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This question is at a very low level. Let $R$ be a commutative ring with unity. I am using the standard definition $\langle x \rangle = \{rx: r \in R\}$ for the ideal generated by $x$. It seems like the polynomial $p(x) = x^2 + x$ should not live in $\langle x \rangle$, since there is no $r \in R$ such that $p(x) = rx$. Does $\langle x \rangle$ automatically include all $\underline{linear}$ $\underline{ combinations}$ of $x$? Because the definition doesn't seem to imply that to me.

Thanks!

sobrio35
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2 Answers2

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It is closed under linear combinations (not more), since you have the unit in $R$. Let $y,z \in <x>$, hence you have elements $r_y,r_z \in R$ s.t. $y=r_y x, z = r_z x$ holds. Since $R$ is additively closed also $r:=r_y + r_z \in R$ holds, so $y+z=(r_y+r_z)x=rx \in <x>$ holds. You need the unit, that also $x = 1 x \in <x>$ holds.

You have to be carefull what you are doing above. $<x>$ as definied as you did it, is not an ideal in the ring of polynomials over $R$, but for arbitrary $x \in R$ an ideal in $R$. Instead $<x> := \{ rx \ | \ r \in R[x]\}$ is an ideal in the ring of polynomials over $R$.

Assuming you are now working with a polynomial ring (to see not every linear combination is in the ideal): E.g. your example $p(x)=x^2+1$ has not a root 0 since $1 \neq 0$ but obviously every $f(x) \in <x>$, so $f(x)=rx (r \in R[X])$, has root $0$.

Edit: Since you changed your polynomial later on, here the answer to your edit: for $x \in R$ we have $p(x):=x^2+x=(x+1)x \in <x>$ since $x,1 \in R$ and hence $x+1 \in R$.

ctst
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  • So $\langle x \rangle$ contains polynomials with more than one term only if $R$ contains polynomials, correct? – sobrio35 Sep 17 '16 at 21:56
  • @sobrio35 I think you have a little bit confusion about polynomials and rings. Either you think of a ring or you speak about the polynoms over that ring (which you normally notate as $R[X]$ and is itself a ring and is hence called ring of polynomials (over $R$) and is a different ring than $R$, which contains $R$ as a subring). Now consider an element $x \in R$ (this can be everything, e.g. if $R=S[X]$ for some ring $S$ this could be even $x=X$). Then it makes sense to speak of $x^2+x= x \cdot x + x$ which is in $R$ as $R$ is additive and multiplicative. – ctst Sep 18 '16 at 00:30
  • It does not make sense to say $p \in R$ for a polynomial $p(X)=X^2+...$ (with $X \notin R$) but $p \in R[X]$. On the other hand it makes absolutly sense to say $p(x)=x^2+... \in R$ for some $x \in R$ as mentioned before. Mark the difference between $x \in R$ as an arbitrary (but fixed) element of $R$ and $X$ as placeholder/variable (or transcendental extension if you have heard that word, if not ignore that part)! – ctst Sep 18 '16 at 00:34
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Talking about the ideal $(x)$ only makes sense if $x$ is an element of $R$, in which case $r:=x+1$ is also an element of R. So

$$x^2+x=(x+1)x=r x \in (x)$$

as desired.

Daniel McLaury
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