I want to show that for a given field $\mathbf{K}$ with chracteristic zero and a fixed element $b\in \mathbf{K}$ there exists another field $\mathbf{L}$ with characteristic zero and an element $c\in \mathbf{L}$ such as, there is no ringhomomorphism between $\mathbf{K}$ and $\mathbf{L}$ which fullfills $\phi(b)=c$.

I already came up with, what I think, is an answer. But I am not quite sure, especially because I don't see why I would need that the characteristic is equal to zero in both cases. So please correct me if you find any mistakes.

First of all, a ringhomomorphism has to follow certain conditions, specially the following ones I will use: $$ \phi(0_K)=0_L$$ $$\phi(1_K)=1_L$$ $$\phi(ab)=\phi(a)\phi(b)$$

Now, if $b$ is equal to zero, I would just choose $\mathbf{K}$ again for $\mathbf{L}$ with $c=1_K$. So lets assume that $b$ isn't zero. Now, just use $\mathbf{L}:=\mathbf{K}$ and $c=0_K$. A ringhomomorphism must fulfill: $$ \phi(1_K)=1_K \Leftrightarrow \phi(bb^{-1})=1_K \Leftrightarrow \phi(bb^{-1})=\phi(b)\phi(b^{-1}) \Leftrightarrow 1_K=0_K*\phi(b^{-1})=0 \ $$ Since we are working with fields this can't be true. So there can't exist a ringhomorphism on $\mathbf{K}$ such as $\phi(b)=0$.

This seems to be too simple, but what did I do wrong? Where did I use that the chracteristic is zero?

  • 1
    $\begingroup$ You are right. Characteristic zero is not needed. $\endgroup$ – Luiz Cordeiro May 26 '16 at 16:27

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