Is this a correct solution to show that the ring $\mathbb{R}[x]$ does not embed in $\mathbb{C}$ I know this is not a typical question and I will delete it later. 
Here is a question together my answer. Please let me know if my answer is correct. The reason that I am posting this is that I believe my answer is correct but my professor refuses to accept that as a correct answer (with no explanation). 
Question: Is it possible to embed the ring $\mathbb{R}[x]$ in $\mathbb{C}$ ?
My Answer: No. Every element of $\mathbb{C}$ is algebraic over $\mathbb{R}$ so there is no copy of $\mathbb{R}[x]$ in $\mathbb{C}$ because $x$ is not algebraic over $\mathbb{R}$.
 A: Just to fill in the details of Tobias Kildetoft's comment. Not sure if they'll be useful to you, but I thought it was interesting.
Let's construct an embedding of $\Bbb{R}[x]$ into $\Bbb{C}$.
Step 1 Observe that the transcendence degree of $\Bbb{C}$ over $\Bbb{Q}$ is $\mathfrak{c}$ (though note that for our argument, it doesn't matter exactly what it is as long as it's infinite). 
This follows from simple cardinality arguments. Any field algebraic over a field with an infinite cardinality has the same cardinality. If a field has cardinality $\alpha$ and we adjoin a cardinality $\beta$'s worth of indeterminates, the result will have cardinality $\max\{\alpha,\beta\}$.
Step 2 Let $T$ be a transcendence basis for $\Bbb{C}$ over $\Bbb{Q}$. Since $\Bbb{C}$ is algebraically closed and is algebraic over the subfield to $\Bbb{Q}(T)$, it is isomorphic to the algebraic closure of $\Bbb{Q}(T)$. Choose a proper subset $S$ of $T$ of the same cardinality as $T$. Then the algebraic closure of $\Bbb{Q}(S)$ in $\Bbb{C}$ is clearly isomorphic to $\Bbb{C}$ as rings, since $\Bbb{Q}(S)\cong \Bbb{Q}(T)$, but it is necessarily a proper subset of $\Bbb{C}$, and doesn't contain any of the elements in $T\setminus S$.
Step 3 By step 2, we have a (necessarily injective) ring homomorphism $\phi :\Bbb{C}\to\Bbb{C}$ which isn't surjective. Let $t \in T\setminus S$. The ring homomorphism $\phi$ extends to an injective ring homomorphism
$\psi : \Bbb{C}[x]\to \Bbb{C}$ sending $x$ to $t$. Restricting to the obvious subset $\Bbb{R}[x]\subset \Bbb{C}[x]$ we have found an embedding of $\Bbb{R}[x]$ into $\Bbb{C}$.
