Is it true that $\mathbb{Z}[\sqrt{79}]/(7, 1+\sqrt{79}) \cong \mathbb{Z}_7[X]/(\hat{1}+x)$?

Question

I am trying to see whether $(7, 1+\sqrt{79})$ is a prime ideal in $\mathbb{Z}[\sqrt{79}]$. I tried doing this by looking at the factor ring $\mathbb{Z}[\sqrt{79}]/(7, 1+\sqrt{79})$. We have $\mathbb{Z}[\sqrt{79}]\cong \mathbb{Z}[X]/(X^2-79)$, so $$\mathbb{Z}[\sqrt{79}]/(7, 1+\sqrt{79}) \cong \frac{\mathbb{Z}[X]/(X^2-79)}{(\hat{7}, \hat{1}+X)}=\frac{\mathbb{Z}[X]/(X^2-79)}{(7, 1+X)/(X^2-79)}\cong \mathbb{Z}[X]/(7, 1+X)\cong \frac{\mathbb{Z}[X]/(7)}{(7, 1+X)/(7)}\cong\frac{\mathbb{Z}_7[X]}{(\hat{1}+X)}.$$ Is this correct? Since $\hat{1}+X$ is irreducible in $\mathbb{Z}_7[X]$, it would follow that my ideal $(7, 1+\sqrt{79})$ is prime in $\mathbb{Z}[\sqrt{79}]$.

Answer 1

You have the right idea to apply the third isomorphism theorem, but you applied it incorrectly. Below I show how to correctly apply it, add some motivation to help avoid such mistakes.

Write $\,I=(x^2-79)\,$ and $J=(7,x+1)\,$ for those ideals in $\,\Bbb Z[x].\,$ By the correspondence and third isomorphism theorems, the ideal $J$ in $\Bbb Z[x]$ corresponds to $J+I$ in $\Bbb Z[x]/I,\,$ and $\,\Bbb Z[x]/(J\!+\!I)\cong (\Bbb Z[x]/I)\:\!/\:\!(J\!+\!I)/I.\,$ But here $J+I = (1)$ as below

$$\begin{align} J+I&=(7,\,x+1, \,x^2-79)\\[.2em] &= (7,\,x+1,\,x^2-79\bmod x+1)\\[.2em] &= (7,\,x+1,\,-78)\ \ \rm by\ \color{#0a0}{PRT}\\[.2em] &= (1),\ \ {\rm by}\ \ (7,\,78)=1\\[.2em] \Rightarrow\ \ \Bbb Z[x]/(J+I) & = \{0\} \end{align}\qquad$$

where we used the fact that $f(a) = f(x)\bmod x\!-\!a\,$ by $\rm\color{#0a0}{PRT} =$ Polynomial Remainder Theorem, while performing ideal modular reduction (an ideal analog of the reduction step in the Euclidean algorith for the gcd).

Otoh if $\,J=(p,x+1)\,$ where $\,p=2,3\,$ or $13$ is a prime factor of $78,\,$ then the same calculation yields $\,J+I = (p,x+1,78) = (p,x+1)\,$ by $\,p\mid 78,\,$ hence

$$\Bbb Z[x]/(J\!+\!I) = \Bbb Z[x]/(x\!+\!1,p) \,\cong\, (\Bbb Z[x]/p)/(x\!+\!1,p)/p\,\cong\, \Bbb Z_p[x]/(x\!+\!1)\,\cong\, \Bbb Z_p\qquad$$ as follows by here and its links.

Note that the remainder $\,x^2-79\bmod x+1$ is just the norm $\, n = \alpha\bar\alpha\,$ of $\,\alpha = 1+\sqrt{79},$ and $\,\alpha\in I\Rightarrow n =\alpha\bar\alpha \in I,\,$ so $\,(a,\alpha) = (a,n,\alpha) = ((a,n),\alpha),\,$ i.e. any integer generator can be replaced by its gcd with the norm of any element in $I$. Lubin's answer is a special case of this.

