Answer to Question #188466 in Abstract Algebra for Ravinder singh

Question

check whether or not q[x]/<8x³+6x²-9x+24> is a field.

Answer 1

It is sufficient to check that the polynomial f(x)=8x³+6x²-9x+24 is irreducible over Q. Check this assertion.

If f(x) is reducible, f(x)=g(x)h(x), then the images of g(x) and h(x) in Q[x]/(x³+3x²-18x+192) would be divisors of zero, since f(x) is a zero in this ring. Therefore this ring contains divisors of zero and cannot be a field.

If f(x) is irreducible then for every polynomial g(x), which is not divisible by f(x), the ideal (f(x),g(x)) contains a unit. Indeed, since Q[x] is a ring of principal ideals, the ideal (f(x),g(x)) must be principal, i.e. (f(x),g(x))=(d(x)). Then d(x) is a divisor of f(x). But f(x) is irreducible. If this divisor is not a unit then it is conjugate to f(x), and then g(x), being divisible by d(x) must be divisible by f(x) which is not the case.

If (f(x),g(x)) contains a unit, therefore f(x)u(x)+g(x)v(x)=1 for some polynomials u(x) and v(x). Since f(x)u(x)=0 in Q[x]/(f(x)), this means that v(x)=1/g(x) in Q[x]/(f(x)), and that g(x) is invertible in this ring.

So, all non-zero elements of Q[x]/(f(x)) are invertible and Q[x]/(f(x)) is a field.

The polynomial f(x) is irreducible over Q if and only if the polynomial 8f(x/4) is irreducible over Q. (Indeed, f(x)=g(x)h(x) is reducible iff 8f(x/4)=8g(x/4)h(x/4) is reducible).

8f(x/4)=x³+3x²-18x+192 is the polynomial over Z, and the leading coefficient of it equals to1. By Gauss's lemma such a polynomial is irreducible over Q if and only if it is irreducible over Z.

If 8f(x/4) is reducible over Z, then one of its divisors (of the least non-zero degree) must be of degree 1. Otherwise, if both the divisors have degree 2 or more, then degree of 8f(x/4) would be 4 or greater. This means 8f(x/4) is reducible over Z if and only if it has an integer root a.

Then a³=-3a²+18a-192 is divisible by 3, hence a is divisible by 3. therefore a³+3a²-18a is divisible by 27, that is 192 is divisible by 27. But this is not true. The contradiction proves that polynomial x³+3x²-18x+192 is irreducible over Z, and the polynomial 8x³+6x²-9x+24 is irreducible over Q. Therefore, Q[x]/(x³+3x²-18x+192) is a field.