Ideal of polynomials in $k[X_1,...X_n]$ vanishing at a point $p$ is $(X_1 - p_1, ...,X_n - p_n)$ I'm having a little trouble following Eisenbud here:

My problem is that I don't see how the isomorphism
$${k[x_1,...,x_n] \over \mathfrak{m}_p} \cong k$$
is constructed. This seems a bit hand-wavey to me. Any clarification would be gladly appreciated!
Thank you.
 A: Modding out by $\mathfrak{m}_p$ is the same as modding out by the congruence $\equiv$ generated by all of the relations $x_i - a_i \equiv 0$ which, in turn, is the same as modding out by the congruence generated by all of the relations $x_i \equiv a_i$.
And since it is a congruence and $f$ is a polynomial function:
$$f(x_1, \ldots, x_n) \equiv f(a_1, \ldots, a_n) \qquad \qquad (*)$$

To spell some things out in excruciating detail....
To save typing, let


*

*$A = k[x_1, \ldots, x_n]$

*$I = \mathfrak{m}_P$

*$R = A / I$

*$a = (a_1, \ldots, a_n)$


From the general properties of polynomial rings, we know there is an "evaluation at a" homomorphism
$$ \phi : A \to k : f \mapsto f(a) $$
There is also the projection map
$$ \pi : A \to R : f \mapsto \overline{f} $$
sending a polynomial to its equivalence class modulo $I$. Because $\phi(x_i - a_i) = 0$,
$$ \bar{\phi} : R \to k : \overline{f} \mapsto f(a) $$
is well-defined, and $\phi = \bar{\phi} \circ \pi$. It is surjective because for any $c \in k$: $\bar{\phi}(\bar{c}) = c$.
The most direct method to see $\bar{\phi}$ is injective is
$$ \bar{\phi}(\bar{f}) = \bar{\phi}(\bar{g}) 
\implies f(a) = g(a) \implies f \equiv g \bmod I \implies \overline{f} = \overline{g} $$
But IMO it's more obvious to observe there is a set function
$$\bar{\rho} : |k| \to |R| : c \mapsto \bar{c}$$
which is the inverse of $\bar{\phi}$ as functions of sets, and therefore $\bar{\phi}$ must be a bijection.
A related, more general idea is that of choosing canonical preimages of the map $\phi$; that is a set function $\rho : |k| \to |A|$ with the property that $\phi \circ \rho$ is the identity function. Such a set function is called a "splitting". In this case I choose
$$\rho : k \to A : c \mapsto c $$
which turns out to actually be a ring homomorphism and not merely a set function. (*) shows that every element of $A$ is congruent, modulo $I$, to an element in the image of $\rho$, and this in turn implies that $\overline{\rho} = \pi \circ \rho$ has the property I used above.

A similar argument that works in this case is to look at the sequence of three homomorphisms
$$ k \xrightarrow{\rho} A \xrightarrow{\pi} R \xrightarrow{\bar{\phi}} k $$
The composite function $k \to k$ is the identity and thus invertible, therefore the composite $k \to R$ must be injective. However, by (*) we know $k \to R$ is surjective and thus bijective, and is directly seen to be an inverse of $\bar{\phi}$.
A "cheat" that streamlines this particular argument is:


*

*$R$ is not the trivial ring (because there is a homomorphism $R \to k$)

*$k \to R$ is surjective by (*)

*Any homomorphism from a field to a nontrivial ring is injective


and thus $k \to R$ is bijective.
A: It's the valuation homomorphism:
$$\phi:k[x_1,...,x_n]\to k\;\;,\;\;\;\phi(f):=f(a_1,...,a_n)$$
The above is a surjective ring homomorphism, and its kernel is precisely $\,\mathfrak  m_p\,$ . Now just apply the first isomorphism theorem...
