# Values of c for which the given quotient ring is a field.

I am stuck with the problem :

Find all values of 'c' in $$F_{5}=\frac{\mathbb{Z}}{5\mathbb{Z}}$$ such that the quotient ring $$\frac{F_{5}}{⟨X^3 + 3X^2 + cX + 3⟩}$$ is a field. Justify your answer.

My approach was, we've got a theorem for commutative ring R that if I is a maximal ideal in R then R/⟨I⟩ is a field. Now to prove $$⟨X^3 + 3X^2 + cX + 3⟩$$ is a maximal ideal in the given field we need to show that this is irreducible. So, I think for the set of values of 'c' for which this polynomial is irreducible, will be the set for which the above quotient ring is a field.

But I don't know how to find all the values of 'c' for which $$⟨X^3 + 3X^2 + cX + 3⟩$$ is irreducible except to try each value of 'c' individually and then use some irreducibility test. Is there a proper and simpler way to find such 'c'. Please help me in finding such values.

• You say $c$ some times and $d$ other times. Are they the same? Also, \langle and \rangle ("Left ANGLE" and "Right ANGLE") look much better than < and > as angle brackets: $\langle X^4\rangle$ versus $<X^4>$ (but to be honest, most people eventually just fall back to using regular parentheses for ideals). – Arthur Jun 15 '20 at 9:07
• @Arthur, I've edited that part, thanks – Adam Warlock Jun 15 '20 at 9:12

For $$f(x)=x^3+bx+c$$ over $$\mathbb F_q$$, where $$q=p^n$$ with $$p>3$$, its discriminant is $$D(f)=-4b^3-27c^2$$. Then $$f$$ is irreducible over $$\mathbb F_q$$ if and only if $$D(f)$$ is a square in $$\mathbb F_q$$, say $$D(f)=81d^2$$, and $$\frac12(-c+dw)$$ is a cube in $$\mathbb F_q$$ if $$q\equiv1\pmod3$$, or in $$\mathbb F_{q^2}$$ if $$q\equiv2\pmod3$$, where $$w$$ is a root of $$x^2+3$$ ($$w\in\mathbb F_q$$ if $$q\equiv1\pmod3$$ and $$w\in\mathbb F_{q^2}$$ if $$q\equiv2\pmod3$$).
In your case, $$f(x)=x^3+3x^2+cx+3=x^3=(x+1)^3+(c-3)(x+1)-c+5$$, so we can consider $$g(x)=x^3+(c-3)x-c$$ WLOG. It is kind of complicated, but there is another criterion which is more pratical in this case.
A polynomial of degree $$2$$ or $$3$$ is irreducible over a field $$F$$ if and only if it has no root in $$F$$.
Therefore we can just let $$c$$ vary over $$\mathbb F_5$$. Then $$f$$ is irreducible, if and only if $$f(\alpha)\ne0$$ for all $$\alpha\in\mathbb F_5$$.