Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The following problem is from Golan's book on linear algebra, chapter 4. I have posted a proposed answer below.

Problem: Let $F$ be a field. Find all nonzero polynomials $P\in F[x]$ satisfying


share|cite|improve this question
up vote 5 down vote accepted

Your approach is generally good, though the execution might perhaps leave something to be desired.

Let $F$ be of characteristic different from $2$; we may assume $p(x)\neq 0$. Let $a_nx^n$ be the term of lowest degree. Then the term of lowest degree of $p(x^2)$ is $a_nx^{2n}$, and that of lowest degree in $p(x)^2$ is $a_n^2x^{2n}$, so $a_n=a_n^2$. The only solutions in a field are $0$ and $1$ (they must be roots of $x^2-x$), and since $a_n\neq 0$, we conclude $a_n=1$. Now assume that there exists $m\gt n$ such that $a_m\neq 0$, and pick $m$ to be the smallest such. Then $p(x) = x^n + a_mx^m + (\text{terms of higher degree})$, so $$p(x^2) = ax^{2n} + a_mx^{2m} + (\text{terms of higher degree}),$$ whereas $$\begin{align*} p(x)^2 &= (x^{n} + a_mx^{m} + (\text{terms of higher degree}))(x^{n} + a_mx^{m} + (\text{terms of higher degree}))\\ &= x^{2n} + 2a_mx^{n+m} + (\text{terms of higher degree}). \end{align*}$$ Since $n+m\lt 2m$ and $2a_m\neq 0$, this is a contradiction. Therefore, $p(x) = x^n$. And since these clearly "work", it follows that the only polynomials with the desired property are the nonnegative powers of $x$ and the zero polynomial.

Now assume that $F$ is of characteristic $2$. If $p(x) = a_0 + a_1x+ \cdots + a_nx^n$, then $$\begin{align*} p(x^2) = a_0 + a_1x^2 + a_2x^4 + \cdots + a_nx^{2n}\\ p(x)^2 = a_0^2 + a_1^2x^2 + a_2^2x^4 + \cdots + a_n^2x^{2n}. \end{align*}$$ Therefore, if $p(x^2) = p(x)^2$, then $a_i^2 = a_i$ for all $i$, so $a_i=0$ or $a_i=1$ for all $i$. Thus, $p(x)\in\mathbb{F}_2[x]$. And all of these work. Thus, for fields of characteristic $2$, the polynomials that work are all polynomials over the prime field (all coefficients equal to either $0$ or $1$).

share|cite|improve this answer

Assume first that $F$ is a field with characteristic not equal to 2. The only ones are 1 and $x^n$, $n\in \mathbb{N}$.

Let $a_n$ denote the coefficient of $x^n$ in $P$. We see immediately that all $a_n=0$ for odd $n>0$. Examining the constant coefficient, we see $a_0=a_0^2\Rightarrow a_0=1$ or $a_0=0$.

Now proceed by induction. Consider the case where $a_0=1$. Assume we have shown $a_n=0$ for all $n<k$, $n\neq 1$. We will show $a_k=0$. If $k$ is odd, we are done. If $k$ is even, the coefficient of $x^{k}$ in $P(x^2)$ is $a_{k/2}$, so it is 0. We evaluate $[P(x)]^2$ and ignore higher order terms, and see


and the only way for the coefficient of $x^k$ to vanish here is for $a_k$ to be 0.

The case with $a_0=0$ is similar. Assume we have shown $a_n=0$ for all $n<k$. The coefficient of $x^{2k}$ in $P(x^2)$ is $a_{k}$. If evaluate $[P(x)]^2=[...a_kx^k]^2$ and ignore higher order terms again, we get $a_k^2x^{2k}$. So $a_k=1$ or $a_k=0$. If $a_k=0$, we continue the induction. If $a_k=1$, we factor $x^k$ out of the original polynomial and are reduced to the first case.

In a field of characteristic 2 however, I believe that any polynomial with all coefficients equal to 0 or 1 works. Just use the "freshman's dream." Further, because equating constants on both sides gives $a_n^2=a_n$, these are the only ones that work.

share|cite|improve this answer
If $p(x) = x$, then $p(x^2) = x^2$, and $(p(x))^2 = x^2$. But you claimed it was only even powers of $x$... So I don't "see immediately" that $a_n=0$ for all odd $n$. In fact, for any $k$, $x^k$ works, since $p(t) = t^k$ means $p(x^2) = x^{2k} = (x^k)^2 = p(x)^2$. – Arturo Magidin May 31 '12 at 3:15
Oops. Let me fix this. – Potato May 31 '12 at 3:16
Also, the answer may depend on the field? If $F=\mathbb{F}_2$, then $(a+b)^2 = a^2+b^2$, so for example $p(x) = x+1$ has the desired property. – Arturo Magidin May 31 '12 at 3:17
You still have the left-over error of thinking the odd coefficients must be zero, when you say (second line, third paragraph) "if $k$ is odd we are done". – Arturo Magidin May 31 '12 at 3:27
Same field $F[\alpha]$. What about $x^2+\alpha x$? Constant term is $0$, so you claim it will work... – Arturo Magidin May 31 '12 at 3:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.