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I am doing a self-study using Artin's algebra. There is a problem which asks us to prove that adjoining an element to a ring doesn't do anything if that element was already in the ring.

More formally, say that $\alpha=a$ where $a\in R$. We want to show that $R\approx R[\alpha]$. (Where $\approx$ means "isomorphic to".)

My proof: We can consider $R[\alpha]\approx R[x]/(x-\alpha)$, which intuitively means "killing" $x-a$ in the ring of polynomials. Any function $f\in R[x]$ can be written as $(x-\alpha)g(x)+r(x)$ for some g,r by the division algorithm. Since the degree of r must be less than the degree of $x-a$, r must be a constant polynomial. Since $(x-\alpha)=0$, the residue of $f$ in $R[x]/(x-\alpha)$ is just r, which is some element of R. We can see that every $s\in R$ has a corresponding constant polynomial $f(x)=s$ in $R[x]$, which means the simple inclusion map $s\mapsto f(x)=s$ is our desired isomorphism.

So, this proof seems fine to me, except that I seem to have proven that adjoining any element keeps the ring the same. This sounds implausible to me. Where did I go wrong?

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HINT $\ $ If $\rm\:\alpha \not\in R\:$ then $\rm\:x\mapsto \alpha$ is an evaluation homomorphism from $\rm\:R[x]\:$ to $\rm\:R[\alpha]\:,\:$ not to $\rm\:R\:.\:$

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So I have a set $\{r(x)=s|s\in R\}$. Why would the homomorphism be an evaluation to $R[\alpha]$? $r(\alpha)\in R$, right, even if $\alpha$ is not? – Xodarap Jun 24 '11 at 14:55
@Xod As I remarked in my comment to Gerry's answer, by the First Isomorphism Theorem and the universal mapping property of polynomial rings, the simple adjunctions $R[a]$ are classified by ideals $I$ in $R[x]$, viz. $R[a] = R[x]/I\:,\:$ where $I$ is the kernel of the evaluation map $x\mapsto a\:.\:$ For the special case $a\in R$ the kernel is $x-a$ hence $R = R[a] = R[x]/(x-a)\:.$ – Bill Dubuque Jun 24 '11 at 15:19
I think I might be misunderstanding because it's too abstract. I've asked the question giving an example, which hopefully will help me to understand: – Xodarap Jun 24 '11 at 15:24

Your proof doesn't show that adjoining any element keeps the ring the same, because if $\alpha$ isn't in $R$, then $x-\alpha$ is not in $R[x]$, so the quotient $R[x]/(x-\alpha)$ doesn't make sense.

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How have you defined $R[\alpha]$? I usually define it as the smallest ring containing $R$ and $\alpha$, in which case, if $\alpha$ is in $R$, it's clear that $R[\alpha]=R$ (and if $\alpha$ is not in $R$, it's clear that $R[\alpha]\ne R$).

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Artin defines it as $R[x]/(f)$, where $f(\alpha)=0$ (in addition to some other ways). Your method does make it much simpler to prove :-) – Xodarap Jun 24 '11 at 1:48
@Xodarap, that can only work when there is a polynomial $f(x)$ in $R[x]$ such that $f(\alpha)=0$. You couldn't define ${\bf Z}[\pi]$ that way (for example). – Gerry Myerson Jun 24 '11 at 2:49
@Xod Despite Gerry's comment, Artin is correct. When $\alpha$ is transcendental over $R$ one has $R[\alpha] \cong R[x]\:,\:$ i.e. the kernel of the evaluation map $\:x\mapsto \alpha\:$ is zero, i.e. $\rm\:f = 0\:$ above. Indeed, by the universal mapping property of the polynomial ring $R[x]\:,\:$ every simple adjunction $R[\alpha]$ is an image of $R[x]$ via the evaluation map $\:x\mapsto \alpha\:.\:$ Hence the hint in my answer. The kernel of the evaluation map is the ideal of polynomials in $R[x]$ that have $\alpha$ as a root. It needn't be a principal ideal if $R[x]$ is not a PID. – Bill Dubuque Jun 24 '11 at 14:26
@Bill, yes, I missed division by (the) zero (ideal). – Gerry Myerson Jun 25 '11 at 3:22

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