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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.

  • Show what you have tried so far, and explain where the problem is. Also please clarify the actual problem. Did I understand correcly that the problem is - "If $\mathbb{C}[x]/(f)$ is a field, then is it true that $f$ must be linear?" ? Units in $\mathbb{C}[x]$ are the non-zero constants. The only difficult part to prove is that $f$ cannot be of a higher degree, and here you can use the fundamental theorem of algebra. – Kolja Apr 12 '23 at 13:15
  • The ring $R/I$ being a field is equivalent $I$ satisfying certain properties, which may be helpful. In any case, you should present whatever proof you have in your solution – Bailey Apr 12 '23 at 13:17
  • Technically, we sometimes call ${0}$ a field, which is what you'd get if $f$ is a unit. So some would say this is true if $R$ is a non-trivial field. – Thomas Andrews Apr 12 '23 at 13:19
  • I apologise for the confusion, however I want to say that the greater problem that I am trying to solve is not relevant to this question whereas the title is. I am trying to progress with as little aid as possible and by using what I have covered so far in the course I am taking. The real thing I am asking here is: given that $f\in \mathbb{C}[x]$ is non-zero and that $\mathbb{C}[x]/(f)$ is a field, can I show that $f$ is a non-unit? –  Apr 12 '23 at 13:23
  • The main reason we sometimes don't call ${0}$ a field is that it breaks the theorem that a homomorphism of fields $K\to L$ is always an inclusion. The main reason we do sometimes call it a field is that the category of (unital) rings is in some sense a "more natural" space to work in, and the ${0}$ ring is more often included. – Thomas Andrews Apr 12 '23 at 13:24
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    In general, if $S$ is a ring and $u\in S$ is a unit, then $S/\langle u\rangle={0}.$ So whether we treat ${0}$ as a field here is the key. I would say, for this problem, we are talking specifically about non-trivial fields, because it is not true otherwise. – Thomas Andrews Apr 12 '23 at 13:27
  • Thank you, I think this is what I needed to see. The definition of a field given in the course is as follows: "A ring $R\neq {0}$ is called a field if every non-zero element of $R$ is a unit." so I must have that $R$ as in the title is not equal to ${0}$ since it is a field. –  Apr 12 '23 at 13:33
  • I removed the useless "$f\ne0$" from your title (since $0$ is a non-unit). You can remove it as well from your last sentence (since $\Bbb C[x]/(0)$ is not a field). – Anne Bauval Apr 12 '23 at 13:40

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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.

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.

Thomas Andrews
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  • Thanks, this is very helpful. I haven't seen the result "for a ring $S$, if $f\in S$ is a unit then $S/\langle f \rangle={0}$". I'm mainly having difficulty seeing this is the case when $f\neq 1$ since for $\mathbb{C}[x]$ the set of units would include all nonzero constant polynomials. If, for example, we took $f=3$ would $S/\langle f \rangle=\mathbb{C}[x]/3\mathbb{C}[x]={0}$ still? –  Apr 12 '23 at 16:38
  • If $f$ is a unit, $gf=1$ for some $g\in S,$ hence $s=(sg)f\in\langle f\rangle$ for any $s\in S.$ – Thomas Andrews Apr 12 '23 at 17:49
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