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Question:

Determine an $\mathbb{R}$-basis for $\mathbb{R}[x,y]/(x^2-x,y^2-y)$.

There is a similar question here Linear basis for a quotient ring, which involves determining a basis for $\mathbb{R}[x]/(x^2+k)$. But I am confused how to apply this method since the ideal we are quotienting by is generated by two polynomials.

Should we do something like: take an $f\in \mathbb{R}[x,y]$ and consider $(f \pmod{x^2-x})\pmod{y^2-y}$? But this seems very messy

user26857
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T. Stark
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2 Answers2

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Well, the monomial standard basis of $R={\Bbb R}[x,y]/\langle x^2-x,y^2-y\rangle$ is given by $\{1,x,y,xy\}$, since $x^2-x=0$ and $y^2-y=0$ in $R$.

Note for T. Stark: Consider all monomials $x^iy^j$ in the polynomial ring ${\Bbb R}[x,y]$ and apply the reduction $x^2=x$ and $y^2=y$ in $R$.

user26857
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Wuestenfux
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  • Sorry why does it follow from $x^2-x=0$ and $y^2-y=0$ in $R$ that ${1,x,y,xy}$ is a basis? – T. Stark Sep 23 '19 at 17:59
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    I believe it is easy to see that $1,x,y,xy$ generate the quotient. (Any polynomial with monomials such as $x^ay^b$ where either $a\ge 2$ or $b\ge 2$ is equivalent to one of a lower degree because $x^2=x$ and $y^2=y$.) @Wuestenfux: How do we prove that they are linearly independent? –  Sep 23 '19 at 18:04
  • @StinkingBishop: If $a + bx + cy +dxy=0$ in the quotient ring, then by the natural surjection $a+bx+cy+dxy=0$ in the polynomial ring. But in the polynomial ring $a=b=c=d=0$. Done. – Wuestenfux Sep 24 '19 at 07:46
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There is a surjective map from $\mathbb{R}[x,y]$ to the quotient, so an $\mathbb{R}$ basis of the polynomial ring will map to a generating set (not necessarily a basis) of the quotient. So take your favorite basis of $\mathbb{R}[x,y]$, and then figure out which of those elements become redundant after the map.

TomGrubb
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  • I am confused in actually figuring out which elements are redundant due to quoitenting by two polynomials.

    Also as an $\mathbb{R}$ vector space $\mathbb{R}[x,y]$ is infinite dimensional so I'm not sure how picking a basis will help

     – T. Stark Sep 23 '19 at 17:57