hypothesis, it would be possible for $g = h$ and so $V(g) = V(h)$ (as in the counterexamples given), and then there is no contradiction to the irreducibility of $V(f)$.