To better understand the correspondence and isomorphism theorems it is illuminating to study them in the special case of a PID, recalling that for principal ideals contains = divides, i.e. $(j)\supseteq (i)\iff j\mid i.\,$ Here the correspondence theorem implies that the ideal $(j)$ in $R$ maps to $(j)+(i) = (\gcd(j,i))$ in $R/i,\,$ and there is an order-preserving bijection between the ideals of $R/i$ and the ideals $(j)$ of $R$ that contain $(i),\,$ i.e. the divisors $j$ of $i$. This means that the ideal lattice of $R$ collapses to a lattice in $R/i$ that is isomorphic to the lattice of divisors of $i$ in $R,\,$ via the map $(j)\to (\gcd(j,i)).\,$ If $\,j\mid i\,$ then $(R/i)(j/i)\cong R/j,\,$ e.g. $\,2\mid 10\Rightarrow (\Bbb Z/10)/(\Bbb 2/10)\cong \Bbb Z/2,\,$ which implies that any arithmetic done $\!\bmod 10\,$ remains valid when specialized $\!\bmod 2,\,$ so e.g. we can compute integer parity from arithmetic on its $\rm\color{#c00}{unit\ digits}$, e.g. $$\begin{align} \bmod 2\!:\ \ 17(19)\, &\equiv\, 1\color{#c00}7(1\color{#c00}9)\bmod 10\\ &\equiv\, \ \ \color{#c00}7(\color{#c00}9)\ \ \bmod 10\\ &\equiv\,\ \ \ \ 63\ \ \ \bmod 10\\ &\equiv\,\ \ \ \ \ \ 3\,\equiv\, 1\\ \end{align}\qquad\qquad\qquad\ \ $$

Less trivially: $\, 41\mid 10^{\large 5}\!-1\,$ thus $\bmod 41\!:\ n\equiv (n \bmod 10^5\!-\!1),\,$ e.g.

$$\begin{align} \bmod 41\!:\,\ \color{#f6f}{9}0000\color{#e00}200030\, &\equiv\, \color{#f6f}{9}(\color{#0a0}{10^{\large 5}})^{\large 2}\! + \color{#e00}2(\color{#0a0}{10^{\large 5}}) + 30\,\bmod\, \color{#0a0}{10^5-1}\\[.0em] &\equiv\, \color{#f6f}{9}(\color{#0a0}1)^{\large 2}\ \ \ +\color{#e00}2 (\color{#0a0}1)\ \ \,+\,\ 30,\ \ \ {\rm by}\ \ \color{#0a0}{10^5\equiv 1}\\[.0em] & \equiv\,\ \ 0\end{align}\ \ $$

which can be viewed as a radix $10^5$ analog of casting out nines in radix $10$ to compute $n\bmod 3$. Such simple modular arithmetic special cases help to motivate the less trivial instances with arbitrary (non-principal) ideals, e.g. as in OP.

Above $\,j\mid i\Rightarrow R/j\cong (R/i)(j/i)\,$ is the universal property of the quotient ring $R/i$, i.e. it is the most general image of $R$ where $\,i=0,\,$ i.e. every image $R/j$ of $R$ with $i=0\:\!$ is an image of $R/i$.

The above PID special case of said theorems is discussed in more "element-ary" language in this post on the method of simpler multiples. You may find these elementary examples useful for providing motivation for the more abstract theorems for general ideals. Such motivation is often glaringly absent from many algebra textbooks. As always, when attempting to understand abstract results one often gains valuable intuition by examining more concrete specializations.

Answer 2

Are you sure you have the problem right? Call $I=(7,\sqrt{79}+1)$. Since it’s an ideal, it also contains $(\sqrt{79}+1)(\sqrt{79}-1)=78$. Since $78$ and $7$ are relatively prime, $1\in I$, so you’re dealing with the unit ideal.

Answer 3

There is a an error. In particular, we do not have $(\hat{7}, \hat{1}+X)=(7, 1+X)/(X^2-79)$. The issue is that $(X^2 -79)$ is not contained in $(7, 1+ X)$, so the notation $(7, 1+X)/(X^2-79)$ doesn't really make sense.

What is true is that $(\hat{7}, \hat{1}+X)=(7, 1+X, X^2 -79)/(X^2-79)$. I will leave it to you to verify this. We end up with $$\mathbb{Z}[\sqrt{79}]/(7, 1+\sqrt{79}) \cong \mathbb{Z}[X]/(7, 1+X, X^2 -79).$$ Finally, we have $$ (7, 1+X, X^2-79) = (7, 1+X, (-1)^2 - 79) = (7, 1+X, -78) =(1).$$ The first equality is due to $1+X$ being in the ideal. (In general, when dealing with generators of an ideal, we may "pretend the generator is zero"; this gives us $X = -1$ in this case.) The final equality comes from the fact that $7$ and $78$ are relatively prime integers. (In particular, this means we can write $1 = 7x + 78y$ for some integers $x$ and $y$, so that $1$ must be in the ideal, too.) We conclude that the original ideal $(7, 1+\sqrt{79})$ of $\mathbb{Z}[\sqrt{79}]$ was also the unit ideal.

As Lubin says, these sorts of things require a bit of experience to see quickly. I was a bit "fast and loose" with my reasoning above. I recommend taking some time to let it digest (i.e. think through why each step was legal).