I've been solving some problems from my Galois Theory course, and I need help with the final step of this one:
Prove $\text{Hom}(K,K)=\{0,id_k\}$ for $K=\mathbb{Q}$.
The work I did so far:
Given $K$ is a field, it's only ideals are the trivial ones, to be said, $0$ and $K$. So, noticing that ring homomorphisms $\varphi$ verify that $ker\varphi$ is ideal of the origin ring, so from that I get that, being $$\varphi: K\to K$$ a ring homomorphism, then $ker\varphi\unlhd K$, so $ker\varphi\in\{0,K\}$. Now, we study both cases separatedly:
- If $ker\varphi = K$, then $\varphi=0$.
- If $ker\varphi = 0$, then $\varphi$ is ring monomorphism, and from the fact that $\varphi$ goes from $K$ to $K$ we get also that it's bijection. My problem is that I don't know how to deduce $\varphi = id_K$ from this.
Is the work I did correct? How can I finish this proof? Any help will be appreciated, thanks in advance.
It's not true for general $K$. E.g. for $K = \Bbb C$, you have a lot of homomorphisms from $\Bbb C$ to itself.
It is true in the case $K = \Bbb Q$. Let $\phi:\Bbb Q \rightarrow \Bbb Q$ be a ring homomorphism, i.e. $\phi(x + y) = \phi(x) + \phi(y)$ and $\phi(xy) = \phi(x)\phi(y)$ for all $x, y \in \Bbb Q$.
We consider $\phi(1)$. Since $\phi(1) = \phi(1^2) = \phi(1)^2$, we see that $\phi(1)$ is either $0$ or $1$.
If $\phi(1) = 0$, then we have $\phi(x) = \phi(x \cdot 1) = \phi(x)\phi(1) = 0$ for any $x\in \Bbb Q$.
Now assume $\phi(1) = 1$. Using $\phi(x + 1) = \phi(x) + \phi(1) = \phi(x) + 1$, we may show by induction that $\phi(n) = n$ for all integers $n \geq 0$. Using $\phi(-x) = -\phi(x)$ (as $\phi$ is a homomorphism of the additive group), we get $\phi(n) = n$ for all $n\in \Bbb Z$.
Therefore for any rational number $q = a/b$ with $a, b \in \Bbb Z$ and $b \neq 0$, we have $$b\phi(q) = \phi(b) \phi(q) = \phi(bq) = \phi(a) = a,$$ which gives $\phi(q) = a/b = q$.