On units in a ring with a polynomial property

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On units in a ring with a polynomial property

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Source: Romania NMO 2025 12.1

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#1 h Apr 2, 2025, 2:21 PM

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We say a ring $(A,+,\cdot)$ has property $(P)$ if :
$\begin{cases} \text{the set } A \text{ has at least } 4 \text{ elements} \\ \text{the element } 1+1 \text{ is invertible}\\ x+x^4=x^2+x^3 \text{ holds for all } x \in A \end{cases}$ a) Prove that if a ring $(A,+,\cdot)$ has property $(P)$ , and $a,b \in A$ are distinct elements, such that $a$ and $a+b$ are units, then $1+ab$ is also a unit, but $b$ is not a unit.
b) Provide an example of a ring with property $(P)$ .

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#2 h Apr 2, 2025, 4:21 PM

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If a ring A has property (P), then for every element x in A, x(x - 1)^2(x + 1) = 0. Let x = ab, where a and a + b are units. This implies (ab - 1)^2(ab + 1) = 0, so either ab = 1 or 1 + ab is a unit. Since ab = 1 would make b = a^(-1), contradicting that a and b are distinct, it must be that 1 + ab is a unit. Now suppose b is a unit. Plug x = b into the identity: b(b - 1)^2(b + 1) = 0. If b + 1 is a unit, then (b - 1)^2 = 0, so b = 1, again contradicting that a and b are distinct. If b + 1 = 0, then b = -1, which implies a + b = a - 1 is a unit, so (a - 1)^2(a + 1) = 0. Since a - 1 is a unit, a + 1 = 0, giving a = -1, so a = b, contradiction. Therefore, b is not a unit.

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#3 h Apr 2, 2025, 5:40 PM

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Put a ring on it

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#4 h Apr 2, 2025, 8:42 PM

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a) We have $2+2^4=2^2+2^3,$ so $18=12,$ i.e. $6=0.$ Since $2$ is invertible, we get $3=0.$

Now for any $x \in A$ we have $(x+1)+(x+1)^4=(x+1)^2+(x+1)^3,$ which after cancellations yields $x^4=x^2.$ From the third condition of $(P)$ we get $x=x^3$ for any $x \in A.$

Thus, for any unit $u \in A$ we have $u^2=1,$ so in particular $a^2=1.$ The element $a(a+b)=1+ab$ is a unit, as a product of units.

We then have $(1+ab)^2=1,$ so $(ab)^2=ab.$ If $b$ is invertible, then $ab$ is invertible, so the last relation yields $ab=1,$ and so $b=aab=a,$ contradiction.

Remark. It is a classical result that if $x^3=x$ holds for any element of a ring, then the ring is commutative. Using this and $(a+b)^2=1$ we get $b^2=ab,$ and it is clear that $b$ cannot be invertible because this would lead to $a=b$ otherwise.

b) For example $\mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ with componentwise addition and multiplication.

This post has been edited 1 time. Last edited by Filipjack, Apr 2, 2025, 8:42 PM

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