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How do we compute the dimension as a $k$-vector space $(\mathrm{char}(k) \neq 2)$ of $k[x,z]/(x^{2}+1,z^{2})$?

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I'm not sure what the notation $k[x,z]$ means. But, for computing the dimension of a quiotient of vector spaces, you can use the following theorem. Let V be a vector space with H a subspace. let $\phi: V\rightarrow \frac{V}{H}$ be given by $\phi(v)=v+H$. Then $\phi$ is an onto homomorphism. So, we have that $nullity(\phi)+rank(\phi)=dim(V)$ and hence As a consequence of you have that $dim(\frac{k[x,z]}{x^2+1, z^2})=dim(V)-nullity(\phi)$ –  Chris Dugale May 22 '12 at 5:47
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Aren't $1,x,z,xz$ linearly independent? –  Gerry Myerson May 22 '12 at 6:30

1 Answer 1

up vote 2 down vote accepted

Succinctly: $$k[x,z]/(x^2 + 1, z^2) \simeq k[x]/(x^2 + 1) \otimes_k k[z]/z^2$$Morally, it's cuz $x$ and $z$ have nothing to do with each other. Formally, these bad boys are isomorphic as $k$ algebras, hence rings, since to map out of either just requires choosing elements of the target $x', z'$ satisfying $x'^2 = -1, z'^2 = 0$.

So the dimension of the whole thing is the product of their individual dimensions, and $k[t]/p(t)$ has dimension degree of $p$, so 4.

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