Is $\mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}[x]$ isomorphic to $\mathbb{Q}[x]$ as a $\mathbb{Z}$-module? as stated in the title, I was wondering whether $ \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}[x] \cong \mathbb{Q}[x]$? 
I think they are and attempt to give a "proof" of this, can you check whether my proof is correct or not and if not, can you point out where I make the mistake?  Thank you. 
To start, we have two diagrams (we'll show they are commutative later) below, 

$\iota$ is the canonical inclusion, $\tau$ is the middle linear map (middle linear can be checked readily) that sends $(\frac{r}{s}, p(x)) \mapsto \frac{r}{s} p(x)$ .  Hence by the Universal Property of tensor product, $\widetilde{\tau}$ exists which makes the left diagram commutes. 
Focusing on the right diagram now, since $\mathbb{Q}[x]$ is a free $\mathbb{Q}$-module with basis $\{1, x, x^2,...\}$, we can define the $\mathbb{Z}$-module homo $\phi: \mathbb{Q}[x] \to \mathbb{Q} \otimes_{\mathbb{Z}} \mathbb{Z}[x]$ by specifying $\phi(\frac{r}{s} x^n ) = \frac{r}{s} \otimes x^n $ where $\frac{r}{s} \in \mathbb{Q}, n \in \mathbb{N}.$ 
To show the right diagram is commutative, it suffices to show that for $\frac{r}{s},\frac{a}{b} \in \mathbb{Q}, n \in \mathbb{N},~~ \iota(\frac{r}{s},\frac{a}{b}x^n) = \phi \tau(\frac{r}{s},\frac{a}{b}x^n) $.  
Now, $\iota(\frac{r}{s},\frac{a}{b}x^n) = \frac{r}{s} \otimes \frac{a}{b}x^n = \frac{ra}{s} \otimes \frac{1}{b}x^n = \frac{rab}{sb} \otimes \frac{1}{b}x^n = \frac{ra}{sb} \otimes \frac{b}{b}x^n = \frac{ra}{sb} \otimes x^n = \phi(\frac{ra}{sb}x^n) = \phi \tau(\frac{r}{s}, \frac{a}{b}x^n)$.  Thus the right diagram is commutative as well.  
Putting the two diagrams together and use uniqueness of Universal Mapping property of tensor product yields $\phi \widetilde{\tau} = Id$.  Similarly, $\widetilde{\tau} \phi = Id $.  Hence, the two are isomorphic.  
Thank you so much.  
Edit: To elaborate on Vincent's point near the end, when we said, "Putting the two diagrams together and use uniqueness of Universal Mapping property of tensor product yields $\phi \widetilde{\tau} = Id$", we meant that we would have a commutative diagram, 

so we have $(\phi \widetilde{\tau} ) \iota = \iota$ but $ Id \circ \iota = \iota$, so by uniqueness of Universal Mapping property of tensor product, $\phi \widetilde{\tau} = Id $ .  
As for $ \widetilde{\tau} \phi = Id$, this can be checked readily on basis of $\mathbb{Q}[x]$.  
 A: I think your proof is correct but if it is not the mistake is at the very end, in the line:
'Putting the two diagrams together and use uniqueness of Universal Mapping property of tensor product yields $\phi \tilde{\tau}=Id$'. 
The proof that the maps exist and that the diagrams are commutative are perfectly fine and easy to follow. The fact that $\phi \tilde{\tau}=Id$ can also be verified directly from the definitions you gave so the conclusion is correct as well. 
However I am not entirely sure how this conclusion would follow from the uniqueness of the UM property. This might however be a result of my own lack of familiarity with category theory.
Still: this uniqueness you are talking about is the thing that gives us $\tilde{\tau}$, right? And so it follows from the diagrams being commutative and the uniqueness of the UM property that if $\phi$ has a (right) inverse $\phi^{-1}$ then this inverse must equal $\tilde{\tau}$. But how do we know that $\phi$ has a right inverse? Perhaps you could clarify this step a bit further.  
EDIT: with the added diagram I believe the proof is watertight
