the leading term of a module

I'm reading CLO and I have a question about the following Prop:

Let I be an ideal in a polynomial ring $k[x_1,\ldots,x_n]$. Then $k[x_1,\ldots,x_n]/I$ is isomorphic as a $k$-vector space to $S = span(x^{\alpha}: x^{\alpha} \not\in \left< LT(I)\right> ).$

Doesn't this mean that $$k[x_1,\ldots,x_n]/I\cong k[x_1,\ldots,x_n]\left< LT(I)\right>$$ where $LT(f)$ is the leading term of $f\in I$ with respect to some monomial order and $$LT(I)= \left< LT(f): f\in I \right>?$$

This appears to be plausible so I would like to double check this with someone else. Thanks.

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Which book is CLO? –  Ted May 20 '12 at 18:38
Cox, Little, and O'Shea: Ideals, Varieties, and Algorithms. –  math-visitor May 20 '12 at 18:40
I think the answer to my question is a yes and the idea is supposed to be simple, but I am relatively new to thinking in terms of computational algebraic geometry. –  math-visitor May 20 '12 at 18:46
Something doesn't seem right here, because the structure of $k[x_1, x_2, \ldots, x_n]/I$ has got to depend on more than just $\langle LT(I) \rangle$. –  Ted May 20 '12 at 18:48
You might be right, Ted.. but at the same time, the quotient module $k[x_1,\ldots, x_n]/I$ is generated by those generators in the complement of $\left< LT(I)\right>$. So that's why this was the next natural question to ask. –  math-visitor May 20 '12 at 18:54