# Polynomial ring over $\mathbb{Z}_2$

As $f(x)$ is an irreducible over $\mathbb{Z}_2[x]$ so $R/(f)$ is an infinite field. Am I right?

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If it's infinite, you should be able to name 5 different elements, right? –  Henning Makholm Jul 21 '12 at 10:50
yah! thank you got it –  La Belle Noiseuse Jul 21 '12 at 10:51
For those wondering where these questions come from: nbhm.dae.gov.in/pdf-docs/phd-sample-2006.pdf (To the OP: Please post your work. It helps us not to repeat what you already know.) –  user21436 Jul 21 '12 at 11:11
Good work, @Kanappan –  Henning Makholm Jul 21 '12 at 11:38
Yes, good job, @Kannappan! "1.5", after all? Betrays a peculiar mindset to just copy... I'd often wondered. –  paul garrett Jul 22 '12 at 16:08

No the quotient ring is a finite field of order 4. The reason is every element in the quotient ring by the division algorithm is a linear polynomial. There are 2 choices for the coefficient of $\bar{1}$ and 2 choices for the coefficient of $\bar{x}$. Hence 4 choices in total and the quotient is a finite field with 4 elements.
Edit: Bill Dubuque suggested that I add why 1 and 2 are ruled out: The polynomial $f(x) = x^2 + x + 1$ has no roots over $\Bbb{Z}/2\Bbb{Z}$ and being a quadratic is thus irreducible over this field, it follows that the ideal generated by $f(x)$ is maximal from which it follows from this fact proved here that $R/(f(x))$ is a field.
The easiest way to see that is a field is by observing that $x(x+1)=1$. Thus, the three non-zero elements are invertible.... –  N. S. Jul 22 '12 at 15:13