# Let $K$ be a field. Prove that the field of all polynomials over $K$ is a vector space over $K$.

Let $K$ be a field. Prove that the field of all polynomials over $K$ is a vector space over $K$.

Let $K[x]$ be the field of all polynomials over $K$. Let $\cdot$ and $+$ be the multiplication and addition defined on $K[x]$.
Let $a,b \in K[x]$ and $s \in K$. Then we can define $a +_{K[x]}b = a+b$ and
$\cdot_{K[x]}: K \times K[x] \to K[x]$ as $sa = a \cdot s(x)$ where $K[x] \ni s(x) = 0x^n \dots0x^1 +s$
Since $K[x]$ is a field:
$$a(bc) = (ab)c) \\a(b+c)=(b+c)a = ab+ac$$ Addition is associative, has neutral element and inverse element.
Identity element of scalar multiplication: $1 = 0x^n + \dots +0x^1+1$
Is it enough or I should add something / change something?

• thank you, yes. Dec 1 '17 at 15:23
• Oh wait, I think I finally get what you're trying to write. I think you mean $sa=s(x)a$ where $s(x)$ is the polynomial $s$. Dec 1 '17 at 15:24
• What do you mean by the field of all polynomials? Dec 1 '17 at 15:28
• @Antonios-AlexandrosRobotis "$K[X]$ is not, in general, a field." The wording is as if you think there is a ring $K$ such that $K[X]$ is a field. Dec 1 '17 at 15:39
• Actually, I meant it the other way... whoops. "It is generally true that $K[X]$ is not a field." Dec 1 '17 at 15:42

If you want to be more precise, maybe show the axioms in a more rigorous manner. For instance, for commutativity under addition, let $$p(X)=\sum_{i=1}^n a_i X^i\:\:\:\:\:q(X)=\sum_{i=1}^n b_i X^i$$ where some of the $a_i, b_i\in K$ may be zero (in particular the degrees may differ). Then $$p(X)+q(X)=\sum_{i=1}^n (a_i+b_i)X^i=\sum_{i=1}^n(b_i+a_i)X^i=q(X)+p(X),$$ so that $K$ is commutative under addition. Indeed, we see here more explicitly that the commutativity follows from the fact that $a_i+b_i=b_i+a_i$ in $K$. Perhaps try recasting your proof in this more explicit manner. It's more insightful, and more rigorous.