Why are prime ideals crucial in algebraic geometry?

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Homework Statement



Why are prime ideals so important?

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The Attempt at a Solution

 
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Hi Metric_Space :smile:

1) Because they form a generalization for prime numbers, and prime numbers are important.
2) Because they correspond to irreducible curves in geometry. For example, the curve y=x2 is irreducible, and indeed, the ideal (y-x2) is prime in \mathbb{C}[x,y]. On the other, the curve xy=0 is not irreducible since it exists out of the pieces x=0 and y=0. And indeed, the ideal (xy) is not prime.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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