Let $I$ be a homogeneous ideal in $R = \mathbb{C}[x_0, \ldots, x_n]$ and let $V := V(I)$ be its set of common zeroes, viewed as a subset of $\mathbb{P}^n$. $I$ being homogeneous means that it is generated by finitely many homogeneous polynomials. However this $I$ is special in that we are promised that
$I$ is generated by a finite set of polynomials that are all homogeneous of the same degree $d$.
First question: do ideals with this property have a name?
Now for the main question.
We define $\mu$ as the smallest number $m$ such that $V$ can be defined as the set of common zeroes of $m$ homogeneous degree $d$ polynomials $f_1, \ldots, f_m$.
[From an a priori perspective it is not clear that given such a minimal set $\{f_1, \ldots, f_\mu\}$ of degree-$d$-polynomials with $V(f_1, \ldots, f_\mu) = V$, the $f_i$ should necessarily lie in the ideal $I$ initially used to define $V$. However I think I can show that in fact they do. Conversely however $I$ might contain some extra elements not strictly necessary in defining $V$ and in particular have more than $\mu$ generators.]
Even if the part in square brackets contains some error it is still clear from the definition of $\mu$ that the value of $\mu$ is a property of the geometric object $V$ which is independent of the precise choice (from possibly multiple options) of the ideal $I$ used to define $V$ in the first place. Hence the following question.
Main question: Given $V$, the promise that it is of the form described in the beginning of the post, and the value of $d$, can we compute/count/deduce the value of $\mu$ by purely geometric means? I.e. by staring intensely at some picture of $V$, counting its number of irreducible components, perhaps intersecting $V$ with a cleverly chosen number of hyperplanes and counting the components of these intersections, or something even cleverer I haven't thought of?
(EDIT: in my example at the end of the post of how to tackle a special case I do allow some algebra as part of the algorithm, but it is really simple algebra that can be carried out by primary-schoolers as long as you don't tell them the back-story.)
Given that $V$ uniquely defines $\mu$, the answer to the question is almost certainly 'yes', so perhaps a better version of the question is how can we construct the value of $\mu$ from $V$ (and knowledge of $d$) geometrically?
I can see how to distinguish the case $\mu = 1$ from the case $\mu > 1$ geometrically (I will type this below the next horizontal line, to show 'research effort' but feel free to ignore if you already can guess how it works). However this hinges on two facts:
$\mu$ is defined as a minimum, and 1 is the smallest number in existence, so if $V$ can be written as $V(f)$ for a single polynomial $f$ of degree $d$, then we will have $\mu = 1$ and
I have a rather good idea of what sets of the form $V(f)$ look like.
Neither fact generalizes easily (for me at least) to the case of higher values of $\mu$ so any help is welcome here.
Concretely my understanding of the $\mu = 1$ case stems from the following lemma. (If I am mistaken and the lemma is false, please let me know!)
Lemma: A projective algebraic set $W$ is irreducible and of co-dimension one if and only if is of the form $V(g)$ for some irreducible homogeneous polynomial $g$.
This is helpful for understanding non-irreducible sets as well.
If indeed $\mu = 1$ we have that $V = V(f)$ for some homogeneous degree-$d$-polynomial $f$. Since $R$ is a UFD we can write $f = g_1^{n_1} \cdot g_2^{n_2} \cdots g_k^{n_k}$ for some distinct, irreducible polynomials $g_i$ and natural numbers $k, n_1, \ldots, n_k \geq 1$. With a little bit of thinking we conclude that the $g_i$ must be homogenous as well and their respective degrees, $d_1, \ldots, d_k$ must satisfy
$$d_1n_1 + \ldots + d_kn_k = d. \qquad (1)$$
Now
$$V = V(f) = \bigcup_{i=1}^k V(g_i) \qquad (2)$$ is the union of the zero-loci of the $g_i$ which by the lemma are irreducible varieties of co-dimension 1.
So what I would do to tell if $\mu = 1$ or $\mu > 1$ is first look at the irreducible components of $V$. If any of them is of co-dimension bigger than 1, we know from the above that $\mu$ must be bigger than 1 as well.
If all of them have co-dimension 1, we look at the irreducible components one by one. By the lemma, each of them is of the form $V(g_i)$ for some irreducible homogeneous polynomial $g_i$ and we can determine the degree of $g_i$ (without ever knowing $g_i$ itself) by shooting the irreducible component by hyperplanes and counting the number of intersections, or so Wikipedia tells me.
This gives a sequence of degrees $d_1, \ldots, d_k$. Now if equation $(1)$ above doesn't have any solution $(n_1, \ldots, n_k)$ in positive integers then the above tells us that $\mu > 1$. And if it does have a solution, we can use this solution to show that $\mu = 1$.
After all, even if we don't know the precise value of the $g_i$ we know that they exist, and using our solution $(n_1, \ldots, n_k)$ to $(1)$ , we can define $f = g_1^{n_1}\cdots g_k^{n_k}$ and conclude that $f$ is homogeneous of degree $d$ and, by construction, satisfies $(2)$, showing that indeed $\mu = 1$.
Very nice of course, but as I wrote above I don't see how to extend this algorithm to the point where if it says that $\mu > 1$, I can next distinguish between $\mu = 2$ and $\mu > 2$ and so on. Any help is appreciated.