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I am more looking for what concepts/techniques I should understand to solve this problem rather than explicit solution (though that definitely does not hurt). It might sound strange but I'm studying for an exam that I don't know what will be tested...

Consider $\mathbb{C}[x,y]$. Consider $I = \langle x^2 + 4x + 4, xy+x+2y+2, y^3 + 3y^2 + 3y + 1\rangle$. What is the dimension of $\mathbb{C}[x,y]/I$ as a vector space over $\mathbb{C}$?

I guess I should first convince myself that $I$ is a subring. But according to a discussion here, it might not be the case, so that's already quite confusing.

Anyway, playing with the polynomials, I get

$$I = \langle (x+2)^2 , (x+2)(y+1), (y+1)^3 \rangle$$

I sense that this could be useful since it looks better now.Then I suppose the question becomes, when we divide some $f$ by $ a(x+2)^2 + b(x+2)(y+1) + c(y+1)^3$, what could be the possible remainders (where $f,a,b,c \in \mathbb{C}[x,y]$)? Because that's how I used to think about quotient ring of one variable.

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3x89g2
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1 Answers1

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  1. $I$ is not a subring of $\mathbb{C}[x, y],$ it is an ideal. But that's the right object you want for taking a quotient of a ring, so that's OK.

  2. $I=\langle (x+2)^2, (x+2)(y+1), (y+1)^3 \rangle$ definitely looks better. What would make it even better is if we were able to get rid of the constants involved. Convince yourself that there is an automorphism of $\varphi:\mathbb{C}[x, y]\rightarrow \mathbb{C}[x, y]$ that is identity on the constant polynomials and under which $(x+2)$ goes to $x$ and $(y+1)$ goes to $y$. Then instead of computing the dimension of $\mathbb{C}[x, y]/I,$ you may as well compute the dimension of $\mathbb{C}[x, y]/\varphi(I)$ (what is $\varphi(I)$?)

  3. Another good trick (which works well together with 2.): Observe that $\mathbb{C}[x,y]$ is a direct sum of subspaces given by homogeneous polynomials of fixed degree $d$ (i.e. $\mathbb{C}[x, y]=\bigoplus_{d \geq 0} V_d,$ where $V_d$ has basis $x^d, x^{d-1}y, \dots, xy^{d-1}, y^d$). If you use 2., you can compute the dimension by computing $\dim_{\mathbb{C}}V_d/\varphi(I)$ degree by degree (this will be more clear after you compute $\varphi(I)$).

Pavel Čoupek
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  • In some books, ideals are actually subrings. Not everyone demands rings to contain $1$. – Arthur Nov 08 '16 at 22:18
  • @Arthur That's true and worth noticing, but in the absence of some specific convention that would say otherwise, I would incline towards requiring rings to have unit, since that is definitely more common. Besides, the point is that from the context it seems important that $I$ is an ideal, regardless of the fact whether ideals are considered subrings or not. – Pavel Čoupek Nov 08 '16 at 22:24
  • I'm afraid that I don't fully understand. For the second part, I intuitively see what we are doing. Changing $x$ to $x+2$ and $y$ to $y+1$ shouldn't change the "structure". I showed that $\phi: \mathbb{C}[x,y] t\ \mathbb{C}[x,y]$ defined by $\phi(x) = x+2$ and $\phi(y) = y +1$ is an automorhpism (though probably not very rigorously). So essentially we are considering $\mathbb{C}[x,y]/\langle x^2, xy, y^3\rangle$. – 3x89g2 Nov 09 '16 at 00:18
  • But I am a little bit lost on the third part. I understand the direct sum part and I am convinced that $\mathbb{C}[x,y]$ can be written like that. Now I suppose that $(\oplus V_d)/ \phi(I) = \oplus ( V_d/ \phi(I))$? Suppose that's true, then are you saying that, we really only need to care about $V_1$ and $V_2$? – 3x89g2 Nov 09 '16 at 00:22
  • By the way, does the existence of such $\phi$ relies on the fact that everything in $\mathbb{C}[x]$ splits uniquely? – 3x89g2 Nov 09 '16 at 00:31
  • @Misakov ad 2: Yes, only if you want $x+2$ go to $x$, you need to send $x$ to $x-2$. Similarly, $y$ should be sent to $y-1$. The map you describe is the inverse of $\varphi$. ad 3: Yes, the equality you suggest is essentially true. Point is that $\varphi(I)$ is generated by homogeneous ideals, from which one can deduce that $\varphi(I)=\bigoplus_{d}\varphi(I)_d,$ where $\varphi(I)_d=V_d \cap \varphi(I)$ (such ideals are called homogeneous). Then $(\bigoplus_dV_d)/(\bigoplus_d\varphi(I)_d=\bigoplus_d(V_d/\varphi(I)_d)$. – Pavel Čoupek Nov 09 '16 at 00:48
  • @Misakov And yes, it turns out that the only nontrivial pieces will be $V_0=\mathbb{C}, V_1/\varphi(I)_1$ and $V_2/\varphi(I)_2$. – Pavel Čoupek Nov 09 '16 at 00:48
  • Let us continue this discussion in chat. – 3x89g2 Nov 09 '16 at 00:56