Usually, we define a polynomial as

$a_n x^n + \cdots + a_1 x + a_0$

where $x$ is called indeterminate.

Would it be better to define it as

$a_n x^n + \cdots + a_1 x + a_0 x^0$

where $x^0$ means the identity element in the structure which $x$ belongs to.

For the study of the polynomial itself, I think these two definition make no difference. But when you treat a polynomial (expression) as a polynomial function, and if we use the second definition, we can just substitute all the $x$ with some variable (eg. some square matrix), rather than define it in a "adhoc" way, i.e. if we are using the first definition, we have to explicitly define the function to add an identity element to the constant term $a_0$ to make $a_1 x$ and $a_0$ addable.

Is the second definition equivalents to the first one?

If yes, then is it true that the authors of those texts actually means def 2 when the define the polynomial using def 1, they just omit the identity element?

If not, why? Would it be nicer to eliminate the non-addable $a_1x + a_0$ with $a_1x + a_0 x^0$, (even though we really don't need to add these two terms together when we are studying the knowledge of polynomial itself)?

  • $\begingroup$ It really doesn't matter. Quite often, when written in sigma notation, we write it as $\sum_{i=0}^n a_ix^i$, even if we otherwise write $a+bx+cx^2$. I often use this notation for one of the reasons we define $0^0=1$ :) $\endgroup$ Commented Aug 7, 2014 at 3:33
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    $\begingroup$ Please give an example of what you mean by the second requires "adding an identity... to make addable" $\endgroup$ Commented Aug 7, 2014 at 3:34
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    $\begingroup$ Did you mean $x^0$ instead of $x_0$ in that last line? $\endgroup$ Commented Aug 7, 2014 at 3:35
  • $\begingroup$ You can also define a polynomial as a sequence $(a_0,a_1,\dots)$ where only finitely many $a_i$ are non-zero. Then $(1,0,0,\dots)$ is not $x^0$, where $x=(0,1,0,0,\dots)$, but rather $(1,0,0,0,0,\dots)$ is just the identity. $\endgroup$ Commented Aug 7, 2014 at 3:37
  • $\begingroup$ @BillDubuque Suppose that $x$ is a matrix, then $a_1 x + a_0$ is not a valid expression, but if you "add" the identity matrix $I$ to $a_0$, you get $a_0I$, which makes $a_1 x + a_0 I$ a valid expression. $\endgroup$
    – Not an ID
    Commented Aug 7, 2014 at 4:09

1 Answer 1


It really doesn't matter which way you define polynomial. Another way is to consider all sequences:

$$(a_0,a_1,\dots,a_n,\dots)$$ where only finitely many $a_i$ are non-zero.

Then we add series point-wise, and we find their products by the Cauchy product.

Then $(1,0,0,0,\dots)$ is the multiplicative identity, and $(0,1,0,0,\dots)$ is $x$. So $1$ isn't really $x^0$, and $x^2$ just means $x\cdot x$.

  • $\begingroup$ For the study of the polynomial itself, it really doesn't matter. It is when you treat it as function that causes a problem, if you think def 2 is equivalent to def 1, then you will run into trouble with this formula $p(\lambda) = det(\lambda I - A)$, where $p$ is the characteristic polynomial of matrix $A$. because substitution is allowed, we can substitute the $\lambda$ in both sides with $A$, then you get "zero (that is the number 0) = ZERO (that is the zero matrix)". (to be continued) $\endgroup$
    – Not an ID
    Commented Aug 7, 2014 at 4:27
  • $\begingroup$ ... Please see this question for detail. $\endgroup$
    – Not an ID
    Commented Aug 7, 2014 at 4:28
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    $\begingroup$ @NotanID No, there is no problem when treating it as a function, as long as you accept that $0^0=1$. $\endgroup$ Commented Aug 7, 2014 at 4:56

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