Intereting Posts

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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.

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

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**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.

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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