# Does $V_1/W_1\cong V_2/W_2$ hold for infinite dimensional vector spaces?

Let $V_1,V_2$ be vector spaces, and $W_1,W_2$ be respective subspaces of $V_1$ and $V_2$.

Suppose $V_1\cong V_2$ and $W_1\cong W_2$.

Main question) Does $V_1/W_1\cong V_2/W_2$ always hold (even for infinite dimensional vector spaces)?

For finite dimensional vector spaces, we can easily see that $\dim(V_1/W_1)=\dim(V_2/W_2)$ so that the isomorphism holds true.

Side question) Just out of curiosity, does the above result hold for any other special classes of modules? I know of this example ($\mathbb{Z}$-modules), $V_1=V_2=V=\mathbb{Z}_2\oplus\mathbb{Z}_4$, $W_1=\langle (1,0)\rangle\cong\mathbb{Z}_2$, $W_2=\langle (0,2)\rangle\cong\mathbb{Z}_2$, but $$V/W_1\cong \mathbb{Z}_4$$ while $$V/W_2\cong\mathbb{Z}_2\oplus\mathbb{Z}_2$$. Hence in general for modules it is false.

Thanks!

## 3 Answers

As you probably suspect, no, this isn't true in general.

Here's a counter-example. Let $V$ have countable dimension over the base field. Then let $V_1 = V_2 = V \times V$. Pick out subspaces $W_1 = V \times V$ and $W_2 = V \times \{ 0 \}$, embedded into their parent spaces in the notationally-suggested way. Then $V_1/W_1 \cong \{ 0 \}$, and $V_2/W_2 \cong V$.

For the second question, it's tough to generalize to modules. Even for one of the nicest classes of modules, free modules, you can still run into trouble. For example, the subgroups $2\Bbb Z$ and $3 \Bbb Z$ are isomorphic, but the quotients $\Bbb Z / 2 \Bbb Z$ and $\Bbb Z / 3 \Bbb Z$ are not.

You might want to look at this question for a better answer: If two submodules are isomorphic, so is their quotient… conditions on the ring!

• Nice. To show $W_1\cong W_2$ are you using the fact that two vector spaces with countable dimension of same cardinality are isomorphic? Or is there a better way to see it? – yoyostein Oct 5 '17 at 4:15
• If two vector space have the same dimension, they're isomorphic. Finite, countable, uncountable, doesn't matter! :) The way to see this is to think of an isomorphism as a bijection between the two bases. – Henry Swanson Oct 5 '17 at 4:28

Take $V_1 = V_2$ to be an infinite vector space (say, the direct product of countably many copies of the ground field) and $W_1, W_2$ to be subspaces of different finite codimension.

For the side question, the first place to look would be free modules, but it's hard to find a situation in which the result you mention would hold. Finite-dimensional vector spaces (over a fixed ground field) are completely characterized by their dimension, which is even additive for short exact sequences.

• Ha. You type faster than I do! – JonathanZ Oct 5 '17 at 3:59

No. Let $$V_1 = V_2 = \mathbb{R}^\infty = \{ (a_0, a_1, ....) | a_i \in \mathbb{R} \},$$ and $$W_1 = \{ (a_0, a_1, ....) | a_0 = 0\},$$ $$W_2 = \{ (a_0, a_1, ....) | a_0 = a_1 = 0\}$$.