How to check whether a $\mathbb{K}$-algebra homomorphism restricts to identity on $\mathbb{K}$

Question

I am working on

Lemma VIII.7.3 If $\mathbb{F}$ is an algebraically closed extension of a field $\mathbb{K}$ and $I<\mathbb{K}[x_{1},\ldots,x_{n}]$ is a proper ideal of $\mathbb{K}[x_{1},\ldots,x_{n}]$, then the affine variety $V(I)\subseteq\mathbb{F}^{n}$ is non-empty.

in Hungerford's Algebra book.

My question is less about the proof of the lemma but about one particular step, which has been bugging me ever since I have had learnt about field extensions and got more used to taking quotients. The question is:

How do you check that a homomorphism of unital $\mathbb{K}$-algebras restricts to the identity on $\mathbb{K}$

What we do in this proof is the following:

1. We define an epimorphism $\Psi:\mathbb{K}[x_{1},\ldots,x_{n}]\twoheadrightarrow R$, where $R$ is an integral domain. It turns out that $\Psi\big|_{\mathbb{K}}$ is an isomorphism onto its image.
2. It turns out that $R$ is integral over $S:=\Psi(\mathbb{K})[t_{1},\ldots,t_{r}]$ where $\{t_{1},\ldots,t_{n}\}\subseteq R$ is an algebraically independent set.
3. We note that $S/J\cong\Psi(\mathbb{K})$ for some (maximal) $J<S$.
4. We apply the going up theorem in order to find a maximal ideal $M<R$ such that $\Psi(\mathbb{K})\cong S/J=(S+M)/M=\pi(S)\subseteq R/M$ where $\pi:R\to R/M$ is the projection.
5. We use integrality of $R$ over $S$ and $S\cong\Psi(\mathbb{K})$ in order to deduce that $R/M$ is algebraic over $\mathbb{K}$ and hence $\overline{\mathbb{K}}\cong\overline{R/M}$, where $\overline{\cdot}$ denotes the algebraic closure.
6. Hence we obtain a monomorphism $\sigma:R/M\to\overline{K}$
7. Now: define $\phi:=\sigma\circ\pi\circ\Psi$. Then the author claims that $\phi\big|_{\mathbb{K}}$ is the identity.

I have no idea why this should be true without making any further explicit identifications before already. It seems that we could happen to loose a lot of information at $\Psi(\mathbb{K})$ already.

It seems to me that I have to show: $\mathbb{K},\mathbb{L}$ fields and $\phi:\mathbb{K}\hookrightarrow\mathbb{L}$ such that $\mathbb{L}/\phi(\mathbb{K})$ algebraic, then there exists $\Phi:\overline{\mathbb{K}}\to\overline{\mathbb{L}}$ an isomorphism of fields such that: $$ \require{AMScd} \begin{CD} \mathbb{K} @>{\phi}>> \mathbb{L}\\ @V{\operatorname{id}_{\mathbb{K}}}VV \cdot @VV{\operatorname{id}_{\mathbb{L}}}V \\ \overline{\mathbb{K}} @>>{\Phi}> \overline{\mathbb{L}} \end{CD} $$

Question: Is this obvious? Have I overseen a basic fact that I actually learnt about?

Answer 1

Indeed I have found an answer to my question on this site. As suggested, I copy the answer her.

Proposition: Let $\mathbb{K},\mathbb{L}$ fields and $\phi:\mathbb{K}\to\mathbb{L}$ a field embedding. Then there is $\Phi:\overline{K}\to\overline{L}$ such that: $$ \require{AMScd} \begin{CD} \mathbb{K} @>{\phi}>> \mathbb{L}\\ @V{\operatorname{id}_{\mathbb{K}}}VV \cdot @VV{\operatorname{id}_{\mathbb{L}}}V \\ \overline{\mathbb{K}} @>>{\Phi}> \overline{\mathbb{L}} \end{CD} $$ The proof is exactly the same as in the usual proof that algebraic closures $\mathbb{A}/\mathbb{K}$ and $\mathbb{B}/\mathbb{K}$ are $\mathbb{K}$-isomorphic. In what follows I denote by $K\hookrightarrow L$ field extensions in the sense that $K\subseteq L$ and by $K\stackrel{\phi|_{\mathbb{K}}}{\hookrightarrow}L$ embeddings of $\mathbb{K}\hookrightarrow K$ in $\mathbb{L}\hookrightarrow L$ such that the restriction to $\mathbb{K}$ agrees with $\phi$ on $\mathbb{K}$. Look at the set $X:=\{\rho:K\stackrel{\phi|_{\mathbb{K}}}{\hookrightarrow}\overline{\mathbb{L}};\mathbb{K}\hookrightarrow K\hookrightarrow\overline{K}\}$. In what follows I denote an element in $\rho:K\stackrel{\phi|_{\mathbb{K}}}{\hookrightarrow}\overline{\mathbb{L}}\in X$ by the symbol $(K,\rho)$. $X$ is non-empty as $(\mathbb{K},\phi)\in X$. Given $(K_{1},\rho_{1})$ and $(K_{2},\rho_{2})$ in $X$, we say $(K_{1},\rho_{1})\leq (K_{2},\rho_{2})$ if $K_{1}\hookrightarrow K_{2}$. This is a partial order and indeed every non-empty chain $C\subseteq X$ has an upper bound given by $L:=\cup_{(K,\rho)\in C}K$ and $\sigma:L\stackrel{\phi|_{\mathbb{K}}}{\hookrightarrow}\overline{L}$ defined by $\sigma(\alpha):=\rho(\alpha)$ for some $(K,\rho)\in C$ such that $\alpha\in K$.

Using Zorn, we deduce that $X$ contains some maximal element $(K,\rho)$. We need to check that $K$ is algebraically closed. If not, we choose $\alpha\in\overline{K}\setminus K$. As $\alpha$ is algebraic over $K$, there exists a minimal monic polynomial $p\in\mathbb{K}[x]$ such that $p(\alpha)=0$. Let $p^{\rho}$ be the image of $p\in\overline{\mathbb{L}}[x]$ via the ring homomorphism $\rho:K[x]\to\overline{L}[x]$ defined by $\rho|_{K}\equiv \rho$ and $\rho(x):=x$. One can check that this extends $\rho$. As $\overline{L}$ is algebraically closed, there exists a root $\beta\in\overline{L}$ of $p^{\rho}$. Then $\rho$ extends to $K(\alpha)$ by $\rho(\alpha)=\beta$ (see and slightly adapt Hungerford, Theorem V.1.8), hence $(K,\rho)$ is not maximal which yields the desired contradiction.

In the situation mentioned above, we just define $\phi(\mathbb{K})$ to be $\pi(\Psi(\mathbb{K}))$ and $\mathbb{L}:=R/M$. A corresponding choice of $\sigma$ then yields the claim.

Of course this does not give an answer to the too general question but this solves most of the cases I bothered with.