Is the kernel of a ring homomorphism a subring? The following link proves that if $f:R \to R'$ is a ring homormorphism, then $\ker(f)$ is a subring of $R:$
https://proofwiki.org/wiki/Kernel_of_Ring_Homomorphism_is_Subring
But an alternative source claims that $\ker(f)$ isn't a ring (See Theorem $3.5$):
http://www.math.uconn.edu/~kconrad/blurbs/ringtheory/ringdefs.pdf
Which is the right convention, and if neither is particularly right, which one is better?
 A: Maybe you should have read the paragraph after Theorem $3.5$:

This last theorem is probably why some people do not insist that rings contain 1. Kernels
  of ring homomorphisms have all the properties of a subring except they almost never contain
  the multiplicative identity. So if we want ring theory to mimic group theory by having
  kernels of ring homomorphisms be subrings, then we should not insist that subrings contain 1
  (and thus perhaps not even insist that rings contain 1). Then kernels of ring homomorphisms
  could be called subrings.

His definition of a ring is not the one used in the first link:

Definition $1.1$. A ring is a set $R$ equipped with two operations $+$ (addition) and $×$
  (multiplication) such that $R$ is an abelian group under addition (with identity denoted $0$
  and the inverse of $a$ denoted $−a$), while multiplication is associative with an identity element
  $1$. Finally, multiplication distributes over addition: $x(y+z) = xy+xz$ and $(x+y)z = xz+yz$
  for all $x, y$, and $z$ in $R.$

A: The kernel is best thought of as an ideal. Every ideal is a subring, but not necessarily with unit. Some authors only consider rings with unit, hence the different conventions.
