Is $f\in \mathbb{C}[x]$ a non-unit if $\mathbb{C}[x]/(f)$ is a field?

Question

I am trying to write a proof showing that, with $f\in \mathbb{C}[x]$ and $R=\mathbb{C}[x]/(f)$, $R$ being a field $\implies$ $f$ linear.

I have written a rough version of my proof however going back through this and writing up a neat version I see that it hinges on $f$ being a non-unit which I am not sure how to show (or even if I can show this). Is this possible?

The proof I am writing is not what I am enquiring about, I simply want to know if I can get "$f$ is a non-unit" or "$\partial f>0$" from $f \in \mathbb{C}[x]$, $f\neq 0$, $R=\mathbb{C}[x]/(f)$ a field. In my course the definition provided for a field requires that we do not take a ring $R=\{0\}$ to be a field so please answer with this constraint in mind.

Answer 1

If $S$ is a (unital) ring (here, $S=\mathbb C[x]$) and $f\in S$ is a unit, then $\langle f\rangle=\langle 1\rangle=S,$ and $S/\langle f\rangle=\{0\}.$

So at heart, the question is whether $\{0\}$ is a field.

Now, there are reasons to define $\{0\}$ as a field. If we do define it to be a field, then your statement isn't true, because $f=1$ is an example of a non-linear polynomial with $\mathbb C[x]/\langle f\rangle$ is a field.

When we define $\{0\}$ to be a field, we usually refer to the fields where $1\neq 0$ as non-trivial fields.

Then the corrected statement would either be:

If $f\in\mathbb C[x]$ and $\mathbb C[x]/\langle f\rangle$ is a non-trivial field, then $f$ is linear.

Or:

If $f\in\mathbb C[x]$ such that $\mathbb C[x]/\langle f\rangle$ then $\deg f\leq 1.$

But I think you can assume in this statement that $\{0\}$ is not considered to be a field.

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The final thing is, of course, that if $F$ is a field, then the units of $F[x]$ are exactly the constant non-zero polynomials.