Dimension of $ k[x,y,z] / (x, y^2, z^2)$ as a $k$-vector space

Question

I have seen that $$k[x] / \langle f \rangle $$ has dimension as a $k$-vector space equal to the degree of $f$.

But how would we extend this definition in the cases such as the one below?

For example, $$k[x,y,z] / \langle x, y^2 , z^2 \rangle \cong k[y,z] / \langle y^2, z^2 \rangle$$

Would it be wrong to think of $k[y,z] / \langle y^2, z^2 \rangle$ as a coordinate ring in which case all elements would look like $p(y) + p(y) + p(y)z^2 + \langle I \rangle ?$

In this case it would then have dimension 3, maybe.

Answer 1

$ k[x,y,z] / \langle x, y^2 , z^2 \rangle \cong k[u,v,w] $ where $u=0$, $v^2=0$, $w^2=0$. So $k[u,v,w]$ is the set of all expressions of the form $a+bv+cw+dvw$, with $a,b,c,d \in k$. So the dimension is $4$.

Answer 2

Unfortunately, as far as I know, there is no criterion for determining the dimension of $k[x_1,\ldots,x_n]/I$ given an ideal $I$, when $n>1$. All one can easily say is that it is infinite-dimensional if $I$ is generated by fewer than $n$ elements.

Your simplification $k[x,y,z] / \langle x, y^2 , z^2 \rangle \cong k[y,z] / \langle y^2, z^2 \rangle$ is a good start. Next you can look where a (standard) basis of $k[y,z]$ over $k$ is mapped to in $k[y,z] / \langle y^2, z^2 \rangle$. For example, the $k$-basis $$\mathcal{B}=\{y^mz^n\mid m,n\geq0\}\qquad\text{ maps to }\qquad \overline{\mathcal{B}}=\{\overline{1},\overline{y},\overline{z},\overline{yz}\},$$ where $\overline{w}:=w+\langle y^2,z^2\rangle\in k[y,z]/\langle y^2,z^2\rangle$. Because the quotient map is surjective this shows that $$\dim_kk[y,z]/\langle y^2,z^2\rangle\leq4,$$ and it is not hard to check that $\overline{\mathcal{B}}$ is linearly independent over $k$, so $\dim_kk[y,z]/\langle y^2,z^2\rangle=4$.