Why is x^3 + 2x + 3 irreducible over Z5[x]?
And how can I express it as a product of polynomials?
And how can I express it as a product of polynomials?
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If x^3 + 2x + 3 factors nontrivially in Z_5[x], one of the factors has to have degree 1. (The only nontrivial way you can get a degree of 3 product is if you're multiplying a degree 1 polynomial by a degree 2 polynomial.)
A degree 1 factor of this polynomial would give rise to a _root_ of this polynomial in Z_5[x]. The reason for this is that if
x^3 + 2x + 3 = (ax + b) q(x) for some a, b in in Z_5[x] with a nonzero and q(x) a quadratic in Z_5[x]
then x^3 + 2x + 3 would be 0 when ax + b = 0, or when x is the element -b/a in Z_5 (the calculation of -b/a, of course, being done in Z_5). And this whole argument is reversible: if you can find an element c of Z_5 for which c^3 + 2c + 3 = 0 in Z_5, then your polynomial has to be divisible by x - c. So to test whether or not this polynomial is irreducible, we need only see if it has roots.
When you evaluate this polynomial at 0, you get 3, which is nonzero.
When you evaluate this polynomial at 1, you get 1 + 2 + 3 = 6, which is the same thing as 1 in Z_5, and still nonzero.
When you evaluate this polynomial at 2, you get 2^3 + 2*2 + 3 = 8 + 4 + 3 = 15, which _is_ 0 in Z_5. So (x - 2) is a factor of our polynomial! And x^3 + 2x + 3 is *not* irreducible over Z_5[x].
When you divide x^3 + 2x + 3 by x - 2 in Z_5[x] you get a quotient of x^2 + 2x + 1, so
x^3 + 2x + 3 = (x^2 + 2x + 1)(x - 2) in Z_5[x].
To do this you just use the usual polynomial long division algorithm, with the arthmetic done in Z_5 (e.g. so that -2 is the same thing as 3, and 6 is the same thing as 1, etc). To see how x^2 + 2x + 1 reduces we just need to look for more roots of it. The values of x^2 + 2x + 1 when x = 0, 1, 2, 3, 4 are 1, 4, 9, 16, and 25 respectively, which are the same as 1, 4, 4, 1, and 0 in Z_5. So 4 is a root of x^2 + 2x + 1 in Z_5[x], and (x - 4) has to divide x^2 + 2x + 1. If you do the polynomial long division of x^2 + 2x + 1 by x - 4 in Z_5[x], you get a quotient of x + 1, so that
x^2 + 2x + 1 = (x - 4) (x + 1) in Z_5[x].
This means your original polynomial factors completely as
x^3 + 2x + 3 = (x - 2) (x - 4) (x + 1).
If you remember that -2 = 3 and -4 = 1 in Z_5 you could rewrite this without any minus signs. This might be the simplest way of writing your original polynomial as a product of polynomials:
x^3 + 2x + 3 = (x + 3)(x + 1)^2 in Z_5[x].
[So in fact x = -1 (= 4) is a double root of this polynomial. As -4 = 1 in Z_5, the factorization x^2 + 2x + 1 = (x - 4)(x + 1) we found in Z_5[x] showed also that x^2 + 2x + 1 = (x + 1)^2 in Z_5. Which you may have noticed earlier anyway, because in fact x^2 + 2x + 1 = (x + 1)^2 in the integers.]
A degree 1 factor of this polynomial would give rise to a _root_ of this polynomial in Z_5[x]. The reason for this is that if
x^3 + 2x + 3 = (ax + b) q(x) for some a, b in in Z_5[x] with a nonzero and q(x) a quadratic in Z_5[x]
then x^3 + 2x + 3 would be 0 when ax + b = 0, or when x is the element -b/a in Z_5 (the calculation of -b/a, of course, being done in Z_5). And this whole argument is reversible: if you can find an element c of Z_5 for which c^3 + 2c + 3 = 0 in Z_5, then your polynomial has to be divisible by x - c. So to test whether or not this polynomial is irreducible, we need only see if it has roots.
When you evaluate this polynomial at 0, you get 3, which is nonzero.
When you evaluate this polynomial at 1, you get 1 + 2 + 3 = 6, which is the same thing as 1 in Z_5, and still nonzero.
When you evaluate this polynomial at 2, you get 2^3 + 2*2 + 3 = 8 + 4 + 3 = 15, which _is_ 0 in Z_5. So (x - 2) is a factor of our polynomial! And x^3 + 2x + 3 is *not* irreducible over Z_5[x].
When you divide x^3 + 2x + 3 by x - 2 in Z_5[x] you get a quotient of x^2 + 2x + 1, so
x^3 + 2x + 3 = (x^2 + 2x + 1)(x - 2) in Z_5[x].
To do this you just use the usual polynomial long division algorithm, with the arthmetic done in Z_5 (e.g. so that -2 is the same thing as 3, and 6 is the same thing as 1, etc). To see how x^2 + 2x + 1 reduces we just need to look for more roots of it. The values of x^2 + 2x + 1 when x = 0, 1, 2, 3, 4 are 1, 4, 9, 16, and 25 respectively, which are the same as 1, 4, 4, 1, and 0 in Z_5. So 4 is a root of x^2 + 2x + 1 in Z_5[x], and (x - 4) has to divide x^2 + 2x + 1. If you do the polynomial long division of x^2 + 2x + 1 by x - 4 in Z_5[x], you get a quotient of x + 1, so that
x^2 + 2x + 1 = (x - 4) (x + 1) in Z_5[x].
This means your original polynomial factors completely as
x^3 + 2x + 3 = (x - 2) (x - 4) (x + 1).
If you remember that -2 = 3 and -4 = 1 in Z_5 you could rewrite this without any minus signs. This might be the simplest way of writing your original polynomial as a product of polynomials:
x^3 + 2x + 3 = (x + 3)(x + 1)^2 in Z_5[x].
[So in fact x = -1 (= 4) is a double root of this polynomial. As -4 = 1 in Z_5, the factorization x^2 + 2x + 1 = (x - 4)(x + 1) we found in Z_5[x] showed also that x^2 + 2x + 1 = (x + 1)^2 in Z_5. Which you may have noticed earlier anyway, because in fact x^2 + 2x + 1 = (x + 1)^2 in the integers.]