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I have a question in Artin's Algebra, about adjoining elements on page 339.

Proposition 11.5.5 says: Let $R$ be a ring, and let $f(x)$ be a monic polynomial of positive degree $n$ with coefficients in $R$. Let $R[\alpha]$ denote the ring $R[x]/(f)$ obtained by adjoining an element satisfying the relation $f(\alpha)=0$.

So the set ($1,\alpha,\dots,\alpha^{n-1}$) is a basis of $R[\alpha]$ over $R$: every elements of $R[\alpha]$ can be written uniquely as a linear combination of this basis, with coefficients in $R$.

My question is: Why the coefficients are in $R$ rather than $R[x]/(f)$? Since we let $\alpha$ denote the residue $\bar x$ of x in $R[x]/(f)$. So the coefficients of polynomial in $R[x]$ will be some $\bar a$ of $R[x]/(f)$.

And on the same page below (11.5.2), there is a sentence: So, dropping bars, $\alpha$ satisfies the relation $f(\alpha)=0$. Why can we drop the bars?

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Consider your ring $A = R[x]/(f)$, inside of which sits the subring $B = \{ r + (f) : r \in R \}$. Consider the new set $$ C = (A \setminus B) \uplus R $$ (that's a disjoint union). Turn $C$ into a ring, by defining operations as in $A$, except that when you have to add or multiply an element $d \in A \setminus B$ with an element $r \in R$, define $$ d + r = d + (r + (f)), \qquad d \cdot r = d \cdot (r + (f)), $$ where the right-hand operations are in $A$.

Clearly $C$ is a ring isomorphic to $A$ (which means they are practically the same, from the point of view of abstract algebra), and it contains $R$ as a subring. So disregard $A$, and continue working with $C$. This is one way to see what is meant here by dropping the bars.

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  • $\begingroup$ You write "$C$ is a ring isomorphic to $A$ (which means they are practically the same, from the point of view of abstract algebra)", and I was wondering if you could explain from what point of view they're not the same? $\endgroup$
    – Carolus
    May 2 '14 at 9:46
  • $\begingroup$ @Carolus, as sets? $\endgroup$ May 2 '14 at 9:47
  • $\begingroup$ Of course. I was just curious if you were thinking of some other specific branch(es) of math in contrast to abstract algebra. Just a stupid question :) $\endgroup$
    – Carolus
    May 2 '14 at 14:22

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