How to tell if algebraic set is a variety?

Question

I've been reading some basic classical algebraic geometry, and some authors choose to define the more general algebraic sets as the locus of points in affine/projective space satisfying a finite collection of polynomials $f_1, \dots, f_m$ in $n$ variables without any more restrictions. Then they define an algebraic variety as an algebraic set where $(f_1, \dots, f_m)$ is a prime ideal in $k[x_1, \dots, x_n]$.

My question has two parts:

1. I'm guessing the distinction is like any other area of math where you try to break things up into the "irreducible" case and deduce the general case from patching those together. How does that happen with varieties and algebraic sets? Is it correct to conclude that every algebraic set is somehow built from algebraic varieties since the ideal $(f_1, \dots, f_m)$ is contained in some prime (maximal) ideal?

2. How can one tell whether or not an algebraic set is a variety intuitively? I know formally you'd have to prove $(f_1, \dots, f_m)$ is prime (or perhaps there are some useful theorems out there?), but many times in texts the author simply states something is a variety without any justification. Is there a way to sort of "eye-ball" varieties in the sense that there are tell-tale signs of algebraic sets which are not varieties?

Perhaps this is all a moot discussion since modern algebraic geometry is done with schemes and this is perhaps a petty discussion in light of that, but nonetheless, I'd like to understand the foundations before pursuing that.

Thanks.

Answer 1

It is true that every algebraic set is a finite union of algebraic varieties (irreducible algebraic sets), and this union is unique up to reordering. These irreducible pieces of an algebraic set are called the irreducible components. This all follows from the fact that a polynomial ring over a field is Noetherian, so that an algebraic set with the Zariski topology is a Noetherian topological space.

As an example, I always think of the algebraic set defined by the ideal $(xz,yz),$ which is not prime. The real picture of this algebraic set is a line through a plane, and these two objects are exactly the irreducible components of the algebraic set. Here is the picture:

$\hspace{2.2in}$

The components are defined by the prime ideals $(z)$ and $(x,y)$ which are the two minimal prime ideals containing $(xz,yz)$. This may be the eyeball test you desire, as most people would look at this set and say it is made of two parts. In general, the irreducible components of an algebraic set defined by an ideal $I$ correspond exactly to the minimal prime ideals containing $I$.

Concerning your second question, it is not easy in general to determine when an ideal is prime. I asked a question here seeking different techniques to detect when ideals are prime. It is often easier to see that an ideal is not prime, as in the example I've given.

Answer 2

a) A useful trick for showing irreducibility of an algebraic set $X$ is to exhibit an open dense subset $X_0\subset X$ which is known to be irreducible.
In particular the closure $X\subset \mathbb P^n$ of any algebraic irreducible subset $X_0\subset \mathbb A^n$ is irreducible.

For example the intersection $C$ of the three quadrics $xw-yz=0, y^2-xz=0, z^2-yw=0$ in $ \mathbb P^3$ (="twisted cubic curve") is irreducible because it is the closure of the intersection $C_0\subset \mathbb A^3$ of the three affine quadrics $x-yz=0, y^2-xz=0, z^2-y=0$ in $ \mathbb A^3$ and $C_0$ is clearly irreducible as it is parametrized by $x=z^3, y=z^2,z=z$ , i.e. is the image of $\mathbb A^1$ under $z\mapsto (z^3,z^2,z)$ [See below].
Note that this is far from trivial: the intersection of any two of the three projective quadrics above is reducible!

b) Another useful trick is that the image of an irreducible algebraic set under a morphism is irreducible too.
For example the above twisted cubic curve $C$ is irreducible because it is the image of $ \mathbb P^1$ under the morphism $\mathbb P^1\to \mathbb P^3: (u:v)\mapsto (u^3:u^2v:uv^2:v^3)$

Answer 3

Here are some personal reflections on the role of schemes in algebraic geometry, commenting on your very interesting remark: "Perhaps this is all a moot discussion since modern algebraic geometry is done with schemes".

Grothendieck introduced scheme theory in the late nineteen fifties and the high level of abstraction of that theory had a discouraging effect on even great mathematicians like Néron.
Soon however the incredible power of these new tools allowed people like Deligne, Grothendieck and Faltings to solve problems untouchable by classical methods: Grothendieck and Deligne solved all of the Weil conjectures and Faltings solved conjectures of Mordell, of Shafarevich and of Tate.
Fields medals were of course awarded to Grothendieck, Deligne and Faltings and several other medals were won by mathematicians whose work involved in a fundamental way scheme theory: Lafforgue, Mumford, Ngô, Voevodski,...

For us lesser mortals scheme theory has now become quite accessible thanks to the didactic efforts of pioniers like Mumford and Hartshorne and then thanks to the excellent textbooks by Eisenbud-Harris, Liu, Görtz-Wedhorn,... and on-line notes like Vakil's splendid Foundations of Algebraic Geometry.

Algebraic varieties retain an enormous place in contemporary algebraic geometry.
However an overwhelming part of the results in that field are obtained with the help of scheme theory.
So every algebraic geometer must at some time learn scheme theory, but the best way to approach algebraic geometry is through one of the many introductory books written in the classical language by experts (who of course also master scheme theory!) like Fulton, Hulek, Perrin, Reid, ...