Gröbner basis is not a vector basis?

Question

We use the same notation for Gröbner basis and vector basis. I recall that $\langle 1\rangle_{GR}$ is the largest Gröbner basis while $\langle 1\rangle_{vector}$ is the smallest vector basis. So for example

$$x^2\not\in \langle 1,x\rangle_{vector}$$

where the vector space satisfy the vector space conditions such as multiplication by scalar and existence of multiplicative identity. Because vector space is not an ideal, the multiplication by variable such as $x\cdot x$ in $\langle 1,x\rangle$ is not allowed. In contrast the Gröbner basis span an ideal so for instance

$$x^2\in \langle 1,x\rangle_{GR}$$

where the Gröbner basis satisfy the ideal conditions such as the multiplication in more general way: $\langle 1,x\rangle_{GR}$ is an ideal so also $x^n\in\langle 1,x\rangle_{GR}\forall n$.

I want to verify the above. So

1. Is Gröbner basis a vector basis?

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2. Do the sizes of the bases work like the below?

GR basis in Algebraic Geometry: which GR space does the $\{ x \}$ span?

1. $\langle 1 \rangle$ is the largest GR space while

2. $\langle 1,x\rangle$ is a little bit smaller, then

3. $\langle 1,x,x^2,\ldots\rangle$ would be the smallest.

Vector basis

1. $\langle 1 \rangle$ is the smallest vector space while

2. $\langle 1,x\rangle$ is a little bit larger, then

3. $\langle 1,x,x^2,\ldots\rangle$ would be the largest.

Answer 1

I can't totally parse the post, but

A Gröbner basis is a special generating set for an ideal in a ring $R$ considered as an $R$-submodule of $R$.

and

A basis for a vector space is a special generating set for the vector space as an $\Bbb F$-module for some field (or even division ring) $\Bbb F$.

Yes, if $1$ were part of a Gröbner basis for an ideal, the ideal would have to be the entire ring since $1\cdot R=R$. The same is true for the $\Bbb F$ vector space $\Bbb F$: $\{1\}$ is a basis. There is no difference between these two examples.

Also in both examples, if you have a basis $\{b_1,b_2,\ldots\}$ of either type, you have an increasing sequence $\{b_1\}\subseteq \{b_1,b_2\}\subseteq\ldots$. There is no difference between the two. Adding elements to the Gröbner basis does not decrease what it generates: after $1$ is added, you get the entire ring, and adding more things makes no difference.

Your whole problem seems to stem from your surprise that the $R$ submodule of $R$ generated by $1_R$ is all of $R$. Of course, if you are simultaneously thinking of $R$ as an $\Bbb F$ algebra over some field, then of course the subspace generated by $\Bbb F$ and $1_R$ may not be the entire ring: you are working with a (typically) smaller coefficient ring, so you won't be able to produce as many multiples.