Integral domain.

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Let k a field.



Let $ A = k+ x^2 k[x] $,



Show that A is integral domain and finite type.



An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $



I know that A is a subalgebra of k[x],
Any hint for integral domain




abstract-algebra ring-theory




share|cite|improve this question

























    up vote
    1
    down vote

    favorite












    Let k a field.



    Let $ A = k+ x^2 k[x] $,



    Show that A is integral domain and finite type.



    An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $



    I know that A is a subalgebra of k[x],
    Any hint for integral domain




    abstract-algebra ring-theory




    share|cite|improve this question























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      Let k a field.



      Let $ A = k+ x^2 k[x] $,



      Show that A is integral domain and finite type.



      An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $



      I know that A is a subalgebra of k[x],
      Any hint for integral domain




      abstract-algebra ring-theory




      share|cite|improve this question













      Let k a field.



      Let $ A = k+ x^2 k[x] $,



      Show that A is integral domain and finite type.



      An element of $A$ is $alpha + x^2. f(x) $ where$ fin k[x] $



      I know that A is a subalgebra of k[x],
      Any hint for integral domain





      abstract-algebra ring-theory





      share|cite|improve this question












      share|cite|improve this question




      share|cite|improve this question








      edited Aug 6 at 20:15
























      asked Aug 6 at 19:56









      Ali NoumSali Traore

      486




      486




















          1 Answer
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          Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.






          share|cite|improve this answer





















          • An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
            – Ali NoumSali Traore
            Aug 6 at 20:06










          • Yes and such an element is a polynomial in $k[x]$.
            – paf
            Aug 6 at 20:07










          • Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
            – misogrumpy
            Aug 6 at 20:36










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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          2
          down vote













          Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.






          share|cite|improve this answer





















          • An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
            – Ali NoumSali Traore
            Aug 6 at 20:06










          • Yes and such an element is a polynomial in $k[x]$.
            – paf
            Aug 6 at 20:07










          • Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
            – misogrumpy
            Aug 6 at 20:36














          up vote
          2
          down vote













          Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.






          share|cite|improve this answer





















          • An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
            – Ali NoumSali Traore
            Aug 6 at 20:06










          • Yes and such an element is a polynomial in $k[x]$.
            – paf
            Aug 6 at 20:07










          • Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
            – misogrumpy
            Aug 6 at 20:36












          up vote
          2
          down vote










          up vote
          2
          down vote









          Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.






          share|cite|improve this answer













          Since $Asubset k[x]$ and since $k[x]$ is an integral domain, we can see that for all $a,bin A$, $(ab = 0$ in $A) Rightarrow (ab = 0$ in $k[x]) Rightarrow (a=0$ or $b=0)$.







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Aug 6 at 20:02









          paf

          3,9391823




          3,9391823











          • An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
            – Ali NoumSali Traore
            Aug 6 at 20:06










          • Yes and such an element is a polynomial in $k[x]$.
            – paf
            Aug 6 at 20:07










          • Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
            – misogrumpy
            Aug 6 at 20:36
















          • An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
            – Ali NoumSali Traore
            Aug 6 at 20:06










          • Yes and such an element is a polynomial in $k[x]$.
            – paf
            Aug 6 at 20:07










          • Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
            – misogrumpy
            Aug 6 at 20:36















          An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
          – Ali NoumSali Traore
          Aug 6 at 20:06




          An element of $A$ is $alpha + x^2. f(x) $ when $ fin k[x] $
          – Ali NoumSali Traore
          Aug 6 at 20:06












          Yes and such an element is a polynomial in $k[x]$.
          – paf
          Aug 6 at 20:07




          Yes and such an element is a polynomial in $k[x]$.
          – paf
          Aug 6 at 20:07












          Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
          – misogrumpy
          Aug 6 at 20:36




          Isn't $(x^2)$ a prime ideal of $A$? Then the quotient $A/(x^2)$ is...?
          – misogrumpy
          Aug 6 at 20:36












           

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