I have difficulties with a task in linear algebra.

R is a ring and R = {a, b, c, d}

These are the tables for + and . in R:

 +| a b c d              .| a b c d
 -------------           -------------   
 a| a b c d              a| a a a a
 b| b a d c              b| a b c d 
 c| c d a b              c| a c c a
 d| d c b a              d| a d a d

I know the definition for ideal and I understand why for example we take c and we look at the . table for the row and column of the element c to find out which elements stand there - because a.x and x.a should stay in the ideal.

But I don't quite understand the role of the + table here. We have that a-x should be in the ideal, too. So what? When we see that multiplying c to whatever element we get a or c, we should look if c+a and a+c gives us a or c, too? I can't quite get this.

Can you please help me? Thanks very much in advance!

  • 1
    $\begingroup$ Some ideas: Start by seeing what ideals you get if you start with one element and add those elements you get by multiplying by the other elements (show that this is an ideal). Then do the same starting with two elements and finally with three elements (consider that you will not necessarily need to consider all possible such combinations and think about which ones you can discard). $\endgroup$ Commented May 29, 2013 at 14:00
  • 1
    $\begingroup$ Ideals are additive subgroups. You can use the addition table to figure out what those are and then see which ones are closed under multiplication with arbitrary elements of the ring. $\endgroup$
    – John Douma
    Commented May 29, 2013 at 14:56

1 Answer 1


The addition table tells us that $a$ is the additive identity, i.e. $a=0$. The multiplication table tells us that $b$ is the multiplicative identity, i.e. $b=1$ so $R$ is a commutative ring with identity and we immediately get that $R$ and $(a)$ are ideals.

What are the non-trivial additive subgroups? We notice that $c+c=a=0$ and $d+d=a=0$ so $\{a,c\}$ and $\{a,d\}$ are additive subgroups. Are they ideals? A quick look at the multiplication table shows they are.

Are there any others? Since $b=1$ we know any ideal containing $b$ is the entire ring and since $c+d=b$ we know any any ideal containing both $c$ and $d$ must be the entire ring.

Therefore, the ideals are $R$, $(a)$, $(c)$ and $(d)$.

  • $\begingroup$ Thank you very much! This was helpful! :) $\endgroup$
    – Faery
    Commented May 29, 2013 at 15:21

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