A basis for $k(X)$ regarded as a vector space over $k$












5












$begingroup$


Can anyone give an explicit basis of the $k$-vector space $k(X) = operatorname{Quot}(k[X])$ of rational functions over $k$?



The dimension is given by $$dim_k k(X) = max(|k|, |mathbb N|).$$



If $k$ is infinite, this follows from $|k| leqslant |k(X)| leqslant |k[X] times k[X]| = |k times k| = |k|$ and the linear independence of $lbrace frac{1}{X - alpha} mid alpha in krbrace.$ If $k$ is finite, the result can be obtained similarly from $|k(X)|=|mathbb N|$ and the linear independence of the monomials $X^n$, $n geqslant 0$.






linear-algebra vector-spaces extension-field







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$endgroup$








  • 2




    $begingroup$
    One way to do this is using partial fractions decompositions.
    $endgroup$
    – Mariano Suárez-Álvarez
    Apr 1 '12 at 1:37
















5












$begingroup$


Can anyone give an explicit basis of the $k$-vector space $k(X) = operatorname{Quot}(k[X])$ of rational functions over $k$?



The dimension is given by $$dim_k k(X) = max(|k|, |mathbb N|).$$



If $k$ is infinite, this follows from $|k| leqslant |k(X)| leqslant |k[X] times k[X]| = |k times k| = |k|$ and the linear independence of $lbrace frac{1}{X - alpha} mid alpha in krbrace.$ If $k$ is finite, the result can be obtained similarly from $|k(X)|=|mathbb N|$ and the linear independence of the monomials $X^n$, $n geqslant 0$.






linear-algebra vector-spaces extension-field







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    One way to do this is using partial fractions decompositions.
    $endgroup$
    – Mariano Suárez-Álvarez
    Apr 1 '12 at 1:37














5












5








5


3



$begingroup$


Can anyone give an explicit basis of the $k$-vector space $k(X) = operatorname{Quot}(k[X])$ of rational functions over $k$?



The dimension is given by $$dim_k k(X) = max(|k|, |mathbb N|).$$



If $k$ is infinite, this follows from $|k| leqslant |k(X)| leqslant |k[X] times k[X]| = |k times k| = |k|$ and the linear independence of $lbrace frac{1}{X - alpha} mid alpha in krbrace.$ If $k$ is finite, the result can be obtained similarly from $|k(X)|=|mathbb N|$ and the linear independence of the monomials $X^n$, $n geqslant 0$.






linear-algebra vector-spaces extension-field







share|cite|improve this question











$endgroup$




Can anyone give an explicit basis of the $k$-vector space $k(X) = operatorname{Quot}(k[X])$ of rational functions over $k$?



The dimension is given by $$dim_k k(X) = max(|k|, |mathbb N|).$$



If $k$ is infinite, this follows from $|k| leqslant |k(X)| leqslant |k[X] times k[X]| = |k times k| = |k|$ and the linear independence of $lbrace frac{1}{X - alpha} mid alpha in krbrace.$ If $k$ is finite, the result can be obtained similarly from $|k(X)|=|mathbb N|$ and the linear independence of the monomials $X^n$, $n geqslant 0$.






linear-algebra vector-spaces extension-field




linear-algebra vector-spaces extension-field






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 31 '16 at 20:01









Tomek Kania

12.2k11945




12.2k11945










asked Apr 1 '12 at 1:35









RalphRalph

995413




995413








  • 2




    $begingroup$
    One way to do this is using partial fractions decompositions.
    $endgroup$
    – Mariano Suárez-Álvarez
    Apr 1 '12 at 1:37














  • 2




    $begingroup$
    One way to do this is using partial fractions decompositions.
    $endgroup$
    – Mariano Suárez-Álvarez
    Apr 1 '12 at 1:37








2




2




$begingroup$
One way to do this is using partial fractions decompositions.
$endgroup$
– Mariano Suárez-Álvarez
Apr 1 '12 at 1:37




$begingroup$
One way to do this is using partial fractions decompositions.
$endgroup$
– Mariano Suárez-Álvarez
Apr 1 '12 at 1:37










1 Answer
1






active

oldest

votes


















8












$begingroup$

A $k$-basis is given by expressions of the form
$$frac{x^i}{[p(x)]^n}$$
where $p(x)$ is a monic irreducible polynomial with coefficients in $k$, $i$ and $n$ are nonnegative integers, and $iltdeg(p)$, together with the positive powers of $x$.



The fact that these elements span $k(x)$ follows from partial fraction decomposition.



To verify linear independence, first note that we cannot have
$$p(x) = frac{f(x)}{g(x)}$$
with $p$, $f$, and $g$ polynomials, $gcd(f,g)=1$, and $gnotin k$ (this is just unique factorization in $k(x)$). Given a linear combination equal to $0$, equating the "polynomial part" and the "rational function part" shows they are both zero; the polynomial part being zero means all coefficients are equal to $0$. So you are reduced to verifying that the "rational function" part is independent. Then can rewrite this part as a sum of the form
$$frac{f_1(x)}{(p_1(x))^{n_1}} + cdots + frac{f_k(x)}{(p_k(x))^{n_k}}$$
where $p_1,ldots,p_k$ are pairwise distinct monic irreducibles, and $deg(f_i)lt n_ideg(p_i)$. Going to an algebra closure and evaluating the numerators you get at roots of the $p_i$ easily yield that $f_i(x)=0$ for all $i$. This reduces to the case of fractions in which the denominators are all powers of the same monic irreducible, which is straightforward.






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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

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    8












    $begingroup$

    A $k$-basis is given by expressions of the form
    $$frac{x^i}{[p(x)]^n}$$
    where $p(x)$ is a monic irreducible polynomial with coefficients in $k$, $i$ and $n$ are nonnegative integers, and $iltdeg(p)$, together with the positive powers of $x$.



