4
$\begingroup$

We are in $R=\mathbb{Z}[x]$ and $I=\langle 4x^3,15 \rangle$ is the ideal of this ring, genereted by $4x^3$ and $15$. We want to prove that $3\not \in \langle 4x^3,15\rangle$.

My answer:

If $3$ was in $I$, then $\exists α(x),β(x)\in R:3=α(x)4x^3+β(x)15$ .

But if $x=0$ then $3=15β(0)\Longrightarrow β(0)\not \in R$ contradiction.

My question: Can we consider $$α(x)=a_0+...+a_nx^n, β(x)=b_0+...+b_mx^m \in R$$ then $$3=4x^3a_0+...+4x^{3+n}a_n+15b_0+...+15b_mx^m=15b_0+...+(15b_3+a_0)x^3+...$$ and then (because factors of two sides must be equal) $3=15b_0\Longrightarrow b_0\not\in \mathbb{Z}$. Contradiction.

Is this proof right?

$\endgroup$
  • 2
    $\begingroup$ Your answer looks good to me. You can do it that other way as well. The result connecting the two solutions is that a polynomial with coefficients in $\Bbb{Z}$ is uniquely determined by its values at integer points. If two (formal) polynomials are equal so are their values (this holds for all commutative rings). The reverse implication holds for $\Bbb{Z}$ as well (and for all infinite integral domains), but not for for example $\Bbb{Z}_p$. $\endgroup$ – Jyrki Lahtonen Aug 6 '16 at 21:37
  • $\begingroup$ So, both ways are correct and "similar"... Is there another way to prove this? For instance, something with degrees? $\endgroup$ – Chris Aug 6 '16 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.