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I have a question which I suspect has the same underlying problem as my question on ring adjuncts yesterday.

Artin asks us to consider adjoining $1/2$ to $R=\mathbb{Z}/12\mathbb{Z}$. So I construct $R[1/2]\cong R[x]/(2x-1)$. For any $f\in R[x]$ we consider $f(x)=(2x-1)g(x)+r(x)$ whose residue in $R[x]/(2x-1)$ is just $r(x)$. $r$ must be a constant polynomial, since its degree must be less than $2x-1$, so it seems like $R[x]/(2x-1)$ is simply isomorphic to $\left\{r(x)=c|c\in R\right\}$ which in turn is isomorphic to $R$.

It seems to me that I've constructed an isomorphism $R\cong R[x]/(2x-1)$, which should indicate that $R\cong R[1/2]$. If I consider $R[1/2]=\left\{\sum r_i (1/2)^i|r_i\in R\right\}$ this isomorphism is patently false, so I think I'm doing something wrong.

Unlike the question I asked yesterday, $(2x-1)$ is certainly in $R[x]$

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The division algorithm acts strangely when there are zero-divisors. See if you can prove that 6*(2x-1) = 0, but that also 6 = 0 in R[1/2]. –  Jack Schmidt Jun 24 '11 at 15:28
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$R$ is a ring where 2 is a zero divisor (and has no multiplicative inverse) so your comment on $r$ being constant is false. Check what happens when you try to apply the division algorithm to the polynomials $x$ or $x^2$ in order to reduce the degree. –  Mark Bennet Jun 24 '11 at 15:31
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The division algorithm you are using requires the leading coefficient to be a unit, but your leading coefficient is not (it's a zero-divisor). –  Bill Dubuque Jun 24 '11 at 15:32
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Unlike yesterday, the leading coefficient of $2x-1$ is not a unit, so not all cosets of the ideal generated by $(2x-1)$ have a representative of degree 1. But here the leading coefficient is a zero divisor, so something else happens. Consider the following: $4x^2-1=(2x-1)(2x+1)$ is zero in $R[x]/(2x-1)$. Therefore also $0=3(4x^2-1)=12x^2-3=-3$. So $\ldots$ ? –  Jyrki Lahtonen Jun 24 '11 at 15:35
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HINT $\ $ The kernel of the evaluation map from $\mathbb Z[x]$ is $(12,2x-1) = (3,x+1)\:,\:$ therefore we infer $R[1/2] \cong \mathbb Z[x]/(3,x+1) \cong \mathbb Z/3\:.$ –  Bill Dubuque Jun 24 '11 at 15:44

2 Answers 2

up vote 7 down vote accepted

HINT $\ 12 = 0\ \Rightarrow\ 0 = 12\:(1/2)^2 = 3\:.\ $ Now you can apply the division algorithm since $\:2\:x-1\: =\: -(x+1)\:$ has unit leading coef. Conclude $R[1/2]\cong (\mathbb Z/3)[x]/(x+1) \cong \mathbb Z/3\:.\:$

Alternatively compute the kernel of the evaluation map from $\mathbb Z[x]$ as $(12,2x-1) = (3,x+1)\:$ hence $R[1/2] \cong \mathbb Z[x]/(3,x+1)\cong \mathbb Z/3\:.$

REMARK $\ $ In the same way one can construct fields of fractions and localizations presented as quotient of (multivariate) polynomial rings, i.e. to force $s$ to be a unit we pass to $R[x]/(s\:x-1)$ and repeat this for all elements to be forced to units. As above, when $R$ isn't a domain, this may force some elements of $R$ to become $0$, i.e. the image of $R$ in the constructed ring generally is not an embedding. See my post here for references.

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The correct way to think about adjoining multiplicative inverses to elements already in the polynomial is localization.

