Prove that $t$ does not have multiplicative inverse in $\mathbb{C}[t]$ In Stewart's Galois Theory Chapter 4.2 he mentioned that we can perform the operations of addition, subtraction, and multiplication in the polynomial ring $\mathbb{C}[t]$ but usually not division because $C[t]$ does not contain an inverse $t^{-1}$ for $t$ and he asked us to prove in the exercise. Isn't it just that $C[t]$ does not contain an inverse $t^{-1}$ for $t$...I'm not sure how he means to "prove" this? Thanks! 
 A: Suppose we have $tp(t) = 1$ for every $t$. Calculate in $0$, you get $0\cdot p(0) = 1$. Contraddiction.
A: Well, how do you know that $t^{-1}$ is not in $C[t]$? Remember that $t^{-1}$ just means "that element of $C[t]$ that if you multiply by $t$ you get $1$." It's true that if you were in the field $C(t)$ (or dealing with functions in general) then the inverse should be the rational function $1/t$, and this doesn't look like a polynomial. But how do you know for sure? After all, how do you know I'm not really, really clever and can write it as a polynomial?
You start with some polynomial $p \in C[t]$ and show that $t p(t) \ne 1$. This is most easily done by noting that the degree of the product is the sum of the degrees.
A: Any element of $\mathbb{C}[t]$ is of the form $p(t)=a_0+a_1 t+...+a_nt^n$ for some $n\in\mathbb{Z}_{\geq 0}$ and $a_i\in \mathbb{C}$ for $i$ from $0,..,n$. If $t$ had an inverse, then we would have $p(t)\in\mathbb{C}[t]$ such that $t.p(t)=1$. But degree$(t.p(t))\geq 1$, which gives us a contradiction.
A: $\mathbb{C}[t]$ is set of polynomials (over ...); so can think only about polynomials in $t$. Since $t^{-1}$ is not a polynomial in $t$, we can not consider directly it in the set $\mathbb{C}[t]$.
We have to proceed like: consider a polynomial $p(t)$ and see why $p(t).t$ can not be $1$. Can you see? [Hint: compare degrees].
