How do I show that this map surjective?
$begingroup$
I am studying the residue field at a point on the affine line.
Suppose we have the homomorphism $ rho: k[t]/(t-a) rightarrow k[t]_{(t-a)}/(t-a)k[t]_{(t-a)} $ such that $ rho $ maps a coset of polynomials in $ k[t] $ to a coset of an equivalence class of a "fraction" $ frac{a}{b} $ in $ k[t]_{(t-a)}. $ These cosets are of course different because the quotients are different, and the equivalence class is induced by the defining equivalence relation of the localisation.
I think that this is an injective map, but I am not entirely sure how to show that it is surjective.
Edit: I can show that this map is injective, but a path to surjectivity eludes me.
abstract-algebra
asked Jan 20 at 0:43
Random StudentRandom Student
62
$endgroup$
add a comment |
$begingroup$
I am studying the residue field at a point on the affine line.
Suppose we have the homomorphism $ rho: k[t]/(t-a) rightarrow k[t]_{(t-a)}/(t-a)k[t]_{(t-a)} $ such that $ rho $ maps a coset of polynomials in $ k[t] $ to a coset of an equivalence class of a "fraction" $ frac{a}{b} $ in $ k[t]_{(t-a)}. $ These cosets are of course different because the quotients are different, and the equivalence class is induced by the defining equivalence relation of the localisation.
I think that this is an injective map, but I am not entirely sure how to show that it is surjective.
Edit: I can show that this map is injective, but a path to surjectivity eludes me.
abstract-algebra
asked Jan 20 at 0:43
Random StudentRandom Student
62
$endgroup$
1
$begingroup$
Observe that both of those rings are in fact isomorphic to $k$. Then use the fact that there is only one $k$-algebra map $krightarrow k$.
$endgroup$
– Levent
Jan 20 at 0:53
$begingroup$
Thanks. This is certainly one way of doing it. I'm thinking of the following line of argument. Please let me know if you think think it's a valid way to show that the map $ rho $ is surjective. Suppose we have some $ A in k[t] $ such that $ Big[frac{A}{1} Big] cong Big[frac{m}{n} Big] text{mod}Big( (t-a)k[t]_{(t-a)}Big). $ Then, $ exists ; s in k[t backslash (t-a) $ such that $ s(An - m) in (t-a). $ This implies that $ An-m in (t-a). $ I am thinking that to show from here that $ A $ is a polynomial in $ t $ would show that the map is surjective, but I'm not sure.
$endgroup$
– Random Student
Jan 22 at 23:37
add a comment |
$begingroup$
I am studying the residue field at a point on the affine line.
Suppose we have the homomorphism $ rho: k[t]/(t-a) rightarrow k[t]_{(t-a)}/(t-a)k[t]_{(t-a)} $ such that $ rho $ maps a coset of polynomials in $ k[t] $ to a coset of an equivalence class of a "fraction" $ frac{a}{b} $ in $ k[t]_{(t-a)}. $ These cosets are of course different because the quotients are different, and the equivalence class is induced by the defining equivalence relation of the localisation.
I think that this is an injective map, but I am not entirely sure how to show that it is surjective.
Edit: I can show that this map is injective, but a path to surjectivity eludes me.
abstract-algebra
asked Jan 20 at 0:43
Random StudentRandom Student
62
$endgroup$
I am studying the residue field at a point on the affine line.
Suppose we have the homomorphism $ rho: k[t]/(t-a) rightarrow k[t]_{(t-a)}/(t-a)k[t]_{(t-a)} $ such that $ rho $ maps a coset of polynomials in $ k[t] $ to a coset of an equivalence class of a "fraction" $ frac{a}{b} $ in $ k[t]_{(t-a)}. $ These cosets are of course different because the quotients are different, and the equivalence class is induced by the defining equivalence relation of the localisation.
I think that this is an injective map, but I am not entirely sure how to show that it is surjective.
Edit: I can show that this map is injective, but a path to surjectivity eludes me.
abstract-algebra
abstract-algebra
asked Jan 20 at 0:43
Random StudentRandom Student
62
asked Jan 20 at 0:43
Random StudentRandom Student
62
edited Jan 22 at 23:41
Random Student
asked Jan 20 at 0:43
Random StudentRandom Student
62
asked Jan 20 at 0:43
Random StudentRandom Student
62
asked Jan 20 at 0:43
Random StudentRandom Student
62
62
1
$begingroup$
Observe that both of those rings are in fact isomorphic to $k$. Then use the fact that there is only one $k$-algebra map $krightarrow k$.
$endgroup$
– Levent
Jan 20 at 0:53
$begingroup$
Thanks. This is certainly one way of doing it. I'm thinking of the following line of argument. Please let me know if you think think it's a valid way to show that the map $ rho $ is surjective. Suppose we have some $ A in k[t] $ such that $ Big[frac{A}{1} Big] cong Big[frac{m}{n} Big] text{mod}Big( (t-a)k[t]_{(t-a)}Big). $ Then, $ exists ; s in k[t backslash (t-a) $ such that $ s(An - m) in (t-a). $ This implies that $ An-m in (t-a). $ I am thinking that to show from here that $ A $ is a polynomial in $ t $ would show that the map is surjective, but I'm not sure.
$endgroup$
– Random Student
Jan 22 at 23:37
add a comment |
1
$begingroup$
Observe that both of those rings are in fact isomorphic to $k$. Then use the fact that there is only one $k$-algebra map $krightarrow k$.
$endgroup$
– Levent
Jan 20 at 0:53
$begingroup$
Thanks. This is certainly one way of doing it. I'm thinking of the following line of argument. Please let me know if you think think it's a valid way to show that the map $ rho $ is surjective. Suppose we have some $ A in k[t] $ such that $ Big[frac{A}{1} Big] cong Big[frac{m}{n} Big] text{mod}Big( (t-a)k[t]_{(t-a)}Big). $ Then, $ exists ; s in k[t backslash (t-a) $ such that $ s(An - m) in (t-a). $ This implies that $ An-m in (t-a). $ I am thinking that to show from here that $ A $ is a polynomial in $ t $ would show that the map is surjective, but I'm not sure.
$endgroup$
– Random Student
Jan 22 at 23:37
1
1
$begingroup$
Observe that both of those rings are in fact isomorphic to $k$. Then use the fact that there is only one $k$-algebra map $krightarrow k$.
$endgroup$
– Levent
Jan 20 at 0:53
$begingroup$
Observe that both of those rings are in fact isomorphic to $k$. Then use the fact that there is only one $k$-algebra map $krightarrow k$.
$endgroup$
– Levent
Jan 20 at 0:53
$begingroup$
Thanks. This is certainly one way of doing it. I'm thinking of the following line of argument. Please let me know if you think think it's a valid way to show that the map $ rho $ is surjective. Suppose we have some $ A in k[t] $ such that $ Big[frac{A}{1} Big] cong Big[frac{m}{n} Big] text{mod}Big( (t-a)k[t]_{(t-a)}Big). $ Then, $ exists ; s in k[t backslash (t-a) $ such that $ s(An - m) in (t-a). $ This implies that $ An-m in (t-a). $ I am thinking that to show from here that $ A $ is a polynomial in $ t $ would show that the map is surjective, but I'm not sure.
$endgroup$
– Random Student
Jan 22 at 23:37
$begingroup$
Thanks. This is certainly one way of doing it. I'm thinking of the following line of argument. Please let me know if you think think it's a valid way to show that the map $ rho $ is surjective. Suppose we have some $ A in k[t] $ such that $ Big[frac{A}{1} Big] cong Big[frac{m}{n} Big] text{mod}Big( (t-a)k[t]_{(t-a)}Big). $ Then, $ exists ; s in k[t backslash (t-a) $ such that $ s(An - m) in (t-a). $ This implies that $ An-m in (t-a). $ I am thinking that to show from here that $ A $ is a polynomial in $ t $ would show that the map is surjective, but I'm not sure.
$endgroup$
– Random Student
Jan 22 at 23:37
add a comment |
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$begingroup$
Observe that both of those rings are in fact isomorphic to $k$. Then use the fact that there is only one $k$-algebra map $krightarrow k$.
$endgroup$
– Levent
Jan 20 at 0:53
$begingroup$
Thanks. This is certainly one way of doing it. I'm thinking of the following line of argument. Please let me know if you think think it's a valid way to show that the map $ rho $ is surjective. Suppose we have some $ A in k[t] $ such that $ Big[frac{A}{1} Big] cong Big[frac{m}{n} Big] text{mod}Big( (t-a)k[t]_{(t-a)}Big). $ Then, $ exists ; s in k[t backslash (t-a) $ such that $ s(An - m) in (t-a). $ This implies that $ An-m in (t-a). $ I am thinking that to show from here that $ A $ is a polynomial in $ t $ would show that the map is surjective, but I'm not sure.
$endgroup$
– Random Student
Jan 22 at 23:37