1. ## Rings of the form R[X] - Ring Adjunction

I am reading R.Y Sharp's book: "Steps in Commutative Algebra".

On page 6 in 1.11 Lemma, we have the following: [see attachment]

"Let S be a subring of the ring R, and let $\Gamma$ be a subset of R.

Then $S[ \Gamma ]$ is defined as the intersection of all subrings of R which contain S and $\Gamma$.

Thus, $S[ \Gamma ]$ is a subring of R which contains both S and $\Gamma$, and it is the smallest such subring of R in the sense that it is contained in every other subring of R that contains S and $\Gamma$.

In the special case in which $\Gamma$ is a finite set $\{ \alpha_1, \alpha_2, ... ... , \alpha_n \}$ we write $S[ \Gamma ]$ as $S [ \alpha_1, \alpha_2, ... ... , \alpha_n ]$.

In the special case in which S is commutative, and $\alpha \in R$ is such that $\alpha s = s \alpha$ for all $s \in S$ we have

$S[ \alpha ] = \{ \ {\sum}_{i = 0}^{t} s_i \alpha^i : t \in {\mathbb{N}}_0 \ s_0, s_1, ... ... , s_t \in S \}$ .............................................. (1)

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Then on page 7 Sharp writes:

Note that when R is a commutative ring and X is an indeterminate, then it follows from 1.11 Lemma that our earlier use of R[X] to denote the polynomial ring is consistent with this new use of R[X] to denote 'ring adjunction'.

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Now in the polynomial ring R[X] we take a subset of ring elements $a_1, a_2, ... ... , a_n \in R$ and use an indeterminate x (whatever that is?) to form sums like the following:

$a_n x^n + a_{n-1} + ... ... + a_1x + a_0$ .................................................. ...... (2)

My problems are as follows:

(a) It looks like (1) and (2) have the same structure BUT $\alpha$ is a member of the ring R, and also the subring S whereas x is not a member of R but is an "indeterminate" [maybe I am overthinking this and it does not matter??] Can someone please clarify this matter?

(b) Again, (1) and (2) seem to have the same structure BUT $a_1, a_2, ... ... , a_n \in R$ is just a subset of R - whereas $s_0, s_1, ... ... , s_t$ are elements of a subring. Does this matter? Can someone please clarify?

(c) Sharp specifies that S has to be commutative - but why? I cannot see how this is needed in his Proof on the bottom of page 6. Can someone help.

I would be grateful if someone can clarify the above.

Peter