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Prime applies to objects in any commutative algebraic system that can't be further decomposed. I don't want to get too technical so I'm speaking in very loose

Prime applies to objects in any commutative algebraic system that can't be further decomposed. I don't want to get too technical so I'm speaking in very loose terms. x^2+1 is prime over the integers but not over the complex numbers. I don't know the French education system but it would be very shocking to me if the discriminant is taught when factoring is first taught. Factoring x^3 - y^3 does not have anything to do with the discriminant and x^3 - y^3 is typically taught when factoring is first introduced. At least in the U.S.

R_Edward 14y ago
x^2 + 1 is prime over the integers? Maybe they've changed math since I was in high school, but I'm pretty sure that, given n, such that n is an integer, if x = (2n + 1), then x^2 + 1 = 4n^2 + 4n + 2, which equals 2(2n^2 + 2n + 1), which is clearly not* prime.
- csense 14y ago
  When speaking of factoring polynomials, any constant factor is considered a unit, which doesn't count toward whether a polynomial is prime (irreducible) or not. In the integers, 1 and -1 are units. So 7 is prime even though it can be "factored" as 7 = 7(-1)(-1) or 7 = (7)(1). Likewise, 4n^2 + 4n + 2 is prime, even though it can be "factored" as (2)(2n^2 + 2n + 1). Also, your method of substituting a variable is suspect. Being irreducible is a property of a particular polynomial; if you substitute a different expression for x and simplify, you have a different polynomial. Your argument is like saying, "7 is prime, but if you add three to it, it becomes 10 which isn't prime, how can this be?" -- you did something to the number 7 which turned it into a different number, so it should be no surprise that its properties may have changed in the process. If you want to know about polynomial factorization over various fields, any good abstract algebra textbook should discuss the topic at great length.
  - R_Edward 14y ago
    Unlikely you'll see this, so long after the fact, but oh well. My "method of substituting a variable" is nothing more than a more mathematically rigorous way of observing that x^2 + 1 is not prime for any of the odd integers, which is roughly half of them. Of course, that's using a definition of "prime" that is at odds with yours, and I'll defer to your more advanced math knowledge in that regard.

R_Edward 14y ago

x^2 + 1 is prime over the integers? Maybe they've changed math since I was in high school, but I'm pretty sure that, given n, such that n is an integer, if x = (2n + 1), then x^2 + 1 = 4n^2 + 4n + 2, which equals 2(2n^2 + 2n + 1), which is clearly not* prime.

csense 14y ago
When speaking of factoring polynomials, any constant factor is considered a unit, which doesn't count toward whether a polynomial is prime (irreducible) or not. In the integers, 1 and -1 are units. So 7 is prime even though it can be "factored" as 7 = 7(-1)(-1) or 7 = (7)(1). Likewise, 4n^2 + 4n + 2 is prime, even though it can be "factored" as (2)(2n^2 + 2n + 1). Also, your method of substituting a variable is suspect. Being irreducible is a property of a particular polynomial; if you substitute a different expression for x and simplify, you have a different polynomial. Your argument is like saying, "7 is prime, but if you add three to it, it becomes 10 which isn't prime, how can this be?" -- you did something to the number 7 which turned it into a different number, so it should be no surprise that its properties may have changed in the process. If you want to know about polynomial factorization over various fields, any good abstract algebra textbook should discuss the topic at great length.
- R_Edward 14y ago
  Unlikely you'll see this, so long after the fact, but oh well. My "method of substituting a variable" is nothing more than a more mathematically rigorous way of observing that x^2 + 1 is not prime for any of the odd integers, which is roughly half of them. Of course, that's using a definition of "prime" that is at odds with yours, and I'll defer to your more advanced math knowledge in that regard.