How to prove that F is extension of K

Question

I need to prove that filed F is extension of K. I know that F is isomorphic with $K[x]/(m)$, K is also some field and polynomial $m \in K[x]$ which is irreducible over K. I think, if F and K is isomorphic they have to be same. So F could be a trivial extension field. Is it right? Any tips how to prove it?

Answer 1

Since $K\subseteq K[x]$ (actually an isomorphic copy), $K\subseteq K[x]/(m)$, for otherwise there are $a,b\in K$ with $a\neq b$ such that $a-b\in (m)$. But $a-b$ is a unit, so $a-b=0$. Since $K[x]/(m)\supseteq K$, and $K[x]\cong F$, it must be the case that $K\cong E$, where $E\subseteq F$. I don't think that it is required for $F\supseteq K$, just from naming of elements, but you could consider it an extension if you view isomorphic fields as the same.

If you are asking if $K[x]/(m)\cong K$, this is not true in general; consider $\mathbb{Z}_2/(x^2)\cong\mathbb{F}_4$.

Sorry if this didn't answer your question (I would have posted in comments if I had 50 reputation), I am kind of confused on why the fields $E,K$ must be equal instead of isomorphic... You could just rename "1,0" to get an isomorphic field which doesn't contain the "same" elements.