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Let $\mathbb{C}(x)$ be the vector space $\mathbb{C}$ of polynomials $p\left(x\right)$ in one variable $x$ with coefficients in $\mathbb{C}$.

Is the set $p(x) \in \mathbb{C}\left(x\right)$ such that $1$ is a root of $p(x)$ a linear subspace?

Also show whether the set $p(x) \in \mathbb{C}(x)$ such that $1$ is not a root of $p(x)$ is a linear subspace or not?

So to my understanding, the set such that $1$ is a root of $p(x)$ is all polynomials with $p(-1)=0$ and I understand we need to determine whether these polynomials are closed under addition and scalar multiplication, but I don't know how to do this without the question defining the polynomials degree.

Also for the set such that $1$ is not a root, I assume this is all polynomials with $p(-1)\neq0$

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    $\begingroup$ Polynomials must (by definition) have finite degree. Moreover the polynomial ring over a ring $R$ is usually denoted by $R[x]$, as $R(x)$ is normally reserved for the ring of fractional functions. $\endgroup$ – b00n heT May 14 '16 at 10:02
  • $\begingroup$ To say that 1 is a root of $p (x)$ means that $p(1)=0$. $\endgroup$ – Simon May 14 '16 at 10:06
  • $\begingroup$ @Simon no, you are mistaken. The OP definition is the standard one $\endgroup$ – b00n heT May 14 '16 at 10:07
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    $\begingroup$ Ninja edit ! How about now ? $\endgroup$ – Simon May 14 '16 at 10:08
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    $\begingroup$ I think you are misreading the three equivalent facts "$1$ is a root of $p$", "$p(1)=0$ and "$(x-1)$ is a factor of $p(x)$ when you write "$p(-1) = 0$".You can use either of the last two correct versions to show that the set of polynomials you care about is closed under the operations you care about. $\endgroup$ – Ethan Bolker May 14 '16 at 10:14
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$1$ is a root of $p(x)$ iff $p(1) = 0$. If $p,q$ are two such polynomials $(p+q)(1) = p(1) + q(1) = 0$ and for $\lambda \in \mathbb{C}, \lambda p(1) = \lambda \cdot 0 = 0$. (Note you need to bound $\deg(p+q)$ above also but that is straightforward) Then since your set is non-empty it is a subspace.

The set of all polynomials over $\mathbb{C}$ where $1$ is not a root isn't a subspace. (edit: to clarify this is the set of polynomials $f$ such that $f(1) \neq 0$) Try considering $p(x) - 1$ and $q(x) + 1$ with $p,q$ as before.

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  • $\begingroup$ Why do we need to bound above the degree of $p+q$ ? $\endgroup$ – Simon May 14 '16 at 10:25
  • $\begingroup$ polynomials have finite degree, so this is checking the sum is in your vector space. This is obvious, either since they are both in $\mathbb{C}[x]$ which is a vector space or by noting a general bound for degrees of sums of polynomials in $R[x]$ for a ring $R$. $\endgroup$ – Rhys Steele May 14 '16 at 10:28
  • $\begingroup$ Thanks. For the set where 1 is not a root, could I simply show this by considering $\lambda=0$ so that $\lambda p\left(1\right)=0.p\left(1\right)=0$ therefore it is not closed under scalar multiplication? $\endgroup$ – Sophie Filer May 14 '16 at 10:41
  • $\begingroup$ @SophieFiler Sure, that works. What you're really using there though is that the zero polynomial isn't the set (to get that it isn't closed under multiplication). This already gives you the result without considering scalars since the zero polynomial must be in any subspace of $\mathbb{C}[x]$. $\endgroup$ – Rhys Steele May 14 '16 at 11:28
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Hint Show that the evaluation map $\Bbb C [x] \to \Bbb C$ defined by $1 \mapsto p(1)$ is a linear map. By definition its kernel is the set of polynomials for which $1$ is a root, and the kernel of any linear map is a (linear) subspace.

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