Csdrhrt

I'm reading Mumford's Red Book. On page 14, I came across this proposition:

If $Sigma_1,Sigma_2$ are two irreducible algebraic sets, then the set of morphisms from $Sigma_1$ to $Sigma_2$ and the set of $k$-homomorphisms from $Gamma(Sigma_2)$ to $Gamma(Sigma_1)$ are canonically isomorphic:

hom($Sigma_1,Sigma_2$) $cong$ hom$_k$($Gamma(Sigma_2),Gamma(Sigma_1)$),
where $Gamma(Sigma_1) = k[x_1,x_2,dots,x_n]/I(Sigma_1)$ is the coordinate ring or affine ring.

I've no problem with this proposition. But, the problem lies with the following corollary:

If $Sigma$ is an irreducible algebraic set, then $Gamma(Sigma)$ is canonically isomorphic to the set of morphisms from $Sigma$ to $k$.

Applying the proposition gives us hom($Sigma,k$) $cong$ hom$_k$($Gamma(k),Gamma(Sigma)$). Note that $Gamma(k) = k[x]$. Therefore, hom($Sigma,k$) $cong$ hom$_k$($k[x],Gamma(Sigma)$).

But how is hom$_k$($k[x],Gamma(Sigma)$) $cong$ $Gamma(Sigma)$ to get the corollary ?

Besides, I've read from Hartshorne's book that $Gamma(Sigma)$ is isomorphic to the regular functions on $Sigma$, therefore, is the morphism from $Sigma$ to $k$ just the regular function on $Sigma$ ?

Any help would be appreciated, thanks.

A $k$-algebra homomorphism from $k[x]$ to a $k$-algebra $A$ is determined uniquely by the image of $x$, and you can easily check that this identification respects the $k$-algebra structure of $Hom(k[x],A)$, giving an isomorphism between $A$ and $Hom(k[x],A)$.

And yes, for a variety over $k$ a "regular function" is the same as a "element of the coordinate ring", is the same as "morphism to $mathbb{A^1_k}$".