$\mathbb{R}[x]$ denotes the ring of polynomials in $x$ with real coefficients. Let $I \subset \mathbb{R}[x]$ be the subset of those polynomials with constant coefficient $0$, and let $J \subset \mathbb{R}[x]$ be the set of polynomials with linear coefficient $0$: $$I = \{a_1x+a_2x^2+\dots+a_nx^n|n\geq 0,a_i\in\mathbb{R}\}$$ $$J = \{b_0+b_2x^2+\dots+b_mx^m|m\geq 0,b_i\in\mathbb{R}\}$$ i) Is $I$ an ideal of $\mathbb{R}[x]$?

ii) Is $ J$ an ideal of $\mathbb{R}[x]$?

iii) Is $I\cap J$ an ideal of $\mathbb{R}[x]$?

My answers:

i)$I = \mathbb{R}[x] / a_0$ so we need only compare $a_0 * I$ and $I * a_0$

$a_0 * I = {a_0(a_1x+a_2x^2+\dots+a_nx^n)\in I}$ $I * a_0= {(a_1x+a_2x^2+\dots+a_nx^n)a_0\in I}$ Hence $I$ is an ideal of $\mathbb{R}[x]$

ii) Proof $J$ is not an ideal by counterexample:

$xJ = {xb_0 + b_2x^3 + b_3x^4+\dots+b_mx^m+1}\not\in J$

iii) $I \cap J$ in an ideal of $\mathbb{R}[x]$

$I \cap J = \{b_2x^2+\dots+b_mx^m|m\geq 0,b_i\in\mathbb{R}\}$

Laptop is lagging to a halt now, but it is an ideal, by same logic as i)

Are these correct?


You nailed the conclusions, but did the proofs wrong.

  1. $I$ is the ideal generated by the polynomial $x$.

  2. $J$ is not an ideal because $1\in J$ and $x\notin J$.

  3. $I\cap J$ is the ideal generated by the polynomial $x^2$.

  • $\begingroup$ Can you explain $2$? $\endgroup$ Jun 24 '14 at 9:10
  • $\begingroup$ @Examin5days You should explain it! ;-) It is the required counterexample, isn't it? Since $1\in J$, for every polynomial $f$ you should have $1\cdot f\in J$. $\endgroup$
    – egreg
    Jun 24 '14 at 9:11

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