# Show that $H = \mathbb{Z}_5[x]/\langle x^4+3x^3+x+4\rangle$ is not a field.

So I am looking over old exams in abstract algebra and I came across this question which seems to be a mistake. (Neither the original teacher who wrote it, nor my own teacher are available to answer)

Let $H = \mathbb{Z}_5[x]/\langle x^4+3x^3+x+4\rangle$. Show that $H$ is not a field.

Letting $p(x) = x^4+3x^3+x+4$, we can see that $p(x)$ has no solutions in $\mathbb{Z}_5$. Therefore $p(x)$ is irreducible over $\mathbb{Z}_5$, and thus $\langle p(x)\rangle$ is a maximal ideal. Now the factor group of a ring and a maximal ideal is a field. Thus H must be a field.

Am I wrong here, and if so how?

• Just because a polynomial has no roots does not mean it is irreducible. If you already know that $H$ shouldn't be a field, you should aim to factor your polynomial into two quadratics: since it has no roots, it cannot factor into a cubic and a linear term. – Pedro Tamaroff May 29 '16 at 11:43
• Hint: That quartic happens to be the square of a quadratic. This is atypical. More typically an irreducible quartic would be the product of two irreducible quadratics. Do note that many a factorization algorithm of polynomials over a finite field begins by eliminating the possibility of multiplicity $\ge2$ factors. This can be done efficiently by calculating $\gcd(f,f')$ with Euclid's algorithm. – Jyrki Lahtonen May 29 '16 at 11:55

By the Berlekamp algorithm we obtain $$x^4+3x^3+x+4=(x^2 + 4x + 2)^2$$ over $\mathbb{F}_5$. Hence the quotient is not a field, because it has zero divisors.
• @Chris $q(x)=x^2+4x+2$ will be a zero divisor, since $(q(x))^2=x^4+3x^3+x+4=0$. – lisyarus May 29 '16 at 12:11