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I'm not sure how the term is being used here:

Let $R$ be a commutative ring and $X_1,\ldots, X_n$ indeterminates over $R$. Set $P = R[X_1, \ldots, X_n]$.

Given a ring homomorphism $\phi: R \rightarrow R'$ and $x_1, \ldots, x_n \in R'$, there is a unique ring homomorphism $\pi: P \rightarrow R'$ with $\pi\restriction_R = \phi$ and $\pi(X_i) = x_i$ for all $i=1,\ldots,n$. Another way to state this is that $P$ is a universal example of an $R$-algebra with $n$ distinguished elements.

How is it used in general? Also, this example was used as an example of a "universal mapping property" and could you help clarify what this means?

Thank you!

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    @modnar [Here](http://math.stackexchange.com/questions/130850/free-group-and-universal-property) is a thread about the free group, one of the easiest examples of a universal property and [here](http://math.stackexchange.com/questions/132438/universal-properties-again)'s a follow up, still about the free group. There is also [this](http://math.stackexchange.com/questions/130950/free-groups-unique-up-to-unique-isomorphism) and [this](http://math.stackexchange.com/questions/132540/universal-properties-and-diagrams). Hope this helps.2012-09-08

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I think "universal example" here is an abuse of language to mean "example of a universal property".

As pointed out by t.b. in the comments:

By definition, an $R$-algebra for a commutative ring $R$ is a pair $(S, f)$ where $f: R \to S$ is a ring homomorphism and $S$ is a commutative ring.

An $R$-algebra $A$ with $n$ distinguished elements is hence a triplet $((a_1, \dots, a_n), A, f)$ for some $a_i \in A$ where $f: R \to A$ is a ring homomorphism. An $R$-algebra homomorphism between two $R$-algebras with $n$ distinguished elements $((a_1, \dots, a_n), A, f)$ , $((b_1, \dots, b_n), B, f^\prime)$ is an $R$-algebra homomorphism $\varphi : A \to B$ such that $\varphi (a_i) = b_i$.

Then the universal property that's stated is:

An $R$-algebra with $n$ distinguished elements is a triplet $((X_1, \dots, X_n), P, \varphi)$ satisfying the following universal property: for every $R$-algebra $((r_1^\prime, \dots, r_n^\prime), R^\prime, \phi)$ with $n$ distinguished elements $r_1^\prime, \dots, r_n^\prime \in R^\prime$ there exists a unique $R$-algebra homomorphism $\pi: P \to R^\prime$ such that $\pi\restriction_R = \pi \circ \varphi = \phi$ and $\pi (X_i) = r_i^\prime$ for the $n$ distinguished elements $r_i^\prime \in R^\prime$ and $n$ distinguished elements in $P$. Can't resist to add the corresponding diagram:

enter image description here

In general, many universal mapping properties read as follows:

The thingamajig is an object $B$ and a morphism $m: A \to B$ such that for every object $C$ and morphism $n: A \to C$ there is a unique morphism $\varphi : B \to C$ such that the diagram of morphisms commutes, that is, such that $\varphi \circ m = n$.

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    Have fun! I mean *the* [polynomial ring](http://en.wikipedia.org/wiki/Polynomial_ring) $R[X_1,\dots,X_n] = P$ in $n$ indeterminates over $R$. As a ring extension of $R$ this $P$ has a natural $R$-algebra structure (inclusion of constants) and it has the distinguished elements $X_1,\dots,X_n$. We set up the category of $R$-algebras with $n$ distinguished elements and inside this category $R[X_1,\dots,X_n]$ has the universal property. Yes, I mean what you'd write $((a_1,\dots,a_n),A,\phi)$ by *arbitrary $R$-algebra* [after we picked(=distinguished) our $a_i$'s]; we extend $\phi$ from $R$ to $P$.2012-09-10