Proof of The Third Isomorphism Theorem

Mathematics Asked by Abhijeet Vats on August 12, 2020

Here’s what I’m trying to prove right now:

Let $V$ be a vector space over $mathbb{F}$. Let $M$ be a linear subspace of $V$ and $N$ be a linear subspace of $M$. Prove that the mapping $x+N mapsto x+M$ between the quotient spaces $V/N to V/M$ is linear with kernel $M/N$. Then, deduce that:

$$frac{V/N}{M/N} cong V/M$$

Proof Attempt:

We have to show that the given relation $T: V/N to V/M$ is a linear mapping. We define:

$$forall x in V: T(x+N) = x+M$$

This is totally-defined. To show well-definedness, let $x+N = y+N$ where $x,y in V$. Then, $x-y in N$. So, $x-y in M$. Hence:

$$x + M = y+M$$

$$iff T(x+N) = T(y+N)$$

To prove that it is linear, we need to show additivity and homogeneity.

  1. Proof of Additivity

Let $u,v in V/N$. Then, $u = x+N$ and $v = y+N$ for some $x,y in V$. Then:

$$T(u+v) = T((x+N)+(y+N)) = T((x+y)+N) = (x+y)+M = (x+M)+(y+M) = T(u) + T(v)$$

This proves additivity.

  1. Proof of Homogeneity

Let $alpha in mathbb{F}$ and $u in V/N$. Then, $u = x+N$ for some $x in V$. So:

$$T(alpha u) = T(alpha(x+N)) = T(alpha x +N) = alpha x + M = alpha (x+M) = alpha T(u)$$

This proves homogeneity. Hence, $T$ is a linear map. To show that the kernel of $T$ is $M/N$, we have:

$$T(x+N) = x+M = theta_V+M$$

$$iff x in M$$

$$iff x+N in M/N$$

$$iff ker(T) = M/N$$

Now, we notice that $T$ is surjective. By the first isomorphism theorem, it follows that:

$$frac{V/N}{M/N} cong V/M$$

That proves the desired result.

Does the proof above work? If it doesn’t, why? How can I fix it?

One Answer

Your approach is absolutely right!

I would change the part concerning $ker T$, writing: begin{align} x + N in ker T & iff T(x+N) = 0 in V/M \ & iff x+M = 0 in V/M \ & iff x in M \ & iff x + N in M/N end{align} which means $ker T = M/N$.

Correct answer by Rodrigo Dias on August 12, 2020

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