    The fact that these elements span $k(x)$ follows from partial fraction decomposition.



    To verify linear independence, first note that we cannot have
    $$p(x) = frac{f(x)}{g(x)}$$
    with $p$, $f$, and $g$ polynomials, $gcd(f,g)=1$, and $gnotin k$ (this is just unique factorization in $k(x)$). Given a linear combination equal to $0$, equating the "polynomial part" and the "rational function part" shows they are both zero; the polynomial part being zero means all coefficients are equal to $0$. So you are reduced to verifying that the "rational function" part is independent. Then can rewrite this part as a sum of the form
    $$frac{f_1(x)}{(p_1(x))^{n_1}} + cdots + frac{f_k(x)}{(p_k(x))^{n_k}}$$
    where $p_1,ldots,p_k$ are pairwise distinct monic irreducibles, and $deg(f_i)lt n_ideg(p_i)$. Going to an algebra closure and evaluating the numerators you get at roots of the $p_i$ easily yield that $f_i(x)=0$ for all $i$. This reduces to the case of fractions in which the denominators are all powers of the same monic irreducible, which is straightforward.






    share|cite|improve this answer









    $endgroup$


















      8












      $begingroup$

      A $k$-basis is given by expressions of the form
      $$frac{x^i}{[p(x)]^n}$$
      where $p(x)$ is a monic irreducible polynomial with coefficients in $k$, $i$ and $n$ are nonnegative integers, and $iltdeg(p)$, together with the positive powers of $x$.



      The fact that these elements span $k(x)$ follows from partial fraction decomposition.



      To verify linear independence, first note that we cannot have
      $$p(x) = frac{f(x)}{g(x)}$$
      with $p$, $f$, and $g$ polynomials, $gcd(f,g)=1$, and $gnotin k$ (this is just unique factorization in $k(x)$). Given a linear combination equal to $0$, equating the "polynomial part" and the "rational function part" shows they are both zero; the polynomial part being zero means all coefficients are equal to $0$. So you are reduced to verifying that the "rational function" part is independent. Then can rewrite this part as a sum of the form
      $$frac{f_1(x)}{(p_1(x))^{n_1}} + cdots + frac{f_k(x)}{(p_k(x))^{n_k}}$$
      where $p_1,ldots,p_k$ are pairwise distinct monic irreducibles, and $deg(f_i)lt n_ideg(p_i)$. Going to an algebra closure and evaluating the numerators you get at roots of the $p_i$ easily yield that $f_i(x)=0$ for all $i$. This reduces to the case of fractions in which the denominators are all powers of the same monic irreducible, which is straightforward.






      share|cite|improve this answer









      $endgroup$
















        8












        8








        8





        $begingroup$

        A $k$-basis is given by expressions of the form
        $$frac{x^i}{[p(x)]^n}$$
        where $p(x)$ is a monic irreducible polynomial with coefficients in $k$, $i$ and $n$ are nonnegative integers, and $iltdeg(p)$, together with the positive powers of $x$.



        The fact that these elements span $k(x)$ follows from partial fraction decomposition.



        To verify linear independence, first note that we cannot have
        $$p(x) = frac{f(x)}{g(x)}$$
        with $p$, $f$, and $g$ polynomials, $gcd(f,g)=1$, and $gnotin k$ (this is just unique factorization in $k(x)$). Given a linear combination equal to $0$, equating the "polynomial part" and the "rational function part" shows they are both zero; the polynomial part being zero means all coefficients are equal to $0$. So you are reduced to verifying that the "rational function" part is independent. Then can rewrite this part as a sum of the form
        $$frac{f_1(x)}{(p_1(x))^{n_1}} + cdots + frac{f_k(x)}{(p_k(x))^{n_k}}$$
        where $p_1,ldots,p_k$ are pairwise distinct monic irreducibles, and $deg(f_i)lt n_ideg(p_i)$. Going to an algebra closure and evaluating the numerators you get at roots of the $p_i$ easily yield that $f_i(x)=0$ for all $i$. This reduces to the case of fractions in which the denominators are all powers of the same monic irreducible, which is straightforward.






        share|cite|improve this answer









        $endgroup$



        A $k$-basis is given by expressions of the form
        $$frac{x^i}{[p(x)]^n}$$
        where $p(x)$ is a monic irreducible polynomial with coefficients in $k$, $i$ and $n$ are nonnegative integers, and $iltdeg(p)$, together with the positive powers of $x$.



        The fact that these elements span $k(x)$ follows from partial fraction decomposition.



        To verify linear independence, first note that we cannot have
        $$p(x) = frac{f(x)}{g(x)}$$
        with $p$, $f$, and $g$ polynomials, $gcd(f,g)=1$, and $gnotin k$ (this is just unique factorization in $k(x)$). Given a linear combination equal to $0$, equating the "polynomial part" and the "rational function part" shows they are both zero; the polynomial part being zero means all coefficients are equal to $0$. So you are reduced to verifying that the "rational function" part is independent. Then can rewrite this part as a sum of the form
        $$frac{f_1(x)}{(p_1(x))^{n_1}} + cdots + frac{f_k(x)}{(p_k(x))^{n_k}}$$
        where $p_1,ldots,p_k$ are pairwise distinct monic irreducibles, and $deg(f_i)lt n_ideg(p_i)$. Going to an algebra closure and evaluating the numerators you get at roots of the $p_i$ easily yield that $f_i(x)=0$ for all $i$. This reduces to the case of fractions in which the denominators are all powers of the same monic irreducible, which is straightforward.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 1 '12 at 1:52









        Arturo MagidinArturo Magidin

        264k34590917




        264k34590917






























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