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I want to show that $x+x^2$ belongs to an ideal $I$ of $\mathbb{Z}_2[x]$ which contains both $1+x^2$ and $1+x^3$. Since $\mathbb{Z}_2[x]$ is a principal ideal domain that contains two polynomials whose greatest common denominator is $1$, is it wrong to assume that $I$ is the ideal generated by $1$ in $\mathbb{Z}_2[x]$ and therefore that any polynomial in the field belongs to the ideal?

Edit: as pointed out, the gcd is $1+x$ in $\mathbb{Z}_2[x]$.

user26857
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Logi
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  • Their GCD is not $1$. Note that $1 + x$ divides both. Try showing that that is the GCD. – Aryaman Maithani Dec 30 '21 at 12:15
  • $1\cdot(1+x^3)+(1+x)\cdot(1+x^2)=(1+x^3)+(1+x+x^2+x^3)=2+x+x^2+2x^3=x+x^2$ in $\mathbb Z_2[x]$. –  Dec 30 '21 at 12:51
  • @AryamanMaithani If two polynomials is in an ideal (in a principal ideal domain), then any multiple of their gcd will also be in the ideal, is that always the case? – Logi Dec 30 '21 at 12:52
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    Sure. First the $\gcd$ will be in the ideal. Then, obviously (as per properties of ideals) all multiplies of the $\gcd$ will be in the ideal too. –  Dec 30 '21 at 12:54
  • @StinkingBishop, great thank you. So I can show that $x+x^2$ is in the ideal by noting that $1+x$ is in the ideal, which means that $x(1+x)$ is in the ideal too. – Logi Dec 30 '21 at 12:56

2 Answers2

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Hint: Compute $x(1+x^3)+x^2(1+x^2)$ in $\mathbb Z_2[x]$.

Alternatively, as mentioned by Aryaman Maithani, compute their greatest common divisor $d(x)$ using the Euclidean algorithm, and observe that $x+x^2$ is a multiple of $d(x)$. In this specific example, you will find $d(x)=x+1$.

Zuy
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  • Thank you. I know I can show it in this way, but I was wondering if I could find an easier (quicker) way, in case I want to show something similar in the future. – Logi Dec 30 '21 at 12:22
  • @JKalle The approach in the answer is probably the easiest. The streamlined method follows a Groebnber basis approach, so you can read about that. – Pedro Dec 30 '21 at 12:24
  • So generally, if two polynomials is in an ideal (in a principal ideal domain), then any multiple of their gcd will also be in the ideal? – Logi Dec 30 '21 at 12:28
  • @JKalle Yes, it is easier to use uniqueness of inverses, as hinted in my answer, i.e. mod $I$ both $x$ and $x^2$ are inverses of $x$ so they are equal by said uniqueness. – Bill Dubuque Dec 30 '21 at 12:28
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Hint: $\bmod I\!: \overbrace{\color{#c00}x\equiv x^{-1}\equiv \color{#0a0}{x^2}}^{\textstyle x\cdot \color{#c00}x\ \equiv\ 1\ \equiv\ x\cdot \color{#0a0}{x^2}}\!\!\! $ (by $\,{-1} = 1\,$ in $\Bbb Z_2,\,$ and uniqueness of inverses)

Alternatively: $\, x^{\large \color{#0a0}2}\equiv 1\equiv x^{\large\color{#0a0}3}\Rightarrow x\,$ has order $\,\color{#c00}{n\!=\!1}\,$ (by $\,n\mid\color{#0a0}{ 2,3})$ so $\,x^{\large \color{#c00}1}\equiv 1\Rightarrow x^2\equiv x $

Remark $ $ Generally the most efficient way to verify ideal membership $\,f\in (g_1,\ldots,g_n) = I\,$ in a Euclidean domain is to compute the generator $\,g\,$ of the ideal $I = (g)$ by the Euclidean algorithm, then $\,f\in (g)\iff g\mid f\,$ so ideal membership reduces to a divisibility test. Doing so here, over the coefficient ring $\,R=\Bbb Z_2,\,$ we compute by Euclid: $\,(g) = (x^2\!-1,\,x^3\!-1) = (x\!-\!1),\,$ hence $\,f\in (g)\!=\!(x\!-\!1)R[x]\!\iff\! x\!-\!1\mid f\,\ {\rm in}\ R[x]$ $\!\iff\! f(1) = 0\,$ in $R=\Bbb Z_2\!\iff 2\mid f(1)\,\ {\rm in}\,\ \Bbb Z$.

In special cases like the OP there may be easier methods, e.g. as above. Without further context it is not clear which of these the exercise was meant to illustrate (but all are useful to know in general)

There are some generalizations to multivariate polynomials, e.g. see the Gröbner basis algorithm (which may be viewed both as a multivariate generalization of the (Euclidean) polynomial division algorithm, as well as a nonlinear generalization of Gaussian elimination for linear systems of equation). A nice example is Gauss's constructive algorithm to rewrite a symmetric polynomial as a polynomial in the elementary symmetric polynomials, yielding a constructive interpretation of the Fundamental Theorem of Symmetric Polynomials.

Bill Dubuque
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