Let $\mathbb{K}$ be a field (not necessarily algebraically closed). Let $f\in \mathbb{K}[x_1,\dots,x_n]$ be an irreducible polynomial. It is true that $I(V(f)) = (f)$? Where as usual $V(f)$ is the algebraic set $V(f)=\{x\in \mathbb{K}^n| \ f(x) = 0 \}$ and $I(V(f))$ is the ideal of all polynomials vanishing on $V(f)$.
It is clearly true if we assume $\mathbb{K}$ algebraically closed, indeed thanks to Nullstellensatz $I(V(f)) = \sqrt{(f)} = (f)$ since $f$ is irreducible.
Unfortunately, in general we only know that $(f)\subset I(V(f))$ and we can assume $I(V(f))$ to be irreducible since $ \mathbb{K}[x_1,\dots,x_n] $ is Noetherian then $I(V(f))$ is intersection of $m\in \mathbb{N}$ irreducible ideals but if $m>1$ then $(f)$ irreducible implies that $I(V(f))=(f)$.
Can anyone help me in finding a counterexample or proving the statement? Thank you in advance.
Edit:
As pointed out by G.Sassatelli the general statement is false. I wonder if there are some basic assumptions under which it is true. For example, what happens if we assume that $V(f)\neq \emptyset$?
It is not.
Usual example: $\Bbb K=\Bbb R$, $n=1$, $f=x^2+1$. $$I(V(f))=I(\emptyset)=(1)\neq (x^2+1)$$
More in general, if $\Bbb K\neq\overline{\Bbb K}$, the claim "$I(V(f))=(f)$" fails for $n=1$, because any irreducible polynomial $f\in\Bbb K[x]$ of degree $>1$ gives the same issue: if $f(\alpha)=0$, then $x-\alpha\mid f$.
After edit: For a counterexample where $V(f)\ne\emptyset$, there is $\Bbb K=\Bbb R$, $f=x_1^2+x_2^2+\cdots +x_n^2$. $V(f)=\{0\}$, hence $IV(f)=(x_1,x_2,\cdots, x_n)$. It's kind of the same example, but with the projective Nullstellensatz.
A similar thing can be done in any field such that $\Bbb K\ne \overline{\Bbb K}$ for $n=2$ in the following way: pick $f\in\Bbb K[x]$. The polynomial $\tilde f\in \Bbb K[x,y]$ defined as $$ \tilde f(x,y)=y^{\deg f} f\left(\frac xy\right)$$ is irreducible in $\Bbb K[x,y]$ if and only if $f$ is irreducible in $\Bbb K[x]$ (and, indeed, the factors of $\tilde f$ are of the form $y^{\deg g}g\left(\frac xy\right)$, where $g\mid f$ ). In that case too, $I(V(f))=(x,y)$.
Once you have a counterexample $f(x_1,x_2)$ with $V(f)\ne \emptyset$ for $n=2$, you can just do the case $n>2$ by considering $f$ as a polynomial in $\Bbb K[x_1,\cdots,x_n]$, and its zero-set will just be $V_{\Bbb K^n}(f)=V_{\Bbb K^2}(f)\times \Bbb K^{n-2}$. It will still hold $$I_{\Bbb K[x_1,\cdots,x_n]}V_{\Bbb K^n}(f)=(x_1,x_2)_{\Bbb K[x_1,\cdots,x_n]}$$