[proofplan] We prove both existence and uniqueness. For uniqueness, we observe that any $R$-algebra homomorphism is determined by its values on the generators $T_1, \ldots, T_n$ (since every polynomial is built from these by ring operations that any homomorphism must preserve). For existence, we define $\varphi$ by evaluation — sending each polynomial $f(T_1, \ldots, T_n)$ to $f(a_1, \ldots, a_n)$ — and verify it is a well-defined $R$-algebra homomorphism. [/proofplan]

[step:Prove uniqueness: an $R$-algebra homomorphism $R[T_1, \ldots, T_n]\to A$ is determined by the images of $T_1, \ldots, T_n$] Let $\varphi, \psi: R[T_1, \ldots, T_n] \to A$ be two $R$-algebra homomorphisms satisfying $\varphi(T_i) = \psi(T_i) = a_i$ for all $1 \leq i \leq n$. Every element of $R[T_1, \ldots, T_n]$ is a finite sum of monomials of the form $r \cdot T_1^{\alpha_1} \cdots T_n^{\alpha_n}$ with $r \in R$ and $\alpha_1, \ldots, \alpha_n \geq 0$. Since both $\varphi$ and $\psi$ are $R$-algebra homomorphisms, they preserve addition, multiplication, and scalar multiplication by elements of $R$. In particular, for any such monomial: \begin{align*} \varphi(r \cdot T_1^{\alpha_1} \cdots T_n^{\alpha_n}) &= r \cdot \varphi(T_1)^{\alpha_1} \cdots \varphi(T_n)^{\alpha_n} = r \cdot a_1^{\alpha_1} \cdots a_n^{\alpha_n}, \end{align*} and the same holds for $\psi$. Since every polynomial is a finite sum of such monomials and both maps are additive, $\varphi = \psi$ on all of $R[T_1, \ldots, T_n]$.[/step]

[guided]Why is an $R$-algebra homomorphism determined by the images of the generators? The key point is that $R[T_1, \ldots, T_n]$ is generated as an $R$-algebra by $T_1, \ldots, T_n$. Every element $f \in R[T_1, \ldots, T_n]$ is a finite $R$-linear combination of monomials $T_1^{\alpha_1} \cdots T_n^{\alpha_n}$: \begin{align*} f = \sum_{\alpha} r_\alpha \, T_1^{\alpha_1} \cdots T_n^{\alpha_n}, \end{align*} where the sum ranges over finitely many multi-indices $\alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbb{N}_0^n$ and $r_\alpha \in R$. Now let $\varphi, \psi: R[T_1, \ldots, T_n] \to A$ be two $R$-algebra homomorphisms with $\varphi(T_i) = \psi(T_i) = a_i$ for all $i$. An $R$-algebra homomorphism must satisfy three properties: it preserves addition, it preserves multiplication, and it fixes every element of $R$ (i.e., $\varphi(r \cdot 1) = r \cdot 1_A$ for all $r \in R$). These three properties together force: \begin{align*} \varphi(f) &= \varphi\Bigl(\sum_\alpha r_\alpha \, T_1^{\alpha_1} \cdots T_n^{\alpha_n}\Bigr) = \sum_\alpha r_\alpha \cdot \varphi(T_1)^{\alpha_1} \cdots \varphi(T_n)^{\alpha_n} = \sum_\alpha r_\alpha \, a_1^{\alpha_1} \cdots a_n^{\alpha_n}. \end{align*} The same computation gives $\psi(f) = \sum_\alpha r_\alpha \, a_1^{\alpha_1} \cdots a_n^{\alpha_n}$. Since this holds for every $f \in R[T_1, \ldots, T_n]$, we conclude $\varphi = \psi$. The essential insight is that the $R$-algebra structure leaves no freedom once the images of the generators are fixed.[/guided]

