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Given an ideal $I = \langle x-y,y^3+y+1 \rangle \subset \mathbb{C}[x,y]$ (this is a Gröbner basis w.r.t. degree-lexicographic order). I want to write $\mathbb{C}[x,y]/I$ as a $\mathbb{C}$-Basis and determire $\operatorname{dim}_{\mathbb{C}}\mathbb{C}[x,y]/I$. I know what a quotient ring is and how it is definied but I have no intuition how $\mathbb{C}[x,y]/I$ looks like. Any hints?

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    $\begingroup$ Hint: $x-y\in I$ means that that in $C[x,y]/I$, $x$ and $y$ are equal. $\endgroup$ – Thomas Andrews Aug 30 '12 at 16:34
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Well, $\mathbb{C}[x,y]/I$ is still spanned as a complex vector space by the monomials $x^i y^j$. However, that spanning set is not linearly independent: any linear combination of monomials that adds up to an element of $I$ is equal to zero!

Thinking of the Groebner basis as a rewrite scheme is useful too: the form of your basis says:

  • Whenever I see an $x$, replace it with $y$
  • Whenever I see a $y^3$, replace it with $-y-1$

which gives you an algorithm to convert any polynomial to a unique normal form... and makes it easy to see what polynomials can be normal forms.

(To be clear, whenever you see a $y^4$, that also means you see a $y^3$, because $y^4 = y^3 \cdot y$)

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  • $\begingroup$ Thanks for the great answer! If I understand you correctly I assume all the equivalence classes are of the form $[a \cdot 1], [a \cdot y], [a \cdot y^2]$ with $a \in \mathbb{C}$. Thus, $\operatorname{dim}_{\mathbb{C}}\mathbb{C}[x,y]/I = 3$? $\endgroup$ – Joachim Aug 30 '12 at 18:11
  • $\begingroup$ That is indeed a basis for $\mathbb{C}[x,y]/I$. (don't forget the quotient ring also contains linear combinations of them!) $\endgroup$ – user14972 Aug 30 '12 at 18:47

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