Basically what says in the top line is :Let $k$ be a filed, $V$ a vector space over $k$ and $T$ an endomorphism of V as a $k$-module.Consider the structure of a $K[x]$-module in V induced by the endomorphism. Can anyone help me prove the following result? Ive tried creating homomorphism with the universal property of the tensor product and then proving they are inverse to one another but im not seeming to get any luck.Thanks in advance.
Let me simply follow what you tried and see where it doesn't work, if it's ever the case.
A first thing to ask yourself is "what is the structure of $k[x]$-module on the right ?". But $k[x]$ is a $k[x]$-module so this is essentially basic commutative algebra : this structure is simply $\lambda\cdot (a\otimes b) =(\lambda a)\otimes b$. Let me call this module $N$.
The second thing to note is that the quotient is a quotient as $k[x]$-modules, so the $\langle \rangle$ denotes the generated sub-$k[x]$-module. Let me call $M$ the quotient module on the right, and $L$ the submodule by which we quotient. In short, $M=N/L$
With these things in mind, let's proceed. By induction, we prove that $x^n\otimes v = 1\otimes T^n v$ in the module on the right.
Indeed $x\otimes v - 1\otimes Tv$ is in the kernel of the quotient map by definition; and then if it's true for $n$ we have $x^{n+1}\otimes v = x\cdot (x^n\otimes v) = x\cdot (1\otimes T^n v) = x\otimes T^nv= 1\otimes T^{n+1}v$.
Now from this it follows by linearity that for any $P\in k[x]$, $P\otimes v = 1\otimes P(T)(v)$.
This proves that $h: V\mapsto M, v\mapsto 1\otimes v$ is $k[x]$-linear.
Let's now define the converse. Define $f: k[x]\times V\to V$, $(P,v)\mapsto P(T)(v)$. It's easy to check that this is $k$-bilinear so factors through $k[x]\otimes_k V = N$. Let's call $g$ the map $N\to V$ thay we get.
But now on $L$, $g(x\otimes v -1\otimes Tv) = g(x\otimes v) - g(1\otimes Tv) = Tv- Tv = 0$. So $g$ factors through $N/L= M$.
What this means is that $k: P\otimes v\mapsto P(f)(v)$ is well-defined on $M$. Now we check that $k$ and $h$ are inverse to eachother, but this is easy as we already proved that $P\otimes v = 1\otimes P(T)(v)$ in $M$.
So we proved that $M\simeq V$ as $k[x]$-modules : what does this have to do with your problem ?
Well for any ring $R$ and any $R$-module $K$, $R\otimes_R K\simeq K$ as $R$-modules via $x\otimes m\mapsto xm$. So $V\simeq k[x]\otimes_{k[x]}V$ as $k[x]$-modules.