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Hungerford's theorem III.5.5 says:

Let $R$ and $S$ be commutative rings with identity and $\phi : R \rightarrow S$ a homomorphism of rings such that $\phi(1_R) = 1_S$. If $s_1, ..., s_n \in S$, then there is a unique homomorphism of rings $\bar{\phi} : R[x_1, ..., x_n \rightarrow S$ such that $\bar{\phi} | R = \phi$ and $\bar{\phi}(x_i) = s_i$ for $i = 1, 2, ..., n$. This property completely determines the polynomial ring $R[x_1, ..., x_n]$ up to isomorphism.

In the proof, he defines $\bar{\phi}$ by $\bar{\phi}(f) = \phi f(s_1, ..., s_n)$ where $\phi f(s_1, ..., s_n)$ is defined as follows: We have $f = \displaystyle\sum_{i = 0}^{m} a_{i} x_{1}^{k_{i1}}...x_{n}^{k_{in}}$. Then we set $\phi f(s_1, ..., s_n) = \displaystyle\sum_{i = 0}^{m} \phi(a_{i}) s_{1}^{k_{i1}}...s_{n}^{k_{in}}$. Then we are supposed to verify that $\bar{\phi}$ is a morphism. I can prove that $\bar{\phi}(f + g) = \bar{\phi}(f) + \bar{\phi}(g)$. But the product is tougher. Apparently I need to use the binomial theorem, but I don't see how. Any hints?

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No need for the binomial theorem - just a Cauchy product [en.wikipedia.org/wiki/Cauchy_product] will do. –  Prahlad Vaidyanathan Sep 2 '13 at 5:30
    
I set $f = \sum_{i = 0}^{m} a_{i} x_{1}^{k_{i1}}...x_{n}^{k_{in}}$ and $g = \sum_{i = 0}^{m} b_{i} x_{1}^{k_{i1}}...x_{n}^{k_{in}}$ (we may WLOG choose them to be that way). I'm not sure how to write the product $fg$ in a way that each monomial appears only once, though. –  Pedro Sep 2 '13 at 5:55
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Write $f = \sum_{i=1}^s f_i$ and $g = \sum_{j=1}^t g_j$ where $$ f_i = a_ix_1^{k_{i1}}x_2^{k_{i2}}\ldots x_n^{k_{in}} $$ $$ g_j = b_jx_1^{l_{j1}}x_2^{l_{j2}}\ldots x_n^{l_{jn}} $$ Then by what you have already proved $$ \overline{\phi}(fg) = \sum_{i,j}\overline{\phi}(f_i g_j) $$ Hence, it suffices to prove that $\overline{\phi}(f_i g_j) = \overline{\phi}(f_i)\overline{\phi}(g_j)$. Now note that $$ f_ig_j = a_ib_jx_1^{k_{i1}+l_{j1}}x_2^{k_{i2}+l_{j2}}\ldots x_n^{k_{in}+l_{jn}} $$ $$ \Rightarrow \overline{\phi}(f_jg_j) = \phi(a_ib_j)s_1^{k_{i1}+l_{j1}}s_2^{k_{i2}+l_{j2}}\ldots s_n^{k_{in}+l_{jn}} $$ And $$ \overline{\phi}(f_i)\overline{\phi}(g_j) = \phi(a_i)s_1^{k_{i1}}s_2^{k_{i2}}\ldots s_n^{k_{in}}\phi(b_j)s_1^{l_{j1}}s_2^{l_{j2}}\ldots s_n^{l_{jn}} $$ Now compare these two expressions and use the fact that $\phi$ is a ring homomorphism.

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