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Let $R$ be a commutative domain.

Prove that the Jacobson radical of $R[X]$, i.e. the intersection of all maximal ideals, is the zero ideal.

Thank you.

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    $\begingroup$ No need for us to prove this; somebody already did that. A searchfor "Jacobson radical" must certainly turn up something. $\endgroup$ – Marc van Leeuwen Jun 3 '12 at 16:47
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Hint $\rm\ f\in J(R[x])\Rightarrow\: f\:$ in all max $\rm M\:\Rightarrow\:1\! +\! x\,f\:$ in no max $\rm M\:\Rightarrow\:1\!+\!x\,f\:$ unit $\rm\:\Rightarrow\:f = 0\ \ $ QED

Remark $\ $ Perhaps the following is of interest, from my post giving a constructive generalization of Euclid's proof of infinitely many primes (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.

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  • $\begingroup$ I'm sorry, but I can't understand the last implication. Why must f be zero if 1+xf is a unit? (BTW, I found this math.stackexchange.com/questions/63851/…, but it's not helping more. $\endgroup$ – Adrian Manea Jun 3 '12 at 17:06
  • $\begingroup$ Hint: $\rm\: gh = 1\:\Rightarrow\: deg\:g = \ldots\:$ but $\rm\:deg(x\:f+1) > \ldots\:$ if $\rm\:f\ne 0,\:$ using $\rm\:R\:$ is domain. $\endgroup$ – Bill Dubuque Jun 3 '12 at 17:33
  • $\begingroup$ This is what I understand: $deg(xf+1)>deg(f)$ and $g(1+xf)=1$ cannot happen unless $f=0$, since $R$ is a domain. Right? Thank you for your patience. $\endgroup$ – Adrian Manea Jun 3 '12 at 18:28
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    $\begingroup$ It's simpler if you follow the hint. If $\rm\:g\:$ is a unit then $\rm\:gh = 1\:$ for some $\rm\:h \in R[x].\:$ Since $\rm\:R\:$ is a domain, $\rm\:deg(gh) = deg\:g + \deg\:h = \deg\:\!1 = 0,\:$ so $\rm\:deg\:g = \ldots\:$ However, if $\rm\:f\ne 0\:$ then $\rm\:deg(xf+1) \ge 1 > \ldots = $ degree of units. So $\rm\:xf+1\:$ is not a unit, having higher degree than units. $\quad$ $\endgroup$ – Bill Dubuque Jun 3 '12 at 18:45
  • $\begingroup$ I had that in mind :) so deg(g)=deg(h)=0 and deg(units)=0, so the conclusion follows (f=0). Thanks again. $\endgroup$ – Adrian Manea Jun 3 '12 at 18:55
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Exercise 1.4 of Atiyah - Macdonald tells you that in any polynomial ring $R[x]$, the Jacobson radical and nilradical are equal. For your problem let us throw in the additional hypothesis that $R$ is an integral domain. Then the nilradical of $R[x]$ is zero because $R[x]$ is an integral domain and hence the Jacobson radical is zero.

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    $\begingroup$ Oh, great solution, thank you! I like it so much that I will present them both (along with the above, thanks to @Bill Dubuque). $\endgroup$ – Adrian Manea Jun 4 '12 at 8:07

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