[step:Construct the evaluation homomorphism $\varphi: R[T_1, \ldots, T_n]\to A$ by $f \mapsto f(a_1, \ldots, a_n)$] Define the map \begin{align*} \varphi: R[T_1, \ldots, T_n] &\to A \\ \sum_\alpha r_\alpha \, T_1^{\alpha_1} \cdots T_n^{\alpha_n} &\mapsto \sum_\alpha r_\alpha \, a_1^{\alpha_1} \cdots a_n^{\alpha_n}, \end{align*} where each sum ranges over finitely many multi-indices $\alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbb{N}_0^n$ with $r_\alpha \in R$. This map is well-defined because the representation of a polynomial as a sum of monomials is unique (two polynomials are equal if and only if all their coefficients agree). We verify that $\varphi$ is an $R$-algebra homomorphism: 1. **Additivity.** For $f = \sum_\alpha r_\alpha T^\alpha$ and $g = \sum_\alpha s_\alpha T^\alpha$ (padding with zero coefficients where needed), $\varphi(f + g) = \sum_\alpha (r_\alpha + s_\alpha) a^\alpha = \sum_\alpha r_\alpha a^\alpha + \sum_\alpha s_\alpha a^\alpha = \varphi(f) + \varphi(g)$. 2. **Multiplicativity.** For the product $fg = \sum_\gamma \bigl(\sum_{\alpha + \beta = \gamma} r_\alpha s_\beta\bigr) T^\gamma$, we have $\varphi(fg) = \sum_\gamma \bigl(\sum_{\alpha + \beta = \gamma} r_\alpha s_\beta\bigr) a^\gamma$. Since $a^\gamma = a^\alpha \cdot a^\beta$ when $\gamma = \alpha + \beta$, this equals $\bigl(\sum_\alpha r_\alpha a^\alpha\bigr)\bigl(\sum_\beta s_\beta a^\beta\bigr) = \varphi(f)\varphi(g)$. 3. **$R$-algebra map.** For $r \in R$, the constant polynomial $r \cdot 1$ maps to $r \cdot 1_A$, since $\varphi(r) = r \cdot a_1^0 \cdots a_n^0 = r \cdot 1_A$. Hence $\varphi$ restricts to the structure map $R \to A$ on constant polynomials. Finally, $\varphi(T_i) = 1 \cdot a_1^0 \cdots a_i^1 \cdots a_n^0 = a_i$ for each $1 \leq i \leq n$, as required.[/step]

[guided]The existence part constructs $\varphi$ as the "evaluate at $(a_1, \ldots, a_n)$" map. The idea is natural: if we want $T_i \mapsto a_i$, then any polynomial $f(T_1, \ldots, T_n)$ must map to $f(a_1, \ldots, a_n)$ by the ring homomorphism properties. So we simply define $\varphi$ to be this evaluation. Define \begin{align*} \varphi: R[T_1, \ldots, T_n] &\to A \\ \sum_\alpha r_\alpha \, T_1^{\alpha_1} \cdots T_n^{\alpha_n} &\mapsto \sum_\alpha r_\alpha \, a_1^{\alpha_1} \cdots a_n^{\alpha_n}, \end{align*} where the sums are over finitely many multi-indices $\alpha \in \mathbb{N}_0^n$ with $r_\alpha \in R$. Well-definedness holds because the coefficient representation of a polynomial is unique. We must check three properties to confirm $\varphi$ is an $R$-algebra homomorphism. **Additivity.** Write $f = \sum_\alpha r_\alpha T^\alpha$ and $g = \sum_\alpha s_\alpha T^\alpha$ (using the shorthand $T^\alpha = T_1^{\alpha_1} \cdots T_n^{\alpha_n}$ and $a^\alpha = a_1^{\alpha_1} \cdots a_n^{\alpha_n}$, and padding with zero coefficients as needed). Then $f + g = \sum_\alpha (r_\alpha + s_\alpha) T^\alpha$, so \begin{align*} \varphi(f + g) = \sum_\alpha (r_\alpha + s_\alpha) a^\alpha = \sum_\alpha r_\alpha a^\alpha + \sum_\alpha s_\alpha a^\alpha = \varphi(f) + \varphi(g). \end{align*} **Multiplicativity.** The product $fg$ has the Cauchy convolution form $fg = \sum_\gamma c_\gamma T^\gamma$ where $c_\gamma = \sum_{\alpha + \beta = \gamma} r_\alpha s_\beta$. We compute: \begin{align*} \varphi(fg) = \sum_\gamma c_\gamma \, a^\gamma = \sum_\gamma \Bigl(\sum_{\alpha + \beta = \gamma} r_\alpha s_\beta\Bigr) a^\gamma. \end{align*} Since $\gamma = \alpha + \beta$ componentwise, we have $a^\gamma = a^{\alpha + \beta} = a^\alpha \cdot a^\beta$ (this uses that the $a_i$ live in the ring $A$, so exponent addition corresponds to multiplication). Hence \begin{align*} \varphi(fg) = \sum_{\alpha, \beta} r_\alpha s_\beta \, a^\alpha a^\beta = \Bigl(\sum_\alpha r_\alpha a^\alpha\Bigr)\Bigl(\sum_\beta s_\beta a^\beta\Bigr) = \varphi(f)\varphi(g). \end{align*} **$R$-algebra map.** For any $r \in R$, the constant polynomial $r$ has the single nonzero coefficient $r_0 = r$ at the zero multi-index. So $\varphi(r) = r \cdot a_1^0 \cdots a_n^0 = r \cdot 1_A$. This shows $\varphi$ restricts to the structure map $R \to A$ on constant polynomials, confirming it is an $R$-algebra homomorphism. Finally, we check the prescribed values: $\varphi(T_i) = 1 \cdot a_i^1 \cdot \prod_{j \neq i} a_j^0 = a_i$ for each $1 \leq i \leq n$. Combining existence with the uniqueness proved in the previous step completes the proof.[/guided]