What is $\mathbb{Z}[x]/(6,2x-1)$ isomorphic to?

Question

What is $\mathbb{Z}[x]/(6,2x-1)$ isomorphic? Since $\mathbb{Z}[x]/(6,2x-1)$ is isomorphic to $\frac{\mathbb{Z}[x]/(6)}{(2x-1)}$ then it's isomorphic to $\mathbb{F}_6[\frac{1}{2}]$. But on the other hand, the ideal $(6,2x-1)$ is equal to $(3,x-1)$ then $\mathbb{Z}[x]/(6,2x-1)$ is isomorphic to $\mathbb{F}_3/(x-1)$ which is isomorphic to $\mathbb{F}_3$? However, $\mathbb{F}_6[\frac{1}{2}]$ is not isomorphic to $\mathbb{F}_3$, so where am I wrong?

Answer 1

First of all, there is no such thing as ${\mathbb F}_6$, the field of six elements. You may mean ${\mathbb Z}_6$, the integers modulo six. Secondly, writing ${\mathbb Z}_6[1/2]$ is at least weird. It means the smallest ring containing the subring ${\mathbb Z}_6$ and the rational number $1/2$, but there is no natural larger ring containing both. Finally, $(6,2x-1) = (3,2x-1) = (3,x+1)$, not $(3,x-1)$.

Now, indeed, $${\mathbb Z}[x]/(6,2x-1) \cong {\mathbb Z}_6[x]/(2x-1).$$

However, in this ring $6x - 3 = 3 \cdot (2x - 1) = 0$ and because $6 = 0$, this means that $3 = 0$. Therefore, you also get $${\mathbb Z}[x]/(6,2x-1) \cong {\mathbb F}_3[x]/(2x-1) = {\mathbb F}_3[x]/(x-2) \cong {\mathbb F}_3.$$

You can also see this by realizing right from the start that $(6,2x-1) = (3,2x-1)$, because $3 = x\cdot 6 - 3 \cdot (2x - 1)$. This immediately gives $${\mathbb Z}[x]/(6,2x-1) = {\mathbb Z}[x]/(3,2x-1) \cong {\mathbb F}_3[x]/(2x-1) = {\mathbb F}_3[x]/(x-2) \cong {\mathbb F}_3.$$

So, apart from the remarks in my first paragraph, both your approaches are correct and give the same result.

Answer 2

You can use, o prove, the following:

Claim: An ideal in $\;\Bbb Z[x]\;$ is maximal iff it is of the form $\;\langle\,p,\,f(x)\,\rangle\;$ , with $\;p\;$ a prime number and $\;f(x)\in\Bbb Z[x]\;$ an element such that $\;f(x)\pmod p\;$ is irreducible in $\;\Bbb F_p[x]\;$ .

We're then almost done when we realize that $\;\langle\,6,\,2x-1\,\rangle=\langle\,3,\,2x-1\,\rangle\;$ and $\;2x-1\in\Bbb F_3[x]\;$ is irreducible and thus the quotient ring is a field...

Finally, the steps from $\;\Bbb Z[x]/\langle6,\,2x-1\rangle\;$ to (isomorphism) $\;\Bbb F_3\;$ are very similar to what was done in the other answer: using the second-third isomorphism theorems and etc.