Exercise 4.3.9 (Uniqueness of homomorphism over K)

Question

Fill in the details of the last paragraph of that proof.

Hassaan Naeem · Accepted Answer

\paragraph{Solution:}
We show that there is at most one homomorphism $\phi: K(t) ightarrow L$ over $K$ such that $\phi(t)= \beta$.
We let $\phi$ and $\phi'$ be two such homomorphisms. Then we have that $\phi(t) = \beta = \phi'(t)$. By 	extbf{Lemma 4.3.1 (ii)}
we have that $t$ generates $K(t)$ over $K$, and hence by 	extbf{Lemma 4.3.6} $\phi = \phi' \quad \square$

Exercise 4.3.9 (Uniqueness of homomorphism over K)

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Exercise 4.3.9 (Uniqueness of homomorphism over K)

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