[proofplan]
The polynomial ring $\mathbb{R}[T_1, \ldots, T_n]$ is a free $\mathbb{R}$-module on the monomial basis $\{T_1^{\alpha_1} \cdots T_n^{\alpha_n} : \alpha \in \mathbb{Z}_{\geq 0}^n\}$. Since tensoring a free module with a ring produces the free module over the new ring on the same basis, the map $\varphi$ is already a $\mathbb{C}$-module isomorphism by the general theory of extension of scalars for free modules. It remains to verify that $\varphi$ is a ring homomorphism (multiplicative and unit-preserving), which then upgrades it to a $\mathbb{C}$-algebra isomorphism.
[/proofplan]
[step:Identify $\mathbb{R}[T_1, \ldots, T_n]$ as a free $\mathbb{R}$-module and invoke the extension of scalars isomorphism]
The polynomial ring $\mathbb{R}[T_1, \ldots, T_n]$ is a free $\mathbb{R}$-module with basis the set of monomials
\begin{align*}
\mathcal{B} := \{ T_1^{\alpha_1} \cdots T_n^{\alpha_n} : (\alpha_1, \ldots, \alpha_n) \in \mathbb{Z}_{\geq 0}^n \}.
\end{align*}
By the general extension of scalars isomorphism for free modules, if $V$ is a free $\mathbb{R}$-module with basis $\mathcal{B}$, then the $\mathbb{C}$-module $\mathbb{C} \otimes_\mathbb{R} V$ is a free $\mathbb{C}$-module with basis $\{1 \otimes b : b \in \mathcal{B}\}$, and the map
\begin{align*}
\varphi: \mathbb{C} \otimes_\mathbb{R} V &\to \mathbb{C}[\mathcal{B}] \\
z \otimes b &\mapsto zb
\end{align*}
is a $\mathbb{C}$-module isomorphism. Applied to $V = \mathbb{R}[T_1, \ldots, T_n]$, this gives a $\mathbb{C}$-linear isomorphism $\varphi: \mathbb{C} \otimes_\mathbb{R} \mathbb{R}[T_1, \ldots, T_n] \xrightarrow{\sim} \mathbb{C}[T_1, \ldots, T_n]$ defined on simple tensors by $\varphi(z \otimes p) = zp$.
[guided]
Why is $\mathbb{R}[T_1, \ldots, T_n]$ free as an $\mathbb{R}$-module? Every polynomial $p \in \mathbb{R}[T_1, \ldots, T_n]$ is a unique finite $\mathbb{R}$-linear combination of monomials $T_1^{\alpha_1} \cdots T_n^{\alpha_n}$. The monomials are $\mathbb{R}$-linearly independent (distinct monomials are distinct formal expressions) and they span the polynomial ring. So $\mathcal{B}$ is a free $\mathbb{R}$-basis.
The general extension of scalars result states: if $V$ is a free $R$-module with basis $\mathcal{B}$, then $S \otimes_R V$ is a free $S$-module with basis $\{1 \otimes b : b \in \mathcal{B}\}$. Here $R = \mathbb{R}$, $S = \mathbb{C}$, and $V = \mathbb{R}[T_1, \ldots, T_n]$. Thus $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}[T_1, \ldots, T_n]$ is a free $\mathbb{C}$-module on $\{1 \otimes T^\alpha : \alpha \in \mathbb{Z}_{\geq 0}^n\}$. Since $\mathbb{C}[T_1, \ldots, T_n]$ is also a free $\mathbb{C}$-module on the same monomial set $\{T^\alpha\}$, the map $1 \otimes T^\alpha \mapsto T^\alpha$ extends to a $\mathbb{C}$-module isomorphism. On simple tensors, this reads $\varphi(z \otimes p) = zp$.
At this stage, $\varphi$ is a $\mathbb{C}$-module isomorphism. To promote it to a $\mathbb{C}$-algebra isomorphism, we must verify multiplicativity and the unit property.
[/guided]
[/step]
[step:Verify that $\varphi$ preserves multiplication]
The ring structure on $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}[T_1, \ldots, T_n]$ is defined on simple tensors by
\begin{align*}
(z_1 \otimes p_1) \cdot (z_2 \otimes p_2) := z_1 z_2 \otimes p_1 p_2,
\end{align*}
where the product $z_1 z_2$ is taken in $\mathbb{C}$ and $p_1 p_2$ in $\mathbb{R}[T_1, \ldots, T_n]$. This is the standard multiplication on a tensor product of algebras.
For any simple tensors $z_1 \otimes p_1$ and $z_2 \otimes p_2$:
\begin{align*}
\varphi\bigl((z_1 \otimes p_1)(z_2 \otimes p_2)\bigr) &= \varphi(z_1 z_2 \otimes p_1 p_2) \\
&= z_1 z_2 \cdot p_1 p_2 \\
&= (z_1 p_1)(z_2 p_2) \\
&= \varphi(z_1 \otimes p_1) \cdot \varphi(z_2 \otimes p_2).
\end{align*}
The third equality uses commutativity: elements of $\mathbb{C}$ commute with polynomials in $\mathbb{C}[T_1, \ldots, T_n]$, so $z_1 z_2 \cdot p_1 p_2 = (z_1 p_1)(z_2 p_2)$.
Since $\varphi$ is multiplicative on simple tensors and every element of $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}[T_1, \ldots, T_n]$ is a finite sum of simple tensors, $\mathbb{C}$-linearity of $\varphi$ and distributivity of multiplication imply that $\varphi$ is multiplicative on all elements.
[/step]
[step:Verify the unit property and conclude]
The multiplicative identity in $\mathbb{C} \otimes_\mathbb{R} \mathbb{R}[T_1, \ldots, T_n]$ is $1_\mathbb{C} \otimes 1_{\mathbb{R}[T_1, \ldots, T_n]}$, and
\begin{align*}
\varphi(1_\mathbb{C} \otimes 1) = 1_\mathbb{C} \cdot 1 = 1 \in \mathbb{C}[T_1, \ldots, T_n],
\end{align*}
which is the multiplicative identity of $\mathbb{C}[T_1, \ldots, T_n]$.
Combining the three properties:
- $\varphi$ is a $\mathbb{C}$-module isomorphism (from the extension of scalars for free modules),
- $\varphi$ is multiplicative,
- $\varphi$ preserves the unit,
we conclude that $\varphi$ is a $\mathbb{C}$-algebra isomorphism.
[/step]
