# Definition of a universal example

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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See en.wikipedia.org/wiki/Universal_property . Universal properties and universal objects are not so easy a thing to describe in words; I think the path to understanding them lies in becoming familiar with enough examples until you've absorbed the underlying idea. Very roughly speaking, they are a strong and surprisingly useful generalization of the idea of a minimal element. Slightly less roughly speaking, at least in some contexts it's reasonable to think of a universal object as the laziest way to accomplish something. – Qiaochu Yuan Sep 8 '12 at 18:36
@QiaochuYuan "minimal element"? I prefer to think of it as "minimal description" or perhaps "definition that gives you the thing with minimal amount of effort". I think Wikipedia calls it "most efficient construction". But minimal element sounds as if we're talking about a poset. Are we? – Rudy the Reindeer Sep 8 '12 at 18:39
@Matt: in some sense, yes. Every universal object is an initial or terminal object in a suitable category. – Qiaochu Yuan Sep 8 '12 at 18:48
@modnar Here is a thread about the free group, one of the easiest examples of a universal property and here's a follow up, still about the free group. There is also this and this. Hope this helps. – Rudy the Reindeer Sep 8 '12 at 18:49

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:

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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Your second paragraph is a bit confusingly phrased: I'd define first what an $R$-algebra with $n$ distinguished elements is, then observe that the $R$-algebra $P = R[X_1,\dots,X_n]$ has $n$ distinguished elements $X_1,\dots,X_n$; and only then state the the universal property of the polynomial algebra. – t.b. Sep 8 '12 at 20:40
My understanding is that an $R$-algebra with $n$ distinguished elements is an $R$-algebra $A$ together with an $n$-tuple of elements $a_1,\dots,a_n \in A$. If $(A,a_1,\dots,a_n)$ and $(B,b_1,\dots,b_n)$ are two algebras, a morphism is an $R$-algebra homomorphism $\varphi\colon A \to B$ such that $\varphi(a_i) = b_i$ for $i = 1,\dots,n$. Your ring homomorphism $\phi\colon R \to R'$ turns $R'$ into an $R$-algebra and it extends uniquely to $\pi:R[X_1,\dots,X_n] \to R'$ such that $\pi X_i = r_i$ witnessing the universal property of $R[X_1,\dots,X_n]$ with distinguished elements $X_1,\dots,X_n$. – t.b. Sep 8 '12 at 20:55
Well, if they're not ordered then the universal example is the subalgebra of symmetric polynomials (and the inclusion corepresents the "forget about the ordering" functor). So, it depends what you care about. – Aaron Mazel-Gee Sep 9 '12 at 11:56
What Aaron said (in other words: yes, they really have to be ordered). The first sentence of what is now the fourth paragraph is still confusing the polynomial algebra $R[X_1,\dots,X_n]$ and its universal property. Try to write this more carefully. Maybe you should also state explicitly that a (commutative) ring $R'$ with a homomorphism $\phi: R \to R'$ is the same thing a commutative $R$-algebra. – t.b. Sep 9 '12 at 12:02
I'm not sure whether the confusion is of a mathematical nature, or not, but what I'm sure of is that your presentation is really clumsy: You write: "An $R$-algebra $A$ with $n$ distinguished elements is a pair $(a_1,\dots,a_n,A)$ for some $a_i \in A$". First of all, I see an $(n+1)$-tuple... Then two paragraphs later, an $R$-algebra with distinguished elements is a pair $(P,\varphi)$ having a certain universal property? Then: do you really write that the $X_i$ are supposed to be the distinguished elements of $P$? – t.b. Sep 9 '12 at 15:34