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The following is a proposition in Artin's Algebra:

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

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

I don't understand the polynomial $f(x)$ in the theorem here. Consider $R={\Bbb Z}$. Let $f(x)=x^2+1$ and $g(x)=(x^2+1)^2$. Then we have $$ f(i)=g(i)=0. $$ Then by the theorem above, ${\Bbb Z}[i]$ is not only ${\Bbb Z}[x]/(f)$ with basis $(1,i)$, but also ${\Bbb Z}[x]/(g)$, with basis $(1,i,i^2,i^3)$. What is wrong here?

When $R$ is a field $F$, and if $\alpha\in E$ is algebraic over $F$ where $E$ is an extension field of $F$, then we do have $$ F[\alpha]\cong F[x]/(f) $$ where $f$ is the minimal polynomial of $\alpha$. But in Proposition 11.5.5, there seems to be no requirement for the degree of $f$. (I thought this might be implicitly in the phrase "the relation $f(\alpha)=0$". But I'm not sure.)

[Added:] Does one have any other references about this proposition? (It seems that some assumption in the proposition is missing according to the comments.)

  • Isn't in the proposition $f$ irreductible? –  Jan 17 '14 at 22:11
  • f should be irreducible, otherwise it is not true that R[alpha] = R[x]/f. – Billy Jan 17 '14 at 22:15
  • @SamiBenRomdhane I copied the proposition exactly the way the author put it. I don't see where the irreducibility is stated explicitly. –  Jan 17 '14 at 22:15
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    Then it was simply forgotten to be written. – anon Jan 17 '14 at 22:15
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    No, nothing is forgotten. Artin means to ajoin an element satisfying the relation and no other (no other which does not follow from the first, that is) In your example, the element i satisfies the relation f(i)=0, which does not follow from g(i)=0, so the ring Z[i] (the subring of C) is not the result of «adjoining an element to Z subject to g(i)=0». – Mariano Suárez-Álvarez Jan 17 '14 at 22:19
  • Of course, they key point here is what precisely the definition of adjoining an element satisfying a relation is. – Mariano Suárez-Álvarez Jan 17 '14 at 22:21
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    I suppose that's a better interpretation. Still seems the wording leaves a bit to be desired though; it simply says "an element satisfying the relation" without further specificity (and certainly $i$ is "an element satisfying the relation $(i^2+1)^2=0$" for example, which ultimately means "the ring $R[\alpha]$" wouldn't be well-defined). – anon Jan 17 '14 at 22:43
  • Well, the point is that it is well defined, just as a group is well defined when given by generators and relations. The group $\langle x:x^4\rangle$ has order four. This is an essentially universally used convention; I can't possibly imagine what more to be desired there is :-) – Mariano Suárez-Álvarez Jan 17 '14 at 23:34
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    @Mariano One could desire an explicit mention of some kind of freeness ("satisfying the relation and no other" in your words), so that the construction is in fact well-defined when the author is taken at face value only for what was explicitly said and not for conventions that more knowledgeable readers will automatically project into the situation in order to continue. – anon Jan 17 '14 at 23:49
  • You are free to avoid the standard convention, as you know. In any case, in this particular situation, in which Artin is in fact introducing the ring which is, by his definition, the result of adjoining an element satisfying a relation, (and therefore it is not a more knowledgeable reader that is needed but an attentive one) your complaint seems quite incomprehensible to me. As this wouldn't be the first one, I don't think we should make much of it. – Mariano Suárez-Álvarez Jan 18 '14 at 00:00
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    (I don't ever define things or introduce notation in the statement of propositions and theorems, but that is a stylistic choice and Artin, as all of us, has ample right to make his own) – Mariano Suárez-Álvarez Jan 18 '14 at 00:05

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Let $I = (f)$. Then "let $R[\alpha]$ denote the ring $R[x]/I$" means that $\alpha := x+I\,$ and $\,R[\alpha]\,$ denotes the subring of $R[x]/I$ generated by $R$ and $\alpha$, i.e. the smallest subring containing both. Clearly this is the whole ring $R[x]/I$ since $\,g(x)+I = g(x+I) = g(\alpha)\in R[\alpha].\ $ Furthermore, notice that $\,0 = g(\alpha) = g(x+I) = g(x)+I$ $\iff$ $g\in I = (f)$ $\iff$ $\,f\mid g\,$ in $R[x].\,$ Thus $\,\alpha\,$ serves as a "generic" root of $\,f\,$ over $R$ since it satisfies $f$ but no smaller degree polynomials. We can view the ring $R[x]/(f)$ as the most general (universal) way of "adjoining" a root of $f$ to $R$. Here "adjoining" has a technical meaning, which I elaborate on below (from a prior answer).

More generally, if $\rm\,R \subset S\,$ are rings and $\rm\,s\in S\,$ then $\rm\,R[s]\,$ denotes the ring-adjunction of $\rm\,s\,$ to $\rm\,R\,,\,$ i.e. the smallest subring of $\rm\,S\,$ containing both $\rm\,R\,$ and $\rm\,s\,.\,$ Equivalently $\,\rm R[s]$ is the image of $\rm\,R[x]\,$ under the evaluation map $\rm\,x\mapsto s.\,$ It is the set of all elements that can be written as polynomials in $\rm\,s\,$ with coefficents in $\rm\,R.\,$ The notation for the polynomial ring $\rm\,R[x]\,$ is the special case where $\rm\,x\,$ is transcendental over $\rm\,R\ $ (an "indeterminate" in old-fashioned language),$\ $ i.e. $\rm\, x\,$ isn't a root of any polynomial with coefficients in $\rm\,R\,$. One may view $\rm\,R[x]\,$ as the adjunction of a universal (or generic) element $\rm\,x\,$, in the sense that any other adjunction $\rm\,R[s]\,$ is a ring-image of $\rm\,R[x]\,$ under the evaluation homomomorphism $\rm\, x\to s\,.\ $ For example, if $\rm\,R \subset S\,$ are fields then $\rm\,R[s]\cong R[x]/(f(x))\,$ where $\rm\,f(x)\,$ is the minimal polynomial of $\rm\,s\,$ over $\rm\,R\,.\,$ Essentially this serves to faithfully ring-theoretically model $\rm\,s\,$ as a "generic" root $\rm\,x\,$ of the minimal polynomial $\rm\,f(x)\,$ for $\rm\,s\,.\,$ Polynomial rings may be characterized by the existence and uniqueness of such evaluation maps ("universal mapping property"), e.g. see any textbook on Universal Algebra, e.g. Bergman.

Bill Dubuque
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