Proving That $ text{Spec}Big( k[x,y,t]/(ty-x^{2}) otimes_{k[t]} k (a) Big) cong text{Spec}Big(...












3












$begingroup$


$ 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.






algebraic-geometry







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    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 $k(a) cong k$.)
    $endgroup$
    – André 3000
    Dec 28 '18 at 20:44












  • $begingroup$
    Okay. I see. I also know that $ kappa (a) = k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}. $
    $endgroup$
    – Addled Student
    Dec 28 '18 at 20:47








  • 3




    $begingroup$
    Did you mean $ k[x,y,t]/(ty-x^{2}) otimes_{k[t]} k[t]/(t-a)$ ? Then it is $cong k[x,y,t]/(ty-x^2,t-a)cong k[x,y]/(ay-x^{2})$
    $endgroup$
    – reuns
    Dec 28 '18 at 20:48






  • 2




    $begingroup$
    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 $R = k[x,y]/(ay-x^2)$
    $endgroup$
    – reuns
    Dec 28 '18 at 21:06








  • 1




    $begingroup$
    You and reuns are talking about the same thing, think of what the residue field of an affine scheme at a point looks like
    $endgroup$
    – Alessandro Codenotti
    Dec 28 '18 at 21:09
















3












$begingroup$


$ 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.






algebraic-geometry







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    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 $k(a) cong k$.)
    $endgroup$
    – André 3000
    Dec 28 '18 at 20:44












  • $begingroup$
    Okay. I see. I also know that $ kappa (a) = k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}. $
    $endgroup$
    – Addled Student
    Dec 28 '18 at 20:47








  • 3




    $begingroup$
    Did you mean $ k[x,y,t]/(ty-x^{2}) otimes_{k[t]} k[t]/(t-a)$ ? Then it is $cong k[x,y,t]/(ty-x^2,t-a)cong k[x,y]/(ay-x^{2})$
    $endgroup$
    – reuns
    Dec 28 '18 at 20:48






  • 2




    $begingroup$
    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 $R = k[x,y]/(ay-x^2)$
    $endgroup$
    – reuns
    Dec 28 '18 at 21:06








  • 1




    $begingroup$
    You and reuns are talking about the same thing, think of what the residue field of an affine scheme at a point looks like
    $endgroup$
    – Alessandro Codenotti
    Dec 28 '18 at 21:09














3












3








3


3



$begingroup$


$ 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.






algebraic-geometry







share|cite|improve this question











$endgroup$




$ 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.






algebraic-geometry




algebraic-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 20:37







Addled Student

















asked Dec 28 '18 at 20:27









Addled StudentAddled Student

528




528








  • 3




    $begingroup$
    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 $k(a) cong k$.)
    $endgroup$
    – André 3000
    Dec 28 '18 at 20:44












  • $begingroup$
    Okay. I see. I also know that $ kappa (a) = k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}. $
    $endgroup$
    – Addled Student
    Dec 28 '18 at 20:47








  • 3




    $begingroup$
    Did you mean $ k[x,y,t]/(ty-x^{2}) otimes_{k[t]} k[t]/(t-a)$ ? Then it is $cong k[x,y,t]/(ty-x^2,t-a)cong k[x,y]/(ay-x^{2})$
    $endgroup$
    – reuns
    Dec 28 '18 at 20:48






  • 2




    $begingroup$
    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 $R = k[x,y]/(ay-x^2)$
    $endgroup$
    – reuns
    Dec 28 '18 at 21:06








  • 1




    $begingroup$
    You and reuns are talking about the same thing, think of what the residue field of an affine scheme at a point looks like
    $endgroup$
    – Alessandro Codenotti
    Dec 28 '18 at 21:09














  • 3




    $begingroup$
    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 $k(a) cong k$.)
    $endgroup$
    – André 3000
    Dec 28 '18 at 20:44












  • $begingroup$
    Okay. I see. I also know that $ kappa (a) = k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}. $
    $endgroup$
    – Addled Student
    Dec 28 '18 at 20:47








  • 3




    $begingroup$
    Did you mean $ k[x,y,t]/(ty-x^{2}) otimes_{k[t]} k[t]/(t-a)$ ? Then it is $cong k[x,y,t]/(ty-x^2,t-a)cong k[x,y]/(ay-x^{2})$
    $endgroup$
    – reuns
    Dec 28 '18 at 20:48






