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.

  • $\begingroup$ 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). $\endgroup$ – Arthur Jun 15 '20 at 9:07
  • $\begingroup$ @Arthur, I've edited that part, thanks $\endgroup$ – Adam Warlock Jun 15 '20 at 9:12

Actually, there is a theorem by Dickson to decide whether a cubic polynomial is irreducible over a finite field.

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 $\alpha\in\mathbb F_5$.


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