2 trivial typo fix

I feel duty-bound to add an answer since I had desired that this question be re-opened. There is nothing much I am able to contribute in addition to the well-known things.

The functorial definition of the polynomial algebra $R[x]$ over a ring $R$ is the following. It is a ring $R[x]$ together with a homomorphism $R \to R[x]$ such that given any ring homomorphism $\phi \colon R \to S$ and any fixed element $a \in S$, there is a unique homomorphism $\phi^\prime \colon R[x] \to S$ such that $\phi^\prime (x) =s$.

This is a proffered way of saying that the indeterminate "x" should be free to vary without any restriction whatsoever except that it is an element of a ring. A concrete manifestation is given by the free $R$-module on the set of symbols $x^k$ where $k \geq 0$ together with a certain multiplication operation. Since we do not a priori know what is "x", we make this precise with a machinery of sequences; and when we are done, we call a particular element of the resulting ring to be "x".

I contend that the ring $\mathbb Z$ and the polynomial rings over it are very important objects in the category of commutative rings with identity. For one, any commutative ring with identity $R$ admits a unique homomorphism $\mathbb Z \rightarrow R$. Moreover, if we let $r$ run through the elements of $R$, then there is an obvious surjective map from $\mathbb Z[(X_r)_{r\in R}]$ to $R$ and this realization of every comm. unital ring as a quotient of some polynomial ring over $\mathbb Z$ can be used to construct the co-product in this category.

I should mention that fixing "x" in the polynomial algebra is in a sense fixing some "co-ordinate". The Symmetric Algebra over a vector space is interesting in this sense.

1

I feel duty-bound to add an answer since I had desired that this question be re-opened. There is nothing much I am able to contribute in addition to the well-known things.

The functorial definition of the polynomial algebra $R[x]$ over a ring $R$ is the following. It is a ring $R[x]$ together with a homomorphism $R \to R[x]$ such that given any ring homomorphism $\phi \colon R \to S$ and any fixed element $a in S$, there is a unique homomorphism $\phi^\prime \colon R[x] \to S$ such that $\phi^\prime (x) =s$.

This is a proffered way of saying that the indeterminate "x" should be free to vary without any restriction whatsoever except that it is an element of a ring. A concrete manifestation is given by the free $R$-module on the set of symbols $x^k$ where $k \geq 0$ together with a certain multiplication operation. Since we do not a priori know what is "x", we make this precise with a machinery of sequences; and when we are done, we call a particular element of the resulting ring to be "x".

I contend that the ring $\mathbb Z$ and the polynomial rings over it are very important objects in the category of commutative rings with identity. For one, any commutative ring with identity $R$ admits a unique homomorphism $\mathbb Z \rightarrow R$. Moreover, if we let $r$ run through the elements of $R$, then there is an obvious surjective map from $\mathbb Z[(X_r)_{r\in R}]$ to $R$ and this realization of every comm. unital ring as a quotient of some polynomial ring over $\mathbb Z$ can be used to construct the co-product in this category.

I should mention that fixing "x" in the polynomial algebra is in a sense fixing some "co-ordinate". The Symmetric Algebra over a vector space is interesting in this sense.