Jtdylktuy

I understand that first i must check J is a subring then prove right & left ideals. My question is the notation on the elements in I. How would i subtract/multiply 2 functions of I.

Consider general arbitrary elements $P,Q in J$, and an element $R in mathbb{Z}[x]$.

Then clearly $P(x)R(x)$ has even coefficients because the product of any integer with an even integer is even, thus $R(x) in J$.

Furthermore, $P(x) + Q(x)$ must also have even coefficients because the sum of any two even numbers is again even.

This is sufficient, and proves that $J$ is an ideal.

(If you're unsure why this is sufficient, consider why these conditions imply that $(J,+)$ is a subgroup of $(mathbb{Z}[x]$,+) and $jr in J$ for all $j in J, r in R$ - these are precisely the requirements of a (two-sided) ideal).

Edit: In terms of notation, because we're talking about polynomials here you consider multiplication of functions to be exactly the multiplication you'd expect, e.g. $(x^2 + 2)(x + 1) = x^3 + x^2 + 2x + 2 in mathbb{Z}[x]$. The same for addition.

Perhaps it would help to see this problem solved more explicitly. We first note that $0 = 2cdot 0 in J$, so $J neq emptyset$. I will consider the case of $f(X) - g(X)$. Consider $f(X),g(X) in J$. Then we have $f(X) = a_nX^n + a_{n-1}X^{n-1} + dotsb + a_0$ and $g(X) = b_mX^m + b_{m-1}X^{m-1} + dotsb + b_0$ with $a_n,a_{n-1},dotsc,a_0 in 2Bbb Z$ and $ b_m,b_{m-1},dotsc,b_0 in 2Bbb{Z}$. That is, $a_i = 2a'_i$ and $b_j = 2b'_j$ for $i = 1,dotsc,n$ and for $j = 1,dotsc,m$. For convenience, suppose $m leq n$ and let $b_j = 0$ for $m < j leq n$ if $m < n$. Thus we may assume $m = n$. We have
$$begin{align}f(X) - g(X) &= left(a_nX^n + a_{n-1}X^{n-1} + dotsb + a_0right) - left(b_nX^n + b_{n-1}X^{n-1} + dotsb + b_0right)\
&= (a_n - b_n)X^n + (a_{n-1} - b_{n-1})X^{n-1} + dotsb + (a_0 - b_0)\
&= 2(a'_n - b'_n)X^n + 2(a'_{n-1} - b'_{n-1})X^{n-1} + dotsb + 2(a'_0 + b'_0) in J.end{align}$$
Thus $f(X) - g(X) in J$ for $f(X),g(X) in J$ and so $J$ has the structure of an abelian group. That $J$ is closed under multiplication follows similarly, and could prove fruitful to write down explicitly such a problem at least once.