Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
No need for us to prove this; somebody already did that. A searchfor "Jacobson radical" must certainly turn up something. – Marc van Leeuwen Jun 3 '12 at 16:47
up vote 7 down vote accepted

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.

share|cite|improve this answer
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…, but it's not helping more. – AdrianM Jun 3 '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 3 '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. – AdrianM Jun 3 '12 at 18:28
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 3 '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. – AdrianM Jun 3 '12 at 18:55

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.

share|cite|improve this answer
Oh, great solution, thank you! I like it so much that I will present them both (along with the above, thanks to @Bill Dubuque). – AdrianM Jun 4 '12 at 8:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.