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No such property exists. The ring isomorphism in part (b)(ii) identifies quotient rings by elements having different irreducible element behaviour.

The element $X^3$ is an irreducible element of $T$. It is nonzero and a nonunit. Every nonzero nonunit of $T$ has polynomial degree at least $2$: the only possible polynomials of smaller polynomial degree are constant polynomials, and every nonzero constant polynomial is a unit. If $X^3=uv$ with $u,v\in T$ both nonunits, additivity of polynomial degree in the integral domain $\mathbb{R}[X]$ would givewhich is impossible. Thus one factor must be a unit.

In $\mathbb{R}[Y]$, however, $Y^3=Y\cdot Y^2$ is a product of two nonunits, so $Y^3$ is not an irreducible element. If a ring isomorphism invariant $P$ characterized irreducible elements by the indicated quotient ring, the irreducible element $X^3\in T$ would imply that $T/(X^3{)}_T$ has $P$. Its isomorphic quotient ring $\mathbb{R}[Y]/(Y^3)$ would then have $P$, incorrectly making $Y^3\in \mathbb{R}[Y]$ an irreducible element.

Consequently,

Set $I=(X^3{)}_T$ and $J=(Y^3{)}_{\mathbb{R}[Y]}$. Part (b)(i) gives a unique representative $a+b{X}^2+c{X}^4$ for each element of the quotient ring $T/I$. By polynomial division, each element of the quotient ring $\mathbb{R}[Y]/J$ likewise has a unique representative $a+bY+c{Y}^2$. Thus the proposed map $\Phi$ is well-defined and a bijection.

It preserves addition and $1$. To verify multiplication, all terms of polynomial degree at least $5$ lie in $I$. HenceIn the other quotient ring, $Y^3=0$, giving exactly the same coefficient formula:Thereforeis a ring isomorphism. Both sides are examples of a truncated polynomial ring; the generator on the left is the class of $X^2$, whose square is nonzero and whose cube vanishes.

The required property isFor a nonzero element $x$ of an integral domain $R$, this follows fromIndeed, $\overline{a}\overline{b}=0$ in the quotient ring exactly when $x\mid ab$. The prime element condition says this forces $x\mid a$ or $x\mid b$, exactly the absence of zero divisors in the quotient ring. The nonunit condition makes the quotient ring nonzero. Being an integral domain is preserved by every ring isomorphism.

With the usual definition of a prime element, $x\ne 0$ is understood. The prime ideal formulation works for every ideal of any commutative ring, including the zero ideal; it avoids this convention about elements.

The polynomial ring ${\mathbb{F}}_2[X]$ is a euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.

A ring isomorphism is a bijective ring homomorphism. It preserves all ring operations and has a ring-homomorphic inverse.

A property of rings is a ring isomorphism invariant when any two rings related by a ring isomorphism either both have the property or both lack it. Being an integral domain is such a property: a ring isomorphism preserves nonzero elements, products, and zero.