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Let $A$ be a commutative ring, and $A[x]$ be the ring of polynomials in an indeterminate x, with coefficients in A.
I found several proofs online that in this case the Jacobson radical equals to the nilradical, and to prove the non-trivial direction (Jacobson radical included in the nilradical), they all used the fact that if $f=a_0+a_1x+...+a_nx^n$ is an element of the Jacobson radical, then $1+fx$ is a unit, but I couldn't see why.
Can anyone please clarify that for me?

I'm not looking for a proof, I'm just looking for an explanation about why $1+fx$ is a unit.

so.very.tired
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  • Commutative ring? – rschwieb Jul 09 '14 at 12:49
  • Agh, stupid superpower. Wanted to mark it as "possible duplicate" and let the votes decide. so.very.tired: if the dupe really isn't a dupe, just ping me and I'll do my best to reverse the closure. In the commutative case (the only one I know it works for sure) this is called Snapper's Theorem. – rschwieb Jul 09 '14 at 12:59

2 Answers2

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The key idea is an ideal-theoretic generalization of $\rm\color{#c00}{Euclid's}$ key idea in his proof that there are infinitely many primes $\rm p.\,$ Namely, if $\,f\,$ is divisible by every prime $\,p\,$ then $\,1+ nf\,$ has no prime divisors so is a unit. In a similar way, replacing "divides" by "contains", and $\,n\,$ by $\rm\,x\,$ we obtain

Hint $\rm\ f\in J(R[x])\Rightarrow\: f\:$ in all max $\rm P\:\color{#c00}\Rightarrow\:1\! +\! x\,f\:$ in no max $\rm P\:\Rightarrow\:1\!+\!x\,f\:$ unit $\ \ $ QED

The proof uses the well-known fact that every nonunit $\,r\,$ has a prime "divisor" (i.e. container), since $\,(r)\ne 1\,\Rightarrow\, (r)\,$ is contained in some maximal (so prime) ideal $\rm\,P.\ $ Hence, $ $ contrapositively, $ $ if $\,r\,$ is not contained in any max ideal, then $\,r\,$ is a unit.

Remark $\ $ Below is a generalization of the key idea, from my post giving a constructive generalization of Euclid's proof (for any ring with fewer units than elements).

Theorem $\ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R)\:.$

$\rm(1)\quad J \subseteq Jac(R),\quad $ i.e. $\rm\:J\:$ lies in every max ideal $\rm\:M\:$ of $\rm\:R\:.$

$\rm(2)\quad 1+J \subseteq U,\quad\ \ $ i.e. $\rm\: 1 + j\:$ is a unit for every $\rm\: j \in J\:.$

$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad\ $ i.e. proper ideals survive in $\rm\:R/J\:.$

$\rm(4)\quad M\:$ max $\rm\:\Rightarrow M+J \ne 1,\quad $ i.e. max ideals survive in $\rm\:R/J\:.$

Proof $\: $ (sketch) $\ $ With $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$

$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.

$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow I = 1\:.$

$\rm(3\Rightarrow 4)\ \:$ Let $\rm\:I = M\:$ max.

$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\:$ by $\rm\:M\:$ max.

Community
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Bill Dubuque
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In general, if $f$ is any element of a ring contained in the jacobson radical, then for every other element $r$ we have that $1+fr$ is a unit. Otherwise it would be contained in a maximal ideal, but $fr$ too, so that $1$ would be contained, a contradiction.

Martin Brandenburg
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