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$ 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.

Confused Student
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    How is $k(a)$ a $k[t]$-module? I suspect it's via the map $k[t] \to k(a)$, $t \mapsto a$, which may answer your question. (Also note that since $k$ is algebraically closed, then $k(a) \cong k$.) – Viktor Vaughn Dec 28 '18 at 20:44
  • Okay. I see. I also know that $ \kappa (a) = k[t]{(t-a)}/(t-a)k[t]{(t-a)}. $ – Confused Student Dec 28 '18 at 20:47
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    Did you mean $ k[x,y,t]/(ty-x^{2}) \otimes_{k[t]} k[t]/(t-a)$ ? Then it is $\cong k[x,y,t]/(ty-x^2,t-a)\cong k[x,y]/(ay-x^{2})$ – reuns Dec 28 '18 at 20:48
  • There are a few facts one should keep in mind for those kind of products that come up all the time when computing fibres of morphisms of schemes, $k[x_1,\dotsc,x_n]\otimes_kk[y_1,\dotsc,y_n]=k[x_1,\dotsc,x_n,y_1,\dotsc,y_n]$, if the tensor product is over $k[z_1,\dotsc,z_r]$ with the maps $z_i\mapsto x_i$ and $z_i\mapsto y_i$ then the tensor product is $k[z_1,\dotsc,z_r,x_{r+1},\dotsc,x_n,y_{r+1},\dotsc,y_n]$ (in general you quotient to identify the image of $z_i$ in the first and second algebra, tensor product commute with quotients etc. Those should all be proved somewhere on MSE already! – Alessandro Codenotti Dec 28 '18 at 20:49
  • @reuns What you have suggested makes sense, but I'm not sure it matches the example in Hartshorne. Perhaps, I'm just being slow. Also, concerning the second isomorphism, I'm not quite seeing why it is true. – Confused Student Dec 28 '18 at 20:55
  • @AlessandroCodenotti Thanks! I'll reflect on this. – Confused Student Dec 28 '18 at 21:02
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    Do you agree that $(ty-x^2,t-a) = (ay-x^2,t-a)$ ? Then $k[x,y,t]/(ay-x^2,t-a) \cong R[t]/(t-a) \cong R$ where $R = k[x,y]/(ay-x^2)$ – reuns Dec 28 '18 at 21:06
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    You and reuns are talking about the same thing, think of what the residue field of an affine scheme at a point looks like – Alessandro Codenotti Dec 28 '18 at 21:09
  • @AlessandroCodenotti You're right. facepalms – Confused Student Dec 28 '18 at 21:24
  • @reuns I agree that $ k[x,y,t]/(ty-x^{2},t-a) = k[x,y,t]/(ay-x^{2},t-a), $ and I think I see how everything you have said follows from this. Thanks. – Confused Student Dec 30 '18 at 01:42

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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 $X=\mbox{Spec}\left(k[x,y,t]/(ty-x^2)\right)$, $Y=\mbox{Spec}\left(k[t]\right)$, $$X_a=X\times_Y k(a)=\mbox{Spec}\left(k[x,y,t]/(ty-x^2)\right)\times_{k[t]}\mbox{Spec}\left(k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}\right)$$ 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 $\sum_i q_i(x,y,t)\otimes_{k[t]} a_i\in k[x,y,t]/(yt-x^2)\otimes_{k[t]}k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}$, 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 $a_i\in k$} \\ \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 $\ker\psi=(ay-x^2)$. 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 $(ay-x^2)\subset\ker\psi$. Similar argument to the equality.

José Alejandro Aburto Araneda
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  • Thanks for this. I was originally attempting to show it by playing with universal properties (though I have trouble with them), but I guess sometimes it helps to think more explicitly as you have done. From what @reuns has said it is not too difficult to show that $ ker (\psi) = (ay-x^2) $ by the first isomorphism theorem. – Confused Student Dec 29 '18 at 15:39
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    Sometime one have to loss the fear to work this things when we are using universal properties and facy stuff in AG, because in some sense this give us intuition like "it's simply replacing those thing here" but writing drown can be cumbersome – José Alejandro Aburto Araneda Dec 29 '18 at 15:41
  • Yes, you are right. I have to develop the habit of writing these things down from "first principles" instead of struggling to find a fancy way. – Confused Student Dec 30 '18 at 00:03