What you are basically doing when adjoining $1/2$ is that you are declaring that $2$ ought to be a unit in $R[1/2]$ and that the rest of the ring ought to be have as close as possible to $R$ (formally, it satisfies a certain universal property; see the wiki for more detail).

In any case, let $S$ be the multiplicative set (monoid really) generated by $2$ in $R$ (in this particular case $S=\{1,2,4,8\}$; we require $1\in S$ for the following to work). Then if $2$ is a unit, units should also be $4$ and $8$. The key property about units is that they are invertible, so all the elements of $R[1/2]$ will be of the form $r/s$ where $r\in R$ and $s\in S$. Just like regular fractions, however, sometimes two elements $r/s$ and $r'/s'$ are actually the same element. Having $r/s=r'/s'$ would imply that $s'r-rs'=0$ in $R[1/2]$.

Multiplication in $R[1/2]$ is not the same as multiplication in $R$ for the following reason. Suppose that in $R$ we have $ts=0$ for some $s\in S$. Then in $R[1/2]$ we should still have $ts=0$, but now, because in $R[1/2]$ we have that $s$ is unit, this implies that $t=0$! In other words, the map $R\to R[1/2]$ given by $r\to r/1$ is not injective since anything that annihilates something in $S$ must now be sent to $0$.

This allows us to write down explicitly the condition of $r/s=r'/s'$ as an equation in $R$: there exists an element $s_0\in S$ annihilated by $(s'r-rs')$, i.e. *$(s'r-rs')s_0=0$ in $R$*. It turns out that this condition is sufficient to describe all of $R[1/2]$.

Now, in the case where $R=\mathbb Z$, $S=\{2^k\colon k\geq 0\}$, $(s'r-rs')s_0=0$ in $R$ implies $s'r-rs'=0$ since $S$ consists only of non-zero divisors (hence we can cancel by them) in $\mathbb Z$, and so $R[1/2]$ is just the ring of the dyadic fractions, that is, fractions with denominator that is a power of $2$.


How does the above answer your question? Well, looking at $R[x]/\left<2x-1\right>$, you must consider the kernel of the map $R\to R[x]/\left<2x-1\right>$. The above gives you intuition that these should be things killed by $2,4$, or $8$ which are $3$, $6$, and $9$, so you expect the kernel to be $\{0,3,6,9\}$.

Lo and behold: it is, since really, that ideal is $R\cap \left<2x-1\right>$ and the only way that a multiple of $2x-1$ is equal to a constant is if all of the coefficients on indeterminates $x$ are $0$. Hence, if $\sum a_ix^i(2x-1)=-a_0$, then we need $2a_i-a_{i+1}=0$, or equivalently $2a_i=a_{i+1}$, which implies that $2^na_0=0$, establishing that $a_0$ is in fact inside the ideal $\{0,3,6,9\}$. That each of $0,3,6,9$ can be achieved is evident from multiplying $2x-1$ by $-\sum 2^ix^ia_0$.

This generalizes to show that in general the kernel of $R\to R[x]/\left<sx-1\right>_{s\in S}$ for a multiplicative set $S$ is the ideal generated by elements that kill at least one element of $S$.

It follows that $R[x]/\left<sx-1\right>_{s\in S}\cong R/I\left<sx-1\right>_{s\in S\setminus I}$, that is, inverting the multiplicative set $S$ in $R$ is the same as inverting the image of the multiplicative set $S$ in $R/I$, which image will have no such ideal. So we end up with showing that if $R\to R[x]/\left<sx-1\right>$ is injective, then the latter is isomorphic to fractions $r/s$ with $r/s=r'/s'$ if $rs'-r's=0$ since the image of $S$ has no zero-divisors, which you can do on your own.


Finally, applying the above to your question, we see that $\mathbb Z/12/\left<2x-1\right>\cong\mathbb Z/3/\left<2x-1\right>\cong\mathbb Z/3$, the latter holding because $2$ is already a unit in $\mathbb Z/3$.

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