I have two rings: $R_1 = \mathbb{Z}_2[x]/\langle x^4+1\rangle$ and $R_2 = \mathbb{Z}_4[x]/\langle x^2+1\rangle$. I've shown that these have the same number of elements. Now I am struggling to show that they are not isomorphic. In general, are there any invariants in rings to check? As in groups, where we could check that the order of elements under an isomorphism is the same etc.
[Math] Showing that two rings are not isomorphic
abstract-algebraring-theory
Related Solutions
By Definition, a ring homomorphism $f: R \rightarrow R'$ must preserve addition and multiplication and must map the multiplicative identity of $R$ to the multiplicative identity of $R'$. In your example, the ring $R'=2\mathbb{Z}$ does not have a multiplicative identity. So the two rings are not isomorphic (there is no isomorphism, or even a homomorphism from one ring to the other, for that matter).
To show that the rings $\mathbb{Z}[\sqrt{2}]$ and $\mathbb{Z}[\sqrt{5}]$ are not isomorphic, you can use your idea that $x^2=2$ has no solution in the latter ring. But you need to justify why this method works. Here is a proof. Suppose there is an isomorphism $f$ between these two rings that takes $a+b\sqrt{2}$ to $a'+b'\sqrt{5}$. Since $f$ must take the identity to the identity, $f$ takes 1 to 1' (here, 1' is the identity in the second ring, and actually equals the integer 1; the primes are just to make things clearer). Since $f$ preserves sums, $f$ must take $1+1$ to $1'+1'$. Now, $(0+1 \sqrt{2})(0+1\sqrt{2}) = 1+1$ in the first ring. We can apply $f$ to both sides. Since $f$ preserves sums and products, we get the equation $(x'+y'\sqrt{5})^2 = 2$, where $x'+y'\sqrt{5}$ is the image of $(0+1\sqrt{2})$ under $f$. This equation has no solutions, and so we get a contradiction. Thus, there does not exist an isomorphism $f$ from the first ring to the second.
Yes, if the cardinalities are different then there is no bijection between the rings, and so no isomorphism.
Over $\mathbb{Q}$ the problem is a bit more interesting. Note that the polynomial $x^2+1$ is irreducible over $\mathbb{Q}$, and so $\mathbb{Q}[x]/(x^2+1)$ is a field. On the other hand, $\mathbb{Q}[x]/(x^3+1)$ is not a field, because $x^3+1$ is reducible over $\mathbb{Q}$.
Best Answer
One invariant of a ring is its characteristic, the smallest natural number $n$ such that $$ \underbrace{1+\ldots+1}_{\text{n times}}=0 $$ The two rings you are considering have different characteristics.