The Basis for the set of cosets (V/U) does not count for (u + U), where u is an element of U (and V)

Question

I havent found any similar questions regarding my interests, feel free to correct me if needed.

My confusion comes from a theorem in my math notes, which goes as follows: Let $B_1$ be a basis for U, and extend $B_1$ to $B$ to form a basis for V, assuming V is finite-dimensional. Then $B_2 = \{e + U | e \in B \setminus B_1 \}$ forms a basis for V/U, where $V/U=\{v+U | v \in V \}$.

With this the notes give an example: Take

$V = F[x]$ with $B = \{1,x,x^2, . . . \}$

$U = $ even polynomials with $B_1 = \{1,x^2, x^4, . . . \}$

Then V/U is similar to the odd polynomials, with $B_2 = \{ x + U, x^3 + U, . . . \}$

My question here is then take $1 + U \in V/U$, and since $1\in U$, essentially $1+U = U$, which denote even polynomials. Then $1 + U = U \in V/U$, but $V/U$ is similar to the odd polynomials, hence does not include any even polymomials, forming a contradiction.

My guess here is that when forming a set of cosets V/U = {v + U}, we dont think about the cases when v is an element of U..? But other then that Iam quite lost, and some help would greatly be appreciated. Thanks a lot!

Answer 1

It is always true that $U\in V/U$ but notice here that $U$ is an element of $V/U$ rather than a subset. If it were a subset then we would have grounds to say that even polynomials are in $V/U$ but what $U\in V/U$ actually expresses is that all even polynomials lie in the same equivalence class in $V/U$ i.e. they are undistinguishable from the zero element in $V/U$.

One way to look at this is to write $[f(x)]$ for the element $f(x) + U$. Then you can note that $[x],[x^{3}]$ and $[x^{5}]$ represent different classes while $[1],[x ^{2}]$ and $[x^{4}]$ are all the same element, therefore $V/U$ "only recognizes" the odd polynomials and acts like it aswell.

Statements such as that which you have been given are true in more general quotients. For instance in group theory we can form similar quotients, you might be familiar with addition modulo $n$, which really comes from the quotient group $\mathbb{Z}/n\mathbb{Z} = \{[0],\dots,[n-1]\}$. Here $[0] = n\mathbb{Z}$ represents all multiples of $n$ and $n\mathbb{Z}\in \mathbb{Z}/n\mathbb{Z}$ but we do not say that any multiples of $n$ are in $\mathbb{Z}/n\mathbb{Z}$ since any multiple of $n$ is just a representative of the single element $[0]$.

Answer 2

As you correctly saw $1 \in U$, therefor $1-0\in U$, so $1 + U = 0 + U$, where $0$ is in $U$ as it is a subvectorspace. This shows that for all $u\in U$ $u+U = 0 +U$.

I think your confusion may arise from the fact that on the one side we have cosets which form a vectorspace which is canonically isomorphic to $B_2$, it is by the first isomorphism theorem the map with kernel $U$. And the image of each element has to lie in a vectorspace isomorphic to our uneven polynomials, but not every preimage has to be an uneven polynomial. In fact in this map every coset has exactly one preimage which is an even polynomial. In formulas we have the following (For W the space of uneven polynomials): $$\phi:F[x] \to F[x]/U, x \mapsto x + U$$ $$f':F[x]/U\to W, x^{2n+1}+U\mapsto x^{2n+1}$$ The composition of those maps gives us $f = f'\circ\phi$: $$f:F[x]\to W, x^n\mapsto \begin{cases} x^n &&for\;n=2k+1\\ 0 &&for\; n = 2k\end{cases}$$

Now you can see that $1$ does lie in the coset $1+U$, by definition, but if you look at the isomorphism $f'$ to $W$ the image of $1$ is an uneven polynomial.