Question about quotients of $\mathbb{Z}[x]$

Question

How can we prove using basic notions of algebra, that there are no quotients of $ \mathbb{Z}[x] $ which are fields with 72 elements, without using for example the fact that a finite field has a power of a prime number as cardinality?

Thank you!

Answer 1

You can avoid linear algebra, though you'll be dancing carefully around it the whole time.

Suppose $\mathbb{Z}[x]/I$ is a field with 72 elements. Consider $J = \mathbb{Z} \cap I$. This is an ideal in $\mathbb{Z}$, so $J = n\mathbb{Z}$ for some $n$. The surjective map $\mathbb{Z}[x] \to \mathbb{Z}[x]/I$ descends to a surjection $\mathbb{Z}[x]/\langle n\rangle \to \mathbb{Z}/I$. Since $\mathbb{Z}[x]/\langle n\rangle \cong \mathbb{Z}/n\mathbb{Z}[x]$, we have a surjection $\mathbb{Z}/n\mathbb{Z}[x] \to \mathbb{Z}[x]/I$. If $n$ were composite, running the two factors through this map would force one of them to be zero, so $n=p$ is prime and $\mathbb{Z}/n\mathbb{Z} = \mathbb{F}_p$ is a field of prime order. Now $\mathbb{F}_p[x]$ is a PID, so the kernel of $\mathbb{F}_p[x] \to \mathbb{Z}[x]/I$ is some principal ideal $\langle f\rangle$. Hence we have that $\mathbb{F}_p[x]/\langle f\rangle$ is a field with 72 elements. We may suppose $f$ is monic of degree $d$.

What is the order of $\mathbb{F}_p[x]/\langle f\rangle$? Consider the polynomials of the form $\{c_0 + c_1 x + \cdots + c_{d-1} x^{d-1} : c_i \in \mathbb{F}_p\}$. There are $|\mathbb{F}_p|^d = p^d$ of these. None of these polynomials are equal in the quotient since their difference has too small of a degree to be divisible by $f$. Moreover, every polynomial of degree $d$ or higher can be reduced to one of lower degree since $f$ is monic of degree $d$. So these polynomials are representatives of the quotient, and $p^d = 72$. This is a contradiction since $72$ is not a power of a prime.