3
$\begingroup$

The explanation I have in my notes is

"The affine line $A^1$ is irreducible because it is infinite.

The real line $R$ is not irreducible because it can be written $R = (-\infty, 0] \cup[0, \infty) $"

I understand that in the affine line we "forget" where the origin is, or where we "are" on the line, but whats to stop be picking a random point on the affine line and writing it as the sum of two disjoint sets, just like the real line?

Thanks!

$\endgroup$
0

1 Answer 1

Reset to default
5
$\begingroup$

That's $\Bbb R$ in the standard topology that's reducible. The Zarisky topology is far from the standard (in fact, on the affine line it is the cofinite topology).

$\endgroup$
2
  • $\begingroup$ I understand that for the affine line, it has the cofinite topology, but I still cant see why does this mean it cannot be "split" into two pieces? Although I will point out I am convinced it is true, by the fact $0 = I(A^n)$ is prime in A, thus $A^n$ is irreducible. But of course being convinced is not the same as understanding! $\endgroup$
    – Max Boxer
    Oct 12, 2016 at 12:56
  • 3
    $\begingroup$ A set is irreducible if it cannot be written as the union of two closed, proper subsets. But all the proper, closed subsets of the affine line are finite, so the union of two such sets is still finite, and therefore cannot be the whole line. That is why $A^1$ is irreducible. $\endgroup$
    – Arthur
    Oct 12, 2016 at 16:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .