Proof that $kappa(A[t])=kappa(A)(t)$

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Let $A$ be an integrally closed ring. Is the fraction field of $A[t]$ just $kappa(A)(t)$, where this denotes the fraction field $(A-0)^-1A$ field extended by $t$.



Why I think this is so: $kappa(A[t])$ easily includes into $kappa(A)(t)$ (because the former consists of elements of the form $f/g$ where $gne 0$ and $f,gin A[t]$, and the latter consists of rational functions in $t$ with coefficients in $Asubset kappa(A)$), where the latter is by definition the smallest field generated by $kappa(A)$ and $t$, and hence they are equal.



Is that a sound argument?




abstract-algebra field-theory extension-field




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  • You can also use the universal property of localization.
    – Andres Mejia
    Aug 27 at 17:46














up vote
5
down vote

favorite
1












Let $A$ be an integrally closed ring. Is the fraction field of $A[t]$ just $kappa(A)(t)$, where this denotes the fraction field $(A-0)^-1A$ field extended by $t$.



Why I think this is so: $kappa(A[t])$ easily includes into $kappa(A)(t)$ (because the former consists of elements of the form $f/g$ where $gne 0$ and $f,gin A[t]$, and the latter consists of rational functions in $t$ with coefficients in $Asubset kappa(A)$), where the latter is by definition the smallest field generated by $kappa(A)$ and $t$, and hence they are equal.



Is that a sound argument?




abstract-algebra field-theory extension-field




share|cite|improve this question




















  • You can also use the universal property of localization.
    – Andres Mejia
    Aug 27 at 17:46












up vote
5
down vote

favorite
1









up vote
5
down vote

favorite
1






1





Let $A$ be an integrally closed ring. Is the fraction field of $A[t]$ just $kappa(A)(t)$, where this denotes the fraction field $(A-0)^-1A$ field extended by $t$.



Why I think this is so: $kappa(A[t])$ easily includes into $kappa(A)(t)$ (because the former consists of elements of the form $f/g$ where $gne 0$ and $f,gin A[t]$, and the latter consists of rational functions in $t$ with coefficients in $Asubset kappa(A)$), where the latter is by definition the smallest field generated by $kappa(A)$ and $t$, and hence they are equal.



Is that a sound argument?




abstract-algebra field-theory extension-field




share|cite|improve this question












Let $A$ be an integrally closed ring. Is the fraction field of $A[t]$ just $kappa(A)(t)$, where this denotes the fraction field $(A-0)^-1A$ field extended by $t$.



Why I think this is so: $kappa(A[t])$ easily includes into $kappa(A)(t)$ (because the former consists of elements of the form $f/g$ where $gne 0$ and $f,gin A[t]$, and the latter consists of rational functions in $t$ with coefficients in $Asubset kappa(A)$), where the latter is by definition the smallest field generated by $kappa(A)$ and $t$, and hence they are equal.



Is that a sound argument?





abstract-algebra field-theory extension-field





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share|cite|improve this question




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asked Jun 16 at 4:44









ant

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  • You can also use the universal property of localization.
    – Andres Mejia
    Aug 27 at 17:46
















  • You can also use the universal property of localization.
    – Andres Mejia
    Aug 27 at 17:46















You can also use the universal property of localization.
– Andres Mejia
Aug 27 at 17:46




You can also use the universal property of localization.
– Andres Mejia
Aug 27 at 17:46










1 Answer
1






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0
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I think your argument is fine. Here is a more explicit approach:



Elements of $kappa(A)(t)$ look like
$$ fracsumfraca_ib_it^isumfracc_jd_jt^j $$
and clearing denominators gives us the following, with $B_i=prod_kneq ib_k$ and similarly for $D_j$:



$$ fracprod d_i sum a_iB_i t^iprod b_i sum c_j D_j t^j $$



which is clearly in $kappa(A[t])$.






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  • @AndresMejia: Not sure what you mean?
    – Steve D
    Aug 27 at 19:18










  • Sorry!$$
    – Andres Mejia
    Aug 27 at 19:19










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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
0
down vote













I think your argument is fine. Here is a more explicit approach:



Elements of $kappa(A)(t)$ look like
$$ fracsumfraca_ib_it^isumfracc_jd_jt^j $$
and clearing denominators gives us the following, with $B_i=prod_kneq ib_k$ and similarly for $D_j$:



$$ fracprod d_i sum a_iB_i t^iprod b_i sum c_j D_j t^j $$



which is clearly in $kappa(A[t])$.






share|cite|improve this answer




















  • @AndresMejia: Not sure what you mean?
    – Steve D
    Aug 27 at 19:18










  • Sorry!$$
    – Andres Mejia
    Aug 27 at 19:19














up vote
0
down vote













I think your argument is fine. Here is a more explicit approach:



Elements of $kappa(A)(t)$ look like
$$ fracsumfraca_ib_it^isumfracc_jd_jt^j $$
and clearing denominators gives us the following, with $B_i=prod_kneq ib_k$ and similarly for $D_j$:



$$ fracprod d_i sum a_iB_i t^iprod b_i sum c_j D_j t^j $$



which is clearly in $kappa(A[t])$.






share|cite|improve this answer




















  • @AndresMejia: Not sure what you mean?
    – Steve D
    Aug 27 at 19:18










  • Sorry!$$
    – Andres Mejia
    Aug 27 at 19:19












up vote
0
down vote










up vote
0
down vote









I think your argument is fine. Here is a more explicit approach:



Elements of $kappa(A)(t)$ look like
$$ fracsumfraca_ib_it^isumfracc_jd_jt^j $$
and clearing denominators gives us the following, with $B_i=prod_kneq ib_k$ and similarly for $D_j$:



$$ fracprod d_i sum a_iB_i t^iprod b_i sum c_j D_j t^j $$



which is clearly in $kappa(A[t])$.






share|cite|improve this answer












I think your argument is fine. Here is a more explicit approach:



Elements of $kappa(A)(t)$ look like
$$ fracsumfraca_ib_it^isumfracc_jd_jt^j $$
and clearing denominators gives us the following, with $B_i=prod_kneq ib_k$ and similarly for $D_j$:



$$ fracprod d_i sum a_iB_i t^iprod b_i sum c_j D_j t^j $$



which is clearly in $kappa(A[t])$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 27 at 17:44









Steve D

2,5971620




2,5971620











  • @AndresMejia: Not sure what you mean?
    – Steve D
    Aug 27 at 19:18










  • Sorry!$$
    – Andres Mejia
    Aug 27 at 19:19
















  • @AndresMejia: Not sure what you mean?
    – Steve D
    Aug 27 at 19:18










  • Sorry!$$
    – Andres Mejia
    Aug 27 at 19:19















@AndresMejia: Not sure what you mean?
– Steve D
Aug 27 at 19:18




@AndresMejia: Not sure what you mean?
– Steve D
Aug 27 at 19:18












Sorry!$$
– Andres Mejia
Aug 27 at 19:19




Sorry!$$
– Andres Mejia
Aug 27 at 19:19

















 

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