When I was trying to prove the following theorem:
Theorem. Let $E$ be an extension field of the field $F$ and let $a \in E$ .If $a$ is algebraic over $F$, then $F(a) \cong F[x]/ \langle p(x)\rangle$ , where $p(x)$ is a polynomial in $F[x]$ of minimum degree such that $p(a)=0$. Moreover, $p(x)$ is irreducible over $F$.
I considered an homomorphism $\phi: F[x] \longrightarrow F(a) $ such that $\text{Ker} \phi \neq 0$, so $\text{ker} \phi=\langle p(x)\rangle$. As $F[x]$ is a Principal Ideal Domain, we have that $p(x)$ is irreducible.
Now by the First Isomorphism Theorem $F[x]/ \langle p(x)\rangle \cong \text{Im} \phi \leq F(a).$ Now it is claimed that $\langle p(x)\rangle$ is a prime ideal of $F[x]$ becuase $\text{Im}\phi$ is an integral domain.
My question comes from this last sentence, how can be proved that:
Theorem. Every subring R of a field F is an integral domain.
So we have to prove R to be commutative with unit. It is clear that it is commutative because if there exist $a,b\in R$ such that $ab\neq ba$, then like $R\subset F$ we have that $a,b\in F$ are no commutative which is a contradiction.
My problem comes when proving that R necessarily has a unit. So I tried to build my counterexample, let $F=\mathbb{Q}$, we consider $R={2\mathbb{Z}}$ wich is a subring because it is closed under subtraction and multiplication.
Can someone tell my why is my counterexample wrong and how it is proved that $1\in R$?
Your counterexample is great.
What happens is that $\text{Im}\phi$ is an integral domain not only because it is a subring of $F(a)$, from which you obtain commutativity, but also from the fact that $1 \in F[x]/ \langle p(x)\rangle$ . This comes from the fact that $ 1 \in F[X]$ as F is a field and because $$\langle p(x)\rangle = \{r(x)p(x):r(x) \in F[x]\}$$ and $$F[x]/ \langle p(x)\rangle = \{f(x) + r(x)p(x):r(x) \in F[x]\}$$ you choose $r(x)=0 \in F[x]$ and you finally have that $\text{Im}\phi$ is an integral domain so $\langle p(x)\rangle$ is a prime ideal.