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Let $k = \mathbb{F}_2$ and $R = k[x,y,z]$, and suppose $I \subseteq R$ is an ideal of R. I want to show that $\dim_k (R/I) = \text{length}_R (R/I)$.

For instance, in my specific example, $I = (x^2,y^2,z^2,xyz)$. Obviously a $k$-basis for $R/I$ is made of all the monomials that are not multiples of $x^2, y^2, z^2$ or $xyz$, i.e. a basis can be $\{1,x,y,z,xy,xz,yz\}$. Since it is of cardinality $7$, we can state that $\dim_k (R/I) = 7$.

My teacher then states that we can deduce from this that $\text{length}_R(R/I)$ is also $7$.

Any idea on how to prove this ?

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    Can you find a maximal submodule in your example? – 2017-01-05
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    The relation holds if $I$ is zero-dimensional. – 2017-01-05
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    In general, the equality you write is not correct. If $I$ is a maximal ideal ideal, then length of $R/I$ is one, but $\dim_k(R/I)$ may be larger. – 2017-01-05
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    Mariano Suarez-Alvarez, I think that $\cfrac{R/I}{(yz)}$ should be a maximal submodule, shouldn't it ? Mohan, I disagree, if $I = (x,y,z)$ is maximal in $R$, $length_R (R/I) = 1$ but a basis of $(x,y,z)$ is $(1)$ and thus $dim_k (R/I)$ is also $1$. – 2017-01-05
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    I did not say any thing about your special case, where indeed you have radical of $I$ is an $\mathbb{F}_2$ rational maximal ideal. But, in general, your equality is wrong. Also, the length in your case is 8, not 7, since $xyz$ is another element which is linearly independent from the seven you write. – 2017-01-05
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    My bad I forgot to pu $xyz$ in the ideal (so it's indeed 7) – 2017-01-06
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    I might have solve my problem, tell me if I'm wrong though. – 2017-01-06
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    On one hand, $dim_k(R/I) = l_k(R/I)$ because $k$ is a field, $\geq l_R(R/I)$ because every chain of $R$-submodules is a chain of $k$-submodules. On the other hand, write $n = dim_k(R/I)$. Then $R/I = R/(x_1, \cdots, x_n)$ for some $x_1, \cdots, x_n \in R = \mathbb{F}_2 [x,y,z]$. Thus we order $x_1, \cdots, x_n$ by polynomial degree (i.e. $deg(x_1) \leq \cdots \leq deg(x_n)$), then a $R$-submodule chain of length $n$ can be : $$0 = \frac{R/I}{(x_1, \cdots, x_n)} \subsetneq \frac{R/I}{(x_2, \cdots, x_n)} \subsetneq \cdots \subsetneq \frac{R/I}{(x_n)} \subsetneq R/I$$ – 2017-01-06

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