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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.

Bill Dubuque
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Adrian Manea
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2 Answers2

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Hint $\rm\ f\in J(R[x])\Rightarrow f\:$ in all max $\rm M$ $\Rightarrow\:\overbrace{\rm 1\! +\! x\:\!f\:\text{ in no max}\ \rm M}^{\small\textstyle \rm 1\!+\!x\:\!f,f\in M\Rightarrow 1\in M}\Rightarrow \rm 1\!+\!x\:\!f\,$ unit $\rm \Rightarrow f = 0$

Remark $ $ More generally this shows that $\rm R[x]$ has Jacobson radical equal to its nilradical: the above shows $\rm\ f\in J(R[x])\Rightarrow 1+x\:\!f\,$ a unit, so by here all coef's of $\rm\,f\,$ are in $\rm\sqrt R$ so $\rm\,f\in \sqrt{R[x]}$

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.

Bill Dubuque
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  • 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 http://math.stackexchange.com/questions/63851/intersection-maximal-ideals-of-a-polynomial-ring, but it's not helping more. – Adrian Manea Jun 03 '12 at 17:06
  • 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. – Bill Dubuque Jun 03 '12 at 17:33
  • 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. – Adrian Manea Jun 03 '12 at 18:28
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    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. $\quad$ – Bill Dubuque Jun 03 '12 at 18:45
  • I had that in mind :) so deg(g)=deg(h)=0 and deg(units)=0, so the conclusion follows (f=0). Thanks again. – Adrian Manea Jun 03 '12 at 18:55
  • @AdrianM Yes, you've got it now. – Bill Dubuque Jun 03 '12 at 19:00
  • why 1+xf is not in any maximal ideal? – user297564 Jan 16 '18 at 17:52
  • @user297564 because if $f$ and $1 + xf$ are in an ideal then $1$ is too, so the ideal is not proper, in particular not maximal. – Badam Baplan Mar 05 '18 at 15:37
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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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    Oh, great solution, thank you! I like it so much that I will present them both (along with the above, thanks to @Bill Dubuque). – Adrian Manea Jun 04 '12 at 08:07