Is it correct to compute that radical ideal in this way? $$\sqrt{(x^2,xz^2-x,y-z)}=\sqrt{(x^2,xz^2-x,y-z,x)}=\sqrt{(y-z,x)}=(x,y-z)$$
In particular, I added $x$ to generators inside the 'root' because there where $x^2$ and the exponent vanishes after applying the radical.
Is it also always true that an ideal in which appear only terms of degree $1$ is a radical ideal?
Thank you for your help.
Your computation is absolutely fine. For the second question, the answer is positive in the following sense:
Let $f_1, \dotsc, f_s \in k[x_1, \dotsc, x_n]$ be linear polynomials (i.e. their total degree is $1$, not necessarily homogeneous). Then $I=(f_1, \dotsc, f_s)$ is radical.
We proceed by induction over $n$, the $n=1$-case being trivial, since $(ax-b) \subset k[x]$ is reduced.
W.l.o.g. we can assume, that $x_n$ occurs within $f_1$ and after dividing by some constant, we can assume $f_1=x_n-f$ with some linear polynomial $f \in k[x_1, \dotsc, x_{n-1}]$. We can compute
$$k[x_1, \dotsc, x_n]/I = k[x_1, \dotsc, x_n]/(x_n-f, f_2 \dotsc, f_s)$$ $$\cong k[x_1, \dotsc, x_{n-1}]/(f_2(x_1, \dotsc, x_{n-1},f), \dotsc, f_s(x_1, \dotsc, x_{n-1},f)).$$
The terms $f_i(x_1, \dotsc, x_{n-1},f)$ are some linear polynomials in the variables $x_1, \dotsc, x_{n-1}$. By induction, the generated ideal is radical, i.e. we have shown that $k[x_1, \dotsc, x_n]/I$ is reduced.
Note that the assertion becomes false if we replace total degree by partial degree, since $(xy,x-y)$ is not radical.