Cfrgtkky

$ k $ is an algebraically closed field, and $ a in k. $

This question stems from Example 3.3.1 in Hartshorne's Algebraic Geometry. There is a surjective morphism $ f: text{Spec}Big(k[x,y,t]/(ty-x^{2}) Big) longrightarrow text{Spec}(k[t]), $ and the closed points of $ k[t] $ are identified with elements of $ k $ because $ text{Spec}(k[t]) = mathbb{A}^{1}_{k}. $ I am trying to understand why the fibre $ X_a $ is the plane curve $ ay = x^{2} $ in $ mathbb{A}^{2}_{k} $ and I think this is the same as establishing the isomorphism above.

Intuituvely, the fiber $X_a$ corresponds with the curve associated to the point $t-a=0$, replacing we obtaing the required fact. But, algebraically, you are right since, writing $X=mbox{Spec}left(k[x,y,t]/(ty-x^2)right)$, $Y=mbox{Spec}left(k[t]right)$,
$$X_a=Xtimes_Y k(a)=mbox{Spec}left(k[x,y,t]/(ty-x^2)right)times_{k[t]}mbox{Spec}left(k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}right)$$
and that is the spectrum of tha tensor product over $k[t]$, if we have an isomorphism between the global sections of affine schemes, we have that the schemes are isomophic, so that is enough to prove the isomorphism of that rings.

Since $k$ is algebraically closed $k[t]_{(t-a)}/(t-a)simeq k$ (think via evaluating in $a$ for intuition). Now, for the correspondence of rings that you require, consider the homomorphism
$$psi:k[x,y]to k[x,y,t]/(yt-x^2)otimes_{k[t]}k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}$$
sending $p(x,y)mapsto p(x,y)otimes_{k[t]} 1$. It is clear that it is an homomorphism of rings. It is an epimorphism since if $sum_i q_i(x,y,t)otimes_{k[t]} a_iin k[x,y,t]/(yt-x^2)otimes_{k[t]}k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}$, then
begin{align}
sum_i q_i(x,y,t)otimes_{k[t]} a_i & = sum_i a_i q_i(x,y,t)otimes_{k[t]}1 \
text{with $a_iin k$} \
sum_i q_i(x,y,t)otimes_{k[t]} a_i & = sum_j f_j(t)g_j(x,y)otimes_{k[t]} 1 \
& = sum_j g_j(x,y)otimes_{k[t]} f_j(t) \
& = sum_j g_j(x,y)otimes_{k[t]} f_j(a) \
& = sum_j g_j(x,y)f_j(a)otimes_{k[t]}1,
end{align}
since $f_j(t)=f_j(a)$ in $k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}$. Now, we have to prove that $kerpsi=(ay-x^2)$. In fact, using the same trick (passing a polynomial over $t$ to the right side of the tensor product and evaluating on $a$) we get easily that $(ay-x^2)subsetkerpsi$. Similar argument to the equality.