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2

First of all, let us find the generator of the (principal) ideal $I_n$ (with obvious dependency on $n$) in small dimensions (for small values of $n$) by making some experiments in sage: sage: R.<t,x1> = PolynomialRing(QQ) sage: J = R.ideal(x1^2-x1, t-x1) Ideal (x1^2 - x1, t - x1) of Multivariate Polynomial Ring in t, x1 over Rational Field sage: J....


3

In $\mathbb F_{47}$one has $x^{46}=1$ then $x^{23}=\pm1$ ► $x^{23}=1\iff x^{11}\cdot x^{12}=1\iff 5\cdot x^{12}=1\Rightarrow x^{12}=\dfrac 15=19$. Then $\begin{cases}x^{12}=19\\x^{11}=5\end{cases}\Rightarrow x=\dfrac{19}{5}=19\cdot19=34$ and $34^{11}=21$ thus $x=34$ is not solution. ►$x^{23}=-1\iff x^{11}\cdot x^{12}=-1\iff 5\cdot x^{12}=-1\Rightarrow x^{...


0

Hint at another related way to think about it: $(x^{11})^5\equiv x^9\equiv 5^5\equiv 23 \bmod 47$ Because of $(a^b)^c=a^{bc}$, $a^{p-1}\equiv 1\bmod p$ for prime p, a not a multiple, and $55\equiv 9\bmod 46$ which is valid by extension of the above rules plus $1^n=1$ and the fact that the first multiple of a number greater than a non multiple is of lower ...


2

Hint $\bmod 47\!:\,\ x^{\large 11}\equiv 5\,\overset{(\ \ )^{\Large 21}}{\Longrightarrow} x\equiv 5^{\large 21}\ $ by $\,11\cdot 21\equiv 1 \pmod{\!46}\,$ and little Fermat. Thus $\,x\equiv \dfrac{\color{#c00}{5^{\large 23}}}{5^{\large 2}}\equiv \dfrac{\color{#c00}{\bf -1}}{25}\equiv \dfrac{-2}{50}\equiv\dfrac{45}3\equiv 15,\,$ by $\, \color{#c00}{5^{\...


2

You claim that $(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/3\Bbb{Z})$ is not a ring, but it is in fact canonically a ring. In general, the product of two rings is again a ring. In this particular case we even have the very nice canonical isomorphism $$(\Bbb{Z}/2\Bbb{Z})\times(\Bbb{Z}/3\Bbb{Z})\cong\Bbb{Z}/6\Bbb{Z},$$ so in stead of representing the coefficients of ...


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