$\mathbb{Z}[X]/(X^2-1)$ is not isomorphic with $\mathbb{Z}\times \mathbb{Z}$

Solution 1:

Okay, let's start with the polynomial ring and try to define an isomorphism to the product ring. The multiplicative identity needs to go to the multiplicative identity, for starters: \[1 \mapsto (1,1)\] The only remaining question is where $X$ goes. Well, we have $X^2 = 1$, so $X$ has to go to an element that squares to the identity. If it's $(1,1)$ or $(-1,-1)$ the homomorphism won't be injective, so it's either $(1,-1)$ or $(-1,1)$, and it doesn't really matter which: \[X \mapsto (1,-1)\] Now a general element of the polynomial ring can be written $a + bX$ for integers $a$ and $b$, and it'll be mapped to: \[(a + b, a - b)\] But the difference between these two entries is $2b$, so we can never make any element of $\mathbb Z \times \mathbb Z$ whose entries differ by an odd number. In particular, the proposed homomorphism does not hit $(1,0)$ and is not surjective.

Solution 2:

$\mathbb{Z}[X]/(X^2-1)$ lacks nontrivial idempotents.

Solution 3:

A homomorphism $\mathbb{Z}[X]\to \mathbb{Z}\times \mathbb{Z}$ is equivalent to specifying an element of $\mathbb{Z}\times \mathbb{Z}$.

A homomorphism $\mathbb{Z}[X]/(X^2-1)\to \mathbb{Z}\times \mathbb{Z}$ is equivalent to specifying an element of $\mathbb{Z}\times \mathbb{Z}$ whose square equals $(1,1)$.

What are the elements of $\mathbb{Z}\times \mathbb{Z}$ whose square equals $(1,1)$? Hint: You've only got $4$ possibilities. So, there are $4$ homomorphisms $\mathbb{Z}[X]/(X^2-1)\to \mathbb{Z}\times \mathbb{Z}$.

Exercise: Check that none of these homomorphisms is surjective.

I hope this helps!

Solution 4:

$\mathbb{Z}[X]/(X^2-1) \cong \mathbb{Z} \times \mathbb{Z}$ would imply (after modding out the ideal $(2)$) that $$\mathbb{F}_2[X]/(X-1)^2 \cong \mathbb{F}_2 \times \mathbb{F}_2.$$The first ring has a nontrivial nilpotent element, whereas the second one does not.