  • 2




    $begingroup$
    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 $R = k[x,y]/(ay-x^2)$
    $endgroup$
    – reuns
    Dec 28 '18 at 21:06








  • 1




    $begingroup$
    You and reuns are talking about the same thing, think of what the residue field of an affine scheme at a point looks like
    $endgroup$
    – Alessandro Codenotti
    Dec 28 '18 at 21:09








3




3




$begingroup$
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 $k(a) cong k$.)
$endgroup$
– André 3000
Dec 28 '18 at 20:44






$begingroup$
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 $k(a) cong k$.)
$endgroup$
– André 3000
Dec 28 '18 at 20:44














$begingroup$
Okay. I see. I also know that $ kappa (a) = k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}. $
$endgroup$
– Addled Student
Dec 28 '18 at 20:47






$begingroup$
Okay. I see. I also know that $ kappa (a) = k[t]_{(t-a)}/(t-a)k[t]_{(t-a)}. $
$endgroup$
– Addled Student
Dec 28 '18 at 20:47






3




3




$begingroup$
Did you mean $ k[x,y,t]/(ty-x^{2}) otimes_{k[t]} k[t]/(t-a)$ ? Then it is $cong k[x,y,t]/(ty-x^2,t-a)cong k[x,y]/(ay-x^{2})$
$endgroup$
– reuns
Dec 28 '18 at 20:48




$begingroup$
Did you mean $ k[x,y,t]/(ty-x^{2}) otimes_{k[t]} k[t]/(t-a)$ ? Then it is $cong k[x,y,t]/(ty-x^2,t-a)cong k[x,y]/(ay-x^{2})$
$endgroup$
– reuns
Dec 28 '18 at 20:48




2




2




$begingroup$
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 $R = k[x,y]/(ay-x^2)$
$endgroup$
– reuns
Dec 28 '18 at 21:06






$begingroup$
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 $R = k[x,y]/(ay-x^2)$
$endgroup$
– reuns
Dec 28 '18 at 21:06






1




1




$begingroup$
You and reuns are talking about the same thing, think of what the residue field of an affine scheme at a point looks like
$endgroup$
– Alessandro Codenotti
Dec 28 '18 at 21:09




$begingroup$
You and reuns are talking about the same thing, think of what the residue field of an affine scheme at a point looks like
$endgroup$
– Alessandro Codenotti
Dec 28 '18 at 21:09










1 Answer
1






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 29 '18 at 15:39








  • 1




    $begingroup$
    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
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 15:41










  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 30 '18 at 0:03













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1 Answer
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1 Answer
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active

oldest

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active

oldest

votes






active

oldest

votes









1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 29 '18 at 15:39








  • 1




    $begingroup$
    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
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 15:41










  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 30 '18 at 0:03


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 29 '18 at 15:39








  • 1




    $begingroup$
    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
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 15:41










  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 30 '18 at 0:03
















1












1








1





$begingroup$

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.






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$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 14:50









José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda

802110




802110












  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 29 '18 at 15:39








  • 1




    $begingroup$
    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
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 15:41










  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 30 '18 at 0:03




















  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 29 '18 at 15:39








  • 1




    $begingroup$
    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
    $endgroup$
    – José Alejandro Aburto Araneda
    Dec 29 '18 at 15:41










  • $begingroup$
    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.
    $endgroup$
    – Addled Student
    Dec 30 '18 at 0:03


















$begingroup$
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.
$endgroup$
– Addled Student
Dec 29 '18 at 15:39






$begingroup$
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.
$endgroup$
– Addled Student
Dec 29 '18 at 15:39






1




1




$begingroup$
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
$endgroup$
– José Alejandro Aburto Araneda
Dec 29 '18 at 15:41




$begingroup$
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
$endgroup$
– José Alejandro Aburto Araneda
Dec 29 '18 at 15:41












$begingroup$
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.
$endgroup$
– Addled Student
Dec 30 '18 at 0:03






$begingroup$
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.
$endgroup$
– Addled Student
Dec 30 '18 at 0:03




















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