The problem is compute the dimension of $\Bbb C[x,y]/I$ over $\Bbb C$ as vector space where $I=\langle(x+2)^2,(x+2)(y+1),(y+1)^3\rangle$. $\Bbb C[x,y]$ is the polynomial ring over $\Bbb C$.

I have tried in this way.

For $f(x,y)$ in $\Bbb C[x,y]$ it has the form $f(x)=(a_{0}+a_{1}(x+2)+…a_{i}(x+2)^i+…a_{n}(x+2)^n)(b_{0}+b_{1}(y+1)+…b_{j}(y+1)^j+…b_{m}(x+2)^m))=[a_{0}+a_{1}(x+2)][(b_{0}+b_{1}(y+1)+b_{2}(y+1)^2]+f_{1}(x,y)(x+2)^2+f_{2}(x,y)(y+1)^3$

And $[a_{0}+a_{1}(x+2)][(b_{0}+b_{1}(y+1)+b_{2}(y+1)^2]=a_{0}b_{0}+a_{1}b_{0}(x+2)+a_{0}b_{1}(y+1)+a_{0}b_{2}(y+1)^2+f_{3}(x,y)(x+2)(y+1)$

So $f(x,y)=a_{0}b_{0}+a_{1}b_{0}(x+2)+a_{0}b_{1}(y+1)+a_{0}b_{2}(y+1)^2$ in $\Bbb C[x,y]/I$ .

The basis are {$1,x+2,y+1,(y+1)^2$}.Any polynomial's coordinate in $\Bbb C[x,y]/I$ is $(a_{0}b_{0},a_{1}b_{0},a_{0}b_{1},a_{0}b_{2})$

The remain is the dimension for $(a_{0}b_{0},a_{1}b_{0},a_{0}b_{1},a_{0}b_{2})$ in $\Bbb C^4$ .I am not very sure for it.4 or 5,or other?

  • $\begingroup$ What is your definition of dimension? Is it Krull dimension? $\endgroup$ – Servaes Dec 25 '16 at 15:15
  • $\begingroup$ @Servaes : I guessed this was the dimension as a vector space but maybe I'm mistaken ... $\endgroup$ – user171326 Dec 25 '16 at 15:18
  • $\begingroup$ That's prefectly possible, I guess if you encountered this in a linear algebra class then this would be the intended dimension. $\endgroup$ – Servaes Dec 25 '16 at 15:20

$\mathbb C[x,y] \cong \mathbb C[x+2,y+1]$. Thus your ideal under this isomorphism is simply $\langle x^2, xy, y^3 \rangle $. In particular you can see easily that as a vector space $\mathbb C[x,y]/I \cong \mathbb C \{1, x, y, y^2\}$, i.e $\dim \mathbb C[x,y]/I = 4$.

  • $\begingroup$ When I use 5 variables for 4 coordinates.Is the dimension also 4? $\endgroup$ – gtx Dec 25 '16 at 15:20
  • $\begingroup$ May be it is trivial,but where can I find a proof for $C[x,y] \cong C[x+2,y+1]$. $\endgroup$ – gtx Dec 25 '16 at 15:25
  • 2
    $\begingroup$ Consider the map $\mathbb C[x,y] \to \mathbb C[x+2,y+3]$, $x \mapsto x+2$, and $y \mapsto y +3$. It has an inverse (namely the map which sends $x$ to $x - 2$ and $y$ to $y-3$. Thus it is an isomorphism. $\endgroup$ – user171326 Dec 25 '16 at 15:33
  • $\begingroup$ (Sorry I did put $y+3$ instead of $y+1$. But the proof is exactly the same.) $\endgroup$ – user171326 Dec 25 '16 at 15:35
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    $\begingroup$ Taking constant polynomial with coeff $1$ gives $(1,0,0,0)$. Similarly you can see that choosing good polynomial you can get a basis of $\mathbb C^4$. But a better way for seeing this is to look simply at the set of the polynomials $\mathbb C\{1, x, y, y^2\}$. By definition of a polynomial there are linearly independant and they generate $\mathbb C[x,y]/I$ by my argument. $\endgroup$ – user171326 Dec 25 '16 at 15:44

A basis of $C[x,y]$ is $\{(x+2)^m(y+1)^n: m,n\ge 0\}$.

In $C[x,y]/I$, $(x+2)^2$ and $(y+1)^3$ are identified with $1$, hence, just because of them the basis reduces to $$ (x+2)^m(y+1)^n, \quad m=0,1,\,\,n=0,1,2. $$ But as $(x+2)(y+1)$ is also identified with $1$, then we are reduced to $$ 1, x+2, y+1, (y+2)^2. $$ Hence $$ \dim C[x,y]/I=4. $$

  • $\begingroup$ So the dimension make no difference with $(a_{0}b_{0},a_{1}b_{0},a_{0}b_{1},a_{0}b_{2})$ ?5 variables describe 4 coordinates. $\endgroup$ – gtx Dec 25 '16 at 15:34

As others have noted, by applying the automorphism \begin{align*} \mathbb{C}[x,y] &\to \mathbb{C}[x,y]\\ x &\mapsto x+2\\ y &\mapsto y+1 \end{align*} we find that $$ \frac{\mathbb{C}[x,y]}{\langle(x+2)^2,(x+2)(y+1),(y+1)^3\rangle} \cong \frac{\mathbb{C}[x,y]}{\langle x^2,xy,y^3\rangle} \, . $$ The ideal $J = \langle x^2,xy,y^3\rangle$ is a monomial ideal, since it is generated by monomials. We can visualize monomial ideals as lattices. For each generator $x^i y^j$ of $J$, we put a star at the point $(i,j)$ in the plane, then shade up and to the right. (This corresponds to the fact that $x^m y^n \in \langle x^i y^j \rangle$ iff $m \geq i$ or $n \geq j$.)

$\hspace{4cm}$enter image description here

The points not contained in the shaded region correspond to monomials that do not belong to $J$. These monomials form a basis for $\mathbb{C}[x,y]/J$, so we can simply count them to find its dimension. For more on this, I recommend Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms.

  • $\begingroup$ So the dimension is 3? $\endgroup$ – gtx Dec 26 '16 at 1:19
  • $\begingroup$ I count $4$ black dots in the unshaded region: $(0,0), (1,0), (0,1)$, and $(0,2)$. $\endgroup$ – Viktor Vaughn Dec 26 '16 at 1:47

Here's an approach you may find simpler: The fact that $(x+2)^2=x^2+4x+4$ is in $I$ means that in $\mathbb{C}[x,y]/I$ you can always replace any occurrence of $x^2$ with $-4x-4$. Similarly, the fact that $(x+2)(y+1)=xy+2y+x+2 \in I$ means that in $\mathbb{C}[x,y]/I$ you can always replace any occurrence of $xy$ with $-2y-x-2$. Finally, the fact that $(y+1)^3 = y^3 + 3y^2 + 3y + 1 \in I$ means that in $\mathbb{C}[x,y]/I$ you can always replace any occurrence of $y^3$ with $-3y^2-3y-1$.

Applying these three "reduction rules" systematically, you can reduce any monomial of the form $x^k y^j$ down to a linear combination of terms in which the highest power of $x$ that appears is $x^1$, the highest power of $y$ that appears is $y^2$, and no term contains both $x$ and $y$.

Any such polynomial can be written as a linear combination of $$1, x, y, y^2.$$ There are no further relations in $I$, so it's impossible to reduce any further than this. So the dimension of $\mathbb{C}[x,y]/I$ as a vector space over $\mathbb{C}$ is $4$, and the monomials listed above form a